Topic 4: Digital Circuits (Integrated Circuits Technology) Part two.
RC and RL Circuits with Piecewise Constant Sources
Transcript of RC and RL Circuits with Piecewise Constant Sources
RC and RL Circuits with Piecewise Constant Sources
M. B. [email protected]
www.ee.iitb.ac.in/~sequel
Department of Electrical EngineeringIndian Institute of Technology Bombay
M. B. Patil, IIT Bombay
RC circuits with DC sources
C
A
B
v
Circuit(resistors,voltage sources,current sources,CCVS, CCCS,VCVS, VCCS)
i
C
A
B
i
v
RTh
≡ VTh
* If all sources are DC (constant), VTh = constant .
* KVL: VTh = RTh i + v → VTh = RThCdv
dt+ v → dv
dt+
v
RThC=
VTh
RThC.
* Homogeneous solution:
dv
dt+
1
τv = 0 , where τ = RTh C is the “time constant.”
(V
Coul/sec× Coul
V
)→ dv
v= − dt
τ→ log v = − t
τ+ K0 → v (h) = exp [(−t/τ) + K0] = K exp(−t/τ) .
* Particular solution is a specific function that satisfies the differential equation. We know that all timederivatives will vanish as t →∞ , making i = 0, and we get v (p) = VTh as a particular solution (whichhappens to be simply a constant).
* v = v (h) + v (p) = K exp(−t/τ) + VTh .
* In general, v(t) = A exp(−t/τ) + B , where A and B can be obtained from known conditions on v .
M. B. Patil, IIT Bombay
RC circuits with DC sources
C
A
B
v
Circuit(resistors,voltage sources,current sources,CCVS, CCCS,VCVS, VCCS)
i
C
A
B
i
v
RTh
≡ VTh
* If all sources are DC (constant), VTh = constant .
* KVL: VTh = RTh i + v → VTh = RThCdv
dt+ v → dv
dt+
v
RThC=
VTh
RThC.
* Homogeneous solution:
dv
dt+
1
τv = 0 , where τ = RTh C is the “time constant.”
(V
Coul/sec× Coul
V
)→ dv
v= − dt
τ→ log v = − t
τ+ K0 → v (h) = exp [(−t/τ) + K0] = K exp(−t/τ) .
* Particular solution is a specific function that satisfies the differential equation. We know that all timederivatives will vanish as t →∞ , making i = 0, and we get v (p) = VTh as a particular solution (whichhappens to be simply a constant).
* v = v (h) + v (p) = K exp(−t/τ) + VTh .
* In general, v(t) = A exp(−t/τ) + B , where A and B can be obtained from known conditions on v .
M. B. Patil, IIT Bombay
RC circuits with DC sources
C
A
B
v
Circuit(resistors,voltage sources,current sources,CCVS, CCCS,VCVS, VCCS)
i
C
A
B
i
v
RTh
≡ VTh
* If all sources are DC (constant), VTh = constant .
* KVL: VTh = RTh i + v → VTh = RThCdv
dt+ v → dv
dt+
v
RThC=
VTh
RThC.
* Homogeneous solution:
dv
dt+
1
τv = 0 , where τ = RTh C is the “time constant.”
(V
Coul/sec× Coul
V
)→ dv
v= − dt
τ→ log v = − t
τ+ K0 → v (h) = exp [(−t/τ) + K0] = K exp(−t/τ) .
* Particular solution is a specific function that satisfies the differential equation. We know that all timederivatives will vanish as t →∞ , making i = 0, and we get v (p) = VTh as a particular solution (whichhappens to be simply a constant).
* v = v (h) + v (p) = K exp(−t/τ) + VTh .
* In general, v(t) = A exp(−t/τ) + B , where A and B can be obtained from known conditions on v .
M. B. Patil, IIT Bombay
RC circuits with DC sources
C
A
B
v
Circuit(resistors,voltage sources,current sources,CCVS, CCCS,VCVS, VCCS)
i
C
A
B
i
v
RTh
≡ VTh
* If all sources are DC (constant), VTh = constant .
* KVL: VTh = RTh i + v → VTh = RThCdv
dt+ v → dv
dt+
v
RThC=
VTh
RThC.
* Homogeneous solution:
dv
dt+
1
τv = 0 , where τ = RTh C is the “time constant.”
(V
Coul/sec× Coul
V
)→ dv
v= − dt
τ→ log v = − t
τ+ K0 → v (h) = exp [(−t/τ) + K0] = K exp(−t/τ) .
* Particular solution is a specific function that satisfies the differential equation. We know that all timederivatives will vanish as t →∞ , making i = 0, and we get v (p) = VTh as a particular solution (whichhappens to be simply a constant).
* v = v (h) + v (p) = K exp(−t/τ) + VTh .
* In general, v(t) = A exp(−t/τ) + B , where A and B can be obtained from known conditions on v .
M. B. Patil, IIT Bombay
RC circuits with DC sources
C
A
B
v
Circuit(resistors,voltage sources,current sources,CCVS, CCCS,VCVS, VCCS)
i
C
A
B
i
v
RTh
≡ VTh
* If all sources are DC (constant), VTh = constant .
* KVL: VTh = RTh i + v → VTh = RThCdv
dt+ v → dv
dt+
v
RThC=
VTh
RThC.
* Homogeneous solution:
dv
dt+
1
τv = 0 , where τ = RTh C is the “time constant.”
(V
Coul/sec× Coul
V
)
→ dv
v= − dt
τ→ log v = − t
τ+ K0 → v (h) = exp [(−t/τ) + K0] = K exp(−t/τ) .
* Particular solution is a specific function that satisfies the differential equation. We know that all timederivatives will vanish as t →∞ , making i = 0, and we get v (p) = VTh as a particular solution (whichhappens to be simply a constant).
* v = v (h) + v (p) = K exp(−t/τ) + VTh .
* In general, v(t) = A exp(−t/τ) + B , where A and B can be obtained from known conditions on v .
M. B. Patil, IIT Bombay
RC circuits with DC sources
C
A
B
v
Circuit(resistors,voltage sources,current sources,CCVS, CCCS,VCVS, VCCS)
i
C
A
B
i
v
RTh
≡ VTh
* If all sources are DC (constant), VTh = constant .
* KVL: VTh = RTh i + v → VTh = RThCdv
dt+ v → dv
dt+
v
RThC=
VTh
RThC.
* Homogeneous solution:
dv
dt+
1
τv = 0 , where τ = RTh C is the “time constant.”
(V
Coul/sec× Coul
V
)→ dv
v= − dt
τ→ log v = − t
τ+ K0 → v (h) = exp [(−t/τ) + K0] = K exp(−t/τ) .
* Particular solution is a specific function that satisfies the differential equation. We know that all timederivatives will vanish as t →∞ , making i = 0, and we get v (p) = VTh as a particular solution (whichhappens to be simply a constant).
* v = v (h) + v (p) = K exp(−t/τ) + VTh .
* In general, v(t) = A exp(−t/τ) + B , where A and B can be obtained from known conditions on v .
M. B. Patil, IIT Bombay
RC circuits with DC sources
C
A
B
v
Circuit(resistors,voltage sources,current sources,CCVS, CCCS,VCVS, VCCS)
i
C
A
B
i
v
RTh
≡ VTh
* If all sources are DC (constant), VTh = constant .
* KVL: VTh = RTh i + v → VTh = RThCdv
dt+ v → dv
dt+
v
RThC=
VTh
RThC.
* Homogeneous solution:
dv
dt+
1
τv = 0 , where τ = RTh C is the “time constant.”
(V
Coul/sec× Coul
V
)→ dv
v= − dt
τ→ log v = − t
τ+ K0 → v (h) = exp [(−t/τ) + K0] = K exp(−t/τ) .
* Particular solution is a specific function that satisfies the differential equation. We know that all timederivatives will vanish as t →∞ , making i = 0, and we get v (p) = VTh as a particular solution (whichhappens to be simply a constant).
* v = v (h) + v (p) = K exp(−t/τ) + VTh .
* In general, v(t) = A exp(−t/τ) + B , where A and B can be obtained from known conditions on v .
M. B. Patil, IIT Bombay
RC circuits with DC sources
C
A
B
v
Circuit(resistors,voltage sources,current sources,CCVS, CCCS,VCVS, VCCS)
i
C
A
B
i
v
RTh
≡ VTh
* If all sources are DC (constant), VTh = constant .
* KVL: VTh = RTh i + v → VTh = RThCdv
dt+ v → dv
dt+
v
RThC=
VTh
RThC.
* Homogeneous solution:
dv
dt+
1
τv = 0 , where τ = RTh C is the “time constant.”
(V
Coul/sec× Coul
V
)→ dv
v= − dt
τ→ log v = − t
τ+ K0 → v (h) = exp [(−t/τ) + K0] = K exp(−t/τ) .
* Particular solution is a specific function that satisfies the differential equation. We know that all timederivatives will vanish as t →∞ , making i = 0, and we get v (p) = VTh as a particular solution (whichhappens to be simply a constant).
* v = v (h) + v (p) = K exp(−t/τ) + VTh .
* In general, v(t) = A exp(−t/τ) + B , where A and B can be obtained from known conditions on v .
M. B. Patil, IIT Bombay
RC circuits with DC sources
C
A
B
v
Circuit(resistors,voltage sources,current sources,CCVS, CCCS,VCVS, VCCS)
i
C
A
B
i
v
RTh
≡ VTh
* If all sources are DC (constant), VTh = constant .
* KVL: VTh = RTh i + v → VTh = RThCdv
dt+ v → dv
dt+
v
RThC=
VTh
RThC.
* Homogeneous solution:
dv
dt+
1
τv = 0 , where τ = RTh C is the “time constant.”
(V
Coul/sec× Coul
V
)→ dv
v= − dt
τ→ log v = − t
τ+ K0 → v (h) = exp [(−t/τ) + K0] = K exp(−t/τ) .
* Particular solution is a specific function that satisfies the differential equation. We know that all timederivatives will vanish as t →∞ , making i = 0, and we get v (p) = VTh as a particular solution (whichhappens to be simply a constant).
* v = v (h) + v (p) = K exp(−t/τ) + VTh .
* In general, v(t) = A exp(−t/τ) + B , where A and B can be obtained from known conditions on v .
M. B. Patil, IIT Bombay
RC circuits with DC sources (continued)
C C
A
B
A
B
i
vv
Circuit(resistors,voltage sources,current sources,CCVS, CCCS,VCVS, VCCS)
i
RTh
≡ VTh
* If all sources are DC (constant), we havev(t) = A exp(−t/τ) + B , τ = RThC .
* i(t) = Cdv
dt= C × A exp(−t/τ)
(− 1
τ
)≡ A′ exp(−t/τ) .
* As t →∞, i → 0, i.e., the capacitor behaves like an open circuit since all derivatives vanish.
* Since the circuit in the black box is linear, any variable (current or voltage) in the circuit can beexpressed asx(t) = K1 exp(−t/τ) + K2 ,where K1 and K2 can be obtained from suitable conditions on x(t).
M. B. Patil, IIT Bombay
RC circuits with DC sources (continued)
C C
A
B
A
B
i
vv
Circuit(resistors,voltage sources,current sources,CCVS, CCCS,VCVS, VCCS)
i
RTh
≡ VTh
* If all sources are DC (constant), we havev(t) = A exp(−t/τ) + B , τ = RThC .
* i(t) = Cdv
dt= C × A exp(−t/τ)
(− 1
τ
)≡ A′ exp(−t/τ) .
* As t →∞, i → 0, i.e., the capacitor behaves like an open circuit since all derivatives vanish.
* Since the circuit in the black box is linear, any variable (current or voltage) in the circuit can beexpressed asx(t) = K1 exp(−t/τ) + K2 ,where K1 and K2 can be obtained from suitable conditions on x(t).
M. B. Patil, IIT Bombay
RC circuits with DC sources (continued)
C C
A
B
A
B
i
vv
Circuit(resistors,voltage sources,current sources,CCVS, CCCS,VCVS, VCCS)
i
RTh
≡ VTh
* If all sources are DC (constant), we havev(t) = A exp(−t/τ) + B , τ = RThC .
* i(t) = Cdv
dt= C × A exp(−t/τ)
(− 1
τ
)≡ A′ exp(−t/τ) .
* As t →∞, i → 0, i.e., the capacitor behaves like an open circuit since all derivatives vanish.
* Since the circuit in the black box is linear, any variable (current or voltage) in the circuit can beexpressed asx(t) = K1 exp(−t/τ) + K2 ,where K1 and K2 can be obtained from suitable conditions on x(t).
M. B. Patil, IIT Bombay
RC circuits with DC sources (continued)
C C
A
B
A
B
i
vv
Circuit(resistors,voltage sources,current sources,CCVS, CCCS,VCVS, VCCS)
i
RTh
≡ VTh
* If all sources are DC (constant), we havev(t) = A exp(−t/τ) + B , τ = RThC .
* i(t) = Cdv
dt= C × A exp(−t/τ)
(− 1
τ
)≡ A′ exp(−t/τ) .
* As t →∞, i → 0, i.e., the capacitor behaves like an open circuit since all derivatives vanish.
* Since the circuit in the black box is linear, any variable (current or voltage) in the circuit can beexpressed asx(t) = K1 exp(−t/τ) + K2 ,where K1 and K2 can be obtained from suitable conditions on x(t).
M. B. Patil, IIT Bombay
Plot of f (t)= e−t/τ
t/τ e−t/τ 1− e−t/τ
0.0 1.0 0.0
1.0 0.3679 0.6321
2.0 0.1353 0.8647
3.0 4.9787×10−2 0.9502
4.0 1.8315×10−2 0.9817
5.0 6.7379×10−3 0.9933
* For t/τ = 5, e−t/τ ' 0, 1− e−t/τ ' 1.
* We can say that the transient lasts for about 5 time constants.
0
1
0 1 2 3 4 5 6
x= t/τ
exp(−x)
1− exp(−x)
M. B. Patil, IIT Bombay
Plot of f (t)= e−t/τ
t/τ e−t/τ 1− e−t/τ
0.0 1.0 0.0
1.0 0.3679 0.6321
2.0 0.1353 0.8647
3.0 4.9787×10−2 0.9502
4.0 1.8315×10−2 0.9817
5.0 6.7379×10−3 0.9933
* For t/τ = 5, e−t/τ ' 0, 1− e−t/τ ' 1.
* We can say that the transient lasts for about 5 time constants.
0
1
0 1 2 3 4 5 6
x= t/τ
exp(−x)
1− exp(−x)
M. B. Patil, IIT Bombay
Plot of f (t)= e−t/τ
t/τ e−t/τ 1− e−t/τ
0.0 1.0 0.0
1.0 0.3679 0.6321
2.0 0.1353 0.8647
3.0 4.9787×10−2 0.9502
4.0 1.8315×10−2 0.9817
5.0 6.7379×10−3 0.9933
* For t/τ = 5, e−t/τ ' 0, 1− e−t/τ ' 1.
* We can say that the transient lasts for about 5 time constants.
0
1
0 1 2 3 4 5 6
x= t/τ
exp(−x)
1− exp(−x)
M. B. Patil, IIT Bombay
Plot of f (t)= e−t/τ
t/τ e−t/τ 1− e−t/τ
0.0 1.0 0.0
1.0 0.3679 0.6321
2.0 0.1353 0.8647
3.0 4.9787×10−2 0.9502
4.0 1.8315×10−2 0.9817
5.0 6.7379×10−3 0.9933
* For t/τ = 5, e−t/τ ' 0, 1− e−t/τ ' 1.
* We can say that the transient lasts for about 5 time constants.
0
1
0 1 2 3 4 5 6
x= t/τ
exp(−x)
1− exp(−x)
M. B. Patil, IIT Bombay
Plot of f (t)=Ae−t/τ + B
* At t = 0, f =A + B.
* As t →∞, f → B.
* The graph of f (t) lies between (A + B) and B.
Note: If A > 0, A + B > B. If A < 0, A + B < B.
* At t = 0,df
dt= Ae−t/τ
(− 1
τ
)= − A
τ.
If A > 0, the derivative (slope) at t = 0 is negative; else, it is positive.
* As t →∞,df
dt→ 0, i.e., f becomes constant (equal to B).
0 5 τ t
A < 0
A+ B
B
0 5 τ t
A > 0
A+ B
B
M. B. Patil, IIT Bombay
Plot of f (t)=Ae−t/τ + B
* At t = 0, f =A + B.
* As t →∞, f → B.
* The graph of f (t) lies between (A + B) and B.
Note: If A > 0, A + B > B. If A < 0, A + B < B.
* At t = 0,df
dt= Ae−t/τ
(− 1
τ
)= − A
τ.
If A > 0, the derivative (slope) at t = 0 is negative; else, it is positive.
* As t →∞,df
dt→ 0, i.e., f becomes constant (equal to B).
0 5 τ t
A < 0
A+ B
B
0 5 τ t
A > 0
A+ B
B
M. B. Patil, IIT Bombay
Plot of f (t)=Ae−t/τ + B
* At t = 0, f =A + B.
* As t →∞, f → B.
* The graph of f (t) lies between (A + B) and B.
Note: If A > 0, A + B > B. If A < 0, A + B < B.
* At t = 0,df
dt= Ae−t/τ
(− 1
τ
)= − A
τ.
If A > 0, the derivative (slope) at t = 0 is negative; else, it is positive.
* As t →∞,df
dt→ 0, i.e., f becomes constant (equal to B).
0 5 τ t
A < 0
A+ B
B
0 5 τ t
A > 0
A+ B
B
M. B. Patil, IIT Bombay
Plot of f (t)=Ae−t/τ + B
* At t = 0, f =A + B.
* As t →∞, f → B.
* The graph of f (t) lies between (A + B) and B.
Note: If A > 0, A + B > B. If A < 0, A + B < B.
* At t = 0,df
dt= Ae−t/τ
(− 1
τ
)= − A
τ.
If A > 0, the derivative (slope) at t = 0 is negative; else, it is positive.
* As t →∞,df
dt→ 0, i.e., f becomes constant (equal to B).
0 5 τ t
A < 0
A+ B
B
0 5 τ t
A > 0
A+ B
B
M. B. Patil, IIT Bombay
Plot of f (t)=Ae−t/τ + B
* At t = 0, f =A + B.
* As t →∞, f → B.
* The graph of f (t) lies between (A + B) and B.
Note: If A > 0, A + B > B. If A < 0, A + B < B.
* At t = 0,df
dt= Ae−t/τ
(− 1
τ
)= − A
τ.
If A > 0, the derivative (slope) at t = 0 is negative; else, it is positive.
* As t →∞,df
dt→ 0, i.e., f becomes constant (equal to B).
0 5 τ t
A < 0
A+ B
B
0 5 τ t
A > 0
A+ B
B
M. B. Patil, IIT Bombay
Plot of f (t)=Ae−t/τ + B
* At t = 0, f =A + B.
* As t →∞, f → B.
* The graph of f (t) lies between (A + B) and B.
Note: If A > 0, A + B > B. If A < 0, A + B < B.
* At t = 0,df
dt= Ae−t/τ
(− 1
τ
)= − A
τ.
If A > 0, the derivative (slope) at t = 0 is negative; else, it is positive.
* As t →∞,df
dt→ 0, i.e., f becomes constant (equal to B).
0 5 τ t
A < 0
A+ B
B
0 5 τ t
A > 0
A+ B
B
M. B. Patil, IIT Bombay
Plot of f (t)=Ae−t/τ + B
* At t = 0, f =A + B.
* As t →∞, f → B.
* The graph of f (t) lies between (A + B) and B.
Note: If A > 0, A + B > B. If A < 0, A + B < B.
* At t = 0,df
dt= Ae−t/τ
(− 1
τ
)= − A
τ.
If A > 0, the derivative (slope) at t = 0 is negative; else, it is positive.
* As t →∞,df
dt→ 0, i.e., f becomes constant (equal to B).
0 5 τ t
A < 0
A+ B
B
0 5 τ t
A > 0
A+ B
B
M. B. Patil, IIT Bombay
Plot of f (t)=Ae−t/τ + B
* At t = 0, f =A + B.
* As t →∞, f → B.
* The graph of f (t) lies between (A + B) and B.
Note: If A > 0, A + B > B. If A < 0, A + B < B.
* At t = 0,df
dt= Ae−t/τ
(− 1
τ
)= − A
τ.
If A > 0, the derivative (slope) at t = 0 is negative; else, it is positive.
* As t →∞,df
dt→ 0, i.e., f becomes constant (equal to B).
0 5 τ t
A < 0
A+ B
B
0 5 τ t
A > 0
A+ B
B
M. B. Patil, IIT Bombay
RL circuits with DC sources
v
CircuitA
B
(resistors,voltage sources,current sources,CCVS, CCCS,VCVS, VCCS)
i
L
i
v
A
B
L
RTh
≡ VTh
* If all sources are DC (constant), VTh = constant .
* KVL: VTh = RTh i + Ldi
dt.
* Homogeneous solution:
di
dt+
1
τi = 0 , where τ = L/RTh
→ i (h) = K exp(−t/τ) .
* Particular solution is a specific function that satisfies the differential equation. We know that all timederivatives will vanish as t →∞ , making v = 0, and we get i (p) = VTh/RTh as a particular solution(which happens to be simply a constant).
* i = i (h) + i (p) = K exp(−t/τ) + VTh/RTh .
* In general, i(t) = A exp(−t/τ) + B , where A and B can be obtained from known conditions on i .
M. B. Patil, IIT Bombay
RL circuits with DC sources
v
CircuitA
B
(resistors,voltage sources,current sources,CCVS, CCCS,VCVS, VCCS)
i
L
i
v
A
B
L
RTh
≡ VTh
* If all sources are DC (constant), VTh = constant .
* KVL: VTh = RTh i + Ldi
dt.
* Homogeneous solution:
di
dt+
1
τi = 0 , where τ = L/RTh
→ i (h) = K exp(−t/τ) .
* Particular solution is a specific function that satisfies the differential equation. We know that all timederivatives will vanish as t →∞ , making v = 0, and we get i (p) = VTh/RTh as a particular solution(which happens to be simply a constant).
* i = i (h) + i (p) = K exp(−t/τ) + VTh/RTh .
* In general, i(t) = A exp(−t/τ) + B , where A and B can be obtained from known conditions on i .
M. B. Patil, IIT Bombay
RL circuits with DC sources
v
CircuitA
B
(resistors,voltage sources,current sources,CCVS, CCCS,VCVS, VCCS)
i
L
i
v
A
B
L
RTh
≡ VTh
* If all sources are DC (constant), VTh = constant .
* KVL: VTh = RTh i + Ldi
dt.
* Homogeneous solution:
di
dt+
1
τi = 0 , where τ = L/RTh
→ i (h) = K exp(−t/τ) .
* Particular solution is a specific function that satisfies the differential equation. We know that all timederivatives will vanish as t →∞ , making v = 0, and we get i (p) = VTh/RTh as a particular solution(which happens to be simply a constant).
* i = i (h) + i (p) = K exp(−t/τ) + VTh/RTh .
* In general, i(t) = A exp(−t/τ) + B , where A and B can be obtained from known conditions on i .
M. B. Patil, IIT Bombay
RL circuits with DC sources
v
CircuitA
B
(resistors,voltage sources,current sources,CCVS, CCCS,VCVS, VCCS)
i
L
i
v
A
B
L
RTh
≡ VTh
* If all sources are DC (constant), VTh = constant .
* KVL: VTh = RTh i + Ldi
dt.
* Homogeneous solution:
di
dt+
1
τi = 0 , where τ = L/RTh
→ i (h) = K exp(−t/τ) .
* Particular solution is a specific function that satisfies the differential equation. We know that all timederivatives will vanish as t →∞ , making v = 0, and we get i (p) = VTh/RTh as a particular solution(which happens to be simply a constant).
* i = i (h) + i (p) = K exp(−t/τ) + VTh/RTh .
* In general, i(t) = A exp(−t/τ) + B , where A and B can be obtained from known conditions on i .
M. B. Patil, IIT Bombay
RL circuits with DC sources
v
CircuitA
B
(resistors,voltage sources,current sources,CCVS, CCCS,VCVS, VCCS)
i
L
i
v
A
B
L
RTh
≡ VTh
* If all sources are DC (constant), VTh = constant .
* KVL: VTh = RTh i + Ldi
dt.
* Homogeneous solution:
di
dt+
1
τi = 0 , where τ = L/RTh
→ i (h) = K exp(−t/τ) .
* Particular solution is a specific function that satisfies the differential equation. We know that all timederivatives will vanish as t →∞ , making v = 0, and we get i (p) = VTh/RTh as a particular solution(which happens to be simply a constant).
* i = i (h) + i (p) = K exp(−t/τ) + VTh/RTh .
* In general, i(t) = A exp(−t/τ) + B , where A and B can be obtained from known conditions on i .
M. B. Patil, IIT Bombay
RL circuits with DC sources
v
CircuitA
B
(resistors,voltage sources,current sources,CCVS, CCCS,VCVS, VCCS)
i
L
i
v
A
B
L
RTh
≡ VTh
* If all sources are DC (constant), VTh = constant .
* KVL: VTh = RTh i + Ldi
dt.
* Homogeneous solution:
di
dt+
1
τi = 0 , where τ = L/RTh
→ i (h) = K exp(−t/τ) .
* Particular solution is a specific function that satisfies the differential equation. We know that all timederivatives will vanish as t →∞ , making v = 0, and we get i (p) = VTh/RTh as a particular solution(which happens to be simply a constant).
* i = i (h) + i (p) = K exp(−t/τ) + VTh/RTh .
* In general, i(t) = A exp(−t/τ) + B , where A and B can be obtained from known conditions on i .
M. B. Patil, IIT Bombay
RL circuits with DC sources
v
CircuitA
B
(resistors,voltage sources,current sources,CCVS, CCCS,VCVS, VCCS)
i
L
i
v
A
B
L
RTh
≡ VTh
* If all sources are DC (constant), VTh = constant .
* KVL: VTh = RTh i + Ldi
dt.
* Homogeneous solution:
di
dt+
1
τi = 0 , where τ = L/RTh
→ i (h) = K exp(−t/τ) .
* Particular solution is a specific function that satisfies the differential equation. We know that all timederivatives will vanish as t →∞ , making v = 0, and we get i (p) = VTh/RTh as a particular solution(which happens to be simply a constant).
* i = i (h) + i (p) = K exp(−t/τ) + VTh/RTh .
* In general, i(t) = A exp(−t/τ) + B , where A and B can be obtained from known conditions on i .
M. B. Patil, IIT Bombay
RL circuits with DC sources
v
CircuitA
B
(resistors,voltage sources,current sources,CCVS, CCCS,VCVS, VCCS)
i
L
i
v
A
B
L
RTh
≡ VTh
* If all sources are DC (constant), VTh = constant .
* KVL: VTh = RTh i + Ldi
dt.
* Homogeneous solution:
di
dt+
1
τi = 0 , where τ = L/RTh
→ i (h) = K exp(−t/τ) .
* Particular solution is a specific function that satisfies the differential equation. We know that all timederivatives will vanish as t →∞ , making v = 0, and we get i (p) = VTh/RTh as a particular solution(which happens to be simply a constant).
* i = i (h) + i (p) = K exp(−t/τ) + VTh/RTh .
* In general, i(t) = A exp(−t/τ) + B , where A and B can be obtained from known conditions on i .
M. B. Patil, IIT Bombay
RL circuits with DC sources (continued)
i
v
A
B
v
CircuitA
B
(resistors,voltage sources,current sources,CCVS, CCCS,VCVS, VCCS)
i
L L
RTh
≡ VTh
* If all sources are DC (constant), we havei(t) = A exp(−t/τ) + B , τ = L/RTh .
* v(t) = Ldi
dt= L× A exp(−t/τ)
(− 1
τ
)≡ A′ exp(−t/τ) .
* As t →∞, v → 0, i.e., the inductor behaves like a short circuit since all derivatives vanish.
* Since the circuit in the black box is linear, any variable (current or voltage) in the circuit can beexpressed asx(t) = K1 exp(−t/τ) + K2 ,where K1 and K2 can be obtained from suitable conditions on x(t).
M. B. Patil, IIT Bombay
RL circuits with DC sources (continued)
i
v
A
B
v
CircuitA
B
(resistors,voltage sources,current sources,CCVS, CCCS,VCVS, VCCS)
i
L L
RTh
≡ VTh
* If all sources are DC (constant), we havei(t) = A exp(−t/τ) + B , τ = L/RTh .
* v(t) = Ldi
dt= L× A exp(−t/τ)
(− 1
τ
)≡ A′ exp(−t/τ) .
* As t →∞, v → 0, i.e., the inductor behaves like a short circuit since all derivatives vanish.
* Since the circuit in the black box is linear, any variable (current or voltage) in the circuit can beexpressed asx(t) = K1 exp(−t/τ) + K2 ,where K1 and K2 can be obtained from suitable conditions on x(t).
M. B. Patil, IIT Bombay
RL circuits with DC sources (continued)
i
v
A
B
v
CircuitA
B
(resistors,voltage sources,current sources,CCVS, CCCS,VCVS, VCCS)
i
L L
RTh
≡ VTh
* If all sources are DC (constant), we havei(t) = A exp(−t/τ) + B , τ = L/RTh .
* v(t) = Ldi
dt= L× A exp(−t/τ)
(− 1
τ
)≡ A′ exp(−t/τ) .
* As t →∞, v → 0, i.e., the inductor behaves like a short circuit since all derivatives vanish.
* Since the circuit in the black box is linear, any variable (current or voltage) in the circuit can beexpressed asx(t) = K1 exp(−t/τ) + K2 ,where K1 and K2 can be obtained from suitable conditions on x(t).
M. B. Patil, IIT Bombay
RL circuits with DC sources (continued)
i
v
A
B
v
CircuitA
B
(resistors,voltage sources,current sources,CCVS, CCCS,VCVS, VCCS)
i
L L
RTh
≡ VTh
* If all sources are DC (constant), we havei(t) = A exp(−t/τ) + B , τ = L/RTh .
* v(t) = Ldi
dt= L× A exp(−t/τ)
(− 1
τ
)≡ A′ exp(−t/τ) .
* As t →∞, v → 0, i.e., the inductor behaves like a short circuit since all derivatives vanish.
* Since the circuit in the black box is linear, any variable (current or voltage) in the circuit can beexpressed asx(t) = K1 exp(−t/τ) + K2 ,where K1 and K2 can be obtained from suitable conditions on x(t).
M. B. Patil, IIT Bombay
RC circuits: Can Vc change “suddenly?”
i
0 V t
5 V
Vs
Vs
C= 1µFVc
R= 1 k
Vc(0)= 0V
* Vs changes from 0V (at t = 0−), to 5V (at t = 0+). As a result of this change, Vc will rise. How fastcan Vc change?
* For example, what would happen if Vc changes by 1V in 1µs at a constant rate of 1V /1µs = 106 V /s?
* i = CdVc
dt= 1µF × 106 V
s= 1A .
* With i = 1A, the voltage drop across R would be 1000V ! Not allowed by KVL.
* We conclude that Vc (0+) =Vc (0−)⇒ A capacitor does not allow abrupt changes in Vc if there is a finiteresistance in the circuit.
* Similarly, an inductor does not allow abrupt changes in iL.
M. B. Patil, IIT Bombay
RC circuits: Can Vc change “suddenly?”
i
0 V t
5 V
Vs
Vs
C= 1µFVc
R= 1 k
Vc(0)= 0V
* Vs changes from 0V (at t = 0−), to 5V (at t = 0+). As a result of this change, Vc will rise. How fastcan Vc change?
* For example, what would happen if Vc changes by 1V in 1µs at a constant rate of 1V /1µs = 106 V /s?
* i = CdVc
dt= 1µF × 106 V
s= 1A .
* With i = 1A, the voltage drop across R would be 1000V ! Not allowed by KVL.
* We conclude that Vc (0+) =Vc (0−)⇒ A capacitor does not allow abrupt changes in Vc if there is a finiteresistance in the circuit.
* Similarly, an inductor does not allow abrupt changes in iL.
M. B. Patil, IIT Bombay
RC circuits: Can Vc change “suddenly?”
i
0 V t
5 V
Vs
Vs
C= 1µFVc
R= 1 k
Vc(0)= 0V
* Vs changes from 0V (at t = 0−), to 5V (at t = 0+). As a result of this change, Vc will rise. How fastcan Vc change?
* For example, what would happen if Vc changes by 1V in 1µs at a constant rate of 1V /1µs = 106 V /s?
* i = CdVc
dt= 1µF × 106 V
s= 1A .
* With i = 1A, the voltage drop across R would be 1000V ! Not allowed by KVL.
* We conclude that Vc (0+) =Vc (0−)⇒ A capacitor does not allow abrupt changes in Vc if there is a finiteresistance in the circuit.
* Similarly, an inductor does not allow abrupt changes in iL.
M. B. Patil, IIT Bombay
RC circuits: Can Vc change “suddenly?”
i
0 V t
5 V
Vs
Vs
C= 1µFVc
R= 1 k
Vc(0)= 0V
* Vs changes from 0V (at t = 0−), to 5V (at t = 0+). As a result of this change, Vc will rise. How fastcan Vc change?
* For example, what would happen if Vc changes by 1V in 1µs at a constant rate of 1V /1µs = 106 V /s?
* i = CdVc
dt= 1µF × 106 V
s= 1A .
* With i = 1A, the voltage drop across R would be 1000V ! Not allowed by KVL.
* We conclude that Vc (0+) =Vc (0−)⇒ A capacitor does not allow abrupt changes in Vc if there is a finiteresistance in the circuit.
* Similarly, an inductor does not allow abrupt changes in iL.
M. B. Patil, IIT Bombay
RC circuits: Can Vc change “suddenly?”
i
0 V t
5 V
Vs
Vs
C= 1µFVc
R= 1 k
Vc(0)= 0V
* Vs changes from 0V (at t = 0−), to 5V (at t = 0+). As a result of this change, Vc will rise. How fastcan Vc change?
* For example, what would happen if Vc changes by 1V in 1µs at a constant rate of 1V /1µs = 106 V /s?
* i = CdVc
dt= 1µF × 106 V
s= 1A .
* With i = 1A, the voltage drop across R would be 1000V ! Not allowed by KVL.
* We conclude that Vc (0+) =Vc (0−)⇒ A capacitor does not allow abrupt changes in Vc if there is a finiteresistance in the circuit.
* Similarly, an inductor does not allow abrupt changes in iL.
M. B. Patil, IIT Bombay
RC circuits: Can Vc change “suddenly?”
i
0 V t
5 V
Vs
Vs
C= 1µFVc
R= 1 k
Vc(0)= 0V
* Vs changes from 0V (at t = 0−), to 5V (at t = 0+). As a result of this change, Vc will rise. How fastcan Vc change?
* For example, what would happen if Vc changes by 1V in 1µs at a constant rate of 1V /1µs = 106 V /s?
* i = CdVc
dt= 1µF × 106 V
s= 1A .
* With i = 1A, the voltage drop across R would be 1000V ! Not allowed by KVL.
* We conclude that Vc (0+) =Vc (0−)⇒ A capacitor does not allow abrupt changes in Vc if there is a finiteresistance in the circuit.
* Similarly, an inductor does not allow abrupt changes in iL.
M. B. Patil, IIT Bombay
RC circuits: Can Vc change “suddenly?”
i
0 V t
5 V
Vs
Vs
C= 1µFVc
R= 1 k
Vc(0)= 0V
* Vs changes from 0V (at t = 0−), to 5V (at t = 0+). As a result of this change, Vc will rise. How fastcan Vc change?
* For example, what would happen if Vc changes by 1V in 1µs at a constant rate of 1V /1µs = 106 V /s?
* i = CdVc
dt= 1µF × 106 V
s= 1A .
* With i = 1A, the voltage drop across R would be 1000V ! Not allowed by KVL.
* We conclude that Vc (0+) =Vc (0−)⇒ A capacitor does not allow abrupt changes in Vc if there is a finiteresistance in the circuit.
* Similarly, an inductor does not allow abrupt changes in iL.
M. B. Patil, IIT Bombay
RC circuits: charging and discharging transients
t0 V
i
Cv
R
Vs
Vs
V0
(A)Let v(t) = A exp(−t/τ) + B, t > 0
(1)
(2)
Conditions on v(t):
v(0−) = Vs(0−) = 0 V
v(0+) ≃ v(0−) = 0 V
Note that we need the condition at 0+ (and not at 0−)
because Eq. (A) applies only for t > 0.
As t → ∞ , i → 0 → v(∞) = Vs(∞) = V0
Imposing (1) and (2) on Eq. (A), we get
i.e., B = V0 ,A = −V0
t = 0+: 0 = A+ B ,
t → ∞: V0 = B .
v(t) = V0 [1− exp(−t/τ)]
0 V
i
t
Cv
RVs
Vs
V0
(A)Let v(t) = A exp(−t/τ) + B, t > 0
(1)
(2)
Conditions on v(t):
Note that we need the condition at 0+ (and not at 0−)
because Eq. (A) applies only for t > 0.
v(0−) = Vs(0−) = V0
v(0+) ≃ v(0−) = V0
As t → ∞ , i → 0 → v(∞) = Vs(∞) = 0 V
Imposing (1) and (2) on Eq. (A), we get
t = 0+: V0 = A+ B ,
i.e., A = V0 ,B = 0
t → ∞: 0 = B .
v(t) = V0 exp(−t/τ)
M. B. Patil, IIT Bombay
RC circuits: charging and discharging transients
t0 V
i
Cv
R
Vs
Vs
V0
(A)Let v(t) = A exp(−t/τ) + B, t > 0
(1)
(2)
Conditions on v(t):
v(0−) = Vs(0−) = 0 V
v(0+) ≃ v(0−) = 0 V
Note that we need the condition at 0+ (and not at 0−)
because Eq. (A) applies only for t > 0.
As t → ∞ , i → 0 → v(∞) = Vs(∞) = V0
Imposing (1) and (2) on Eq. (A), we get
i.e., B = V0 ,A = −V0
t = 0+: 0 = A+ B ,
t → ∞: V0 = B .
v(t) = V0 [1− exp(−t/τ)]
0 V
i
t
Cv
RVs
Vs
V0
(A)Let v(t) = A exp(−t/τ) + B, t > 0
(1)
(2)
Conditions on v(t):
Note that we need the condition at 0+ (and not at 0−)
because Eq. (A) applies only for t > 0.
v(0−) = Vs(0−) = V0
v(0+) ≃ v(0−) = V0
As t → ∞ , i → 0 → v(∞) = Vs(∞) = 0 V
Imposing (1) and (2) on Eq. (A), we get
t = 0+: V0 = A+ B ,
i.e., A = V0 ,B = 0
t → ∞: 0 = B .
v(t) = V0 exp(−t/τ)
M. B. Patil, IIT Bombay
RC circuits: charging and discharging transients
t0 V
i
Cv
R
Vs
Vs
V0
(A)Let v(t) = A exp(−t/τ) + B, t > 0
(1)
(2)
Conditions on v(t):
v(0−) = Vs(0−) = 0 V
v(0+) ≃ v(0−) = 0 V
Note that we need the condition at 0+ (and not at 0−)
because Eq. (A) applies only for t > 0.
As t → ∞ , i → 0 → v(∞) = Vs(∞) = V0
Imposing (1) and (2) on Eq. (A), we get
i.e., B = V0 ,A = −V0
t = 0+: 0 = A+ B ,
t → ∞: V0 = B .
v(t) = V0 [1− exp(−t/τ)]
0 V
i
t
Cv
RVs
Vs
V0
(A)Let v(t) = A exp(−t/τ) + B, t > 0
(1)
(2)
Conditions on v(t):
Note that we need the condition at 0+ (and not at 0−)
because Eq. (A) applies only for t > 0.
v(0−) = Vs(0−) = V0
v(0+) ≃ v(0−) = V0
As t → ∞ , i → 0 → v(∞) = Vs(∞) = 0 V
Imposing (1) and (2) on Eq. (A), we get
t = 0+: V0 = A+ B ,
i.e., A = V0 ,B = 0
t → ∞: 0 = B .
v(t) = V0 exp(−t/τ)
M. B. Patil, IIT Bombay
RC circuits: charging and discharging transients
t0 V
i
Cv
R
Vs
Vs
V0
(A)Let v(t) = A exp(−t/τ) + B, t > 0
(1)
(2)
Conditions on v(t):
v(0−) = Vs(0−) = 0 V
v(0+) ≃ v(0−) = 0 V
Note that we need the condition at 0+ (and not at 0−)
because Eq. (A) applies only for t > 0.
As t → ∞ , i → 0 → v(∞) = Vs(∞) = V0
Imposing (1) and (2) on Eq. (A), we get
i.e., B = V0 ,A = −V0
t = 0+: 0 = A+ B ,
t → ∞: V0 = B .
v(t) = V0 [1− exp(−t/τ)]
0 V
i
t
Cv
RVs
Vs
V0
(A)Let v(t) = A exp(−t/τ) + B, t > 0
(1)
(2)
Conditions on v(t):
Note that we need the condition at 0+ (and not at 0−)
because Eq. (A) applies only for t > 0.
v(0−) = Vs(0−) = V0
v(0+) ≃ v(0−) = V0
As t → ∞ , i → 0 → v(∞) = Vs(∞) = 0 V
Imposing (1) and (2) on Eq. (A), we get
t = 0+: V0 = A+ B ,
i.e., A = V0 ,B = 0
t → ∞: 0 = B .
v(t) = V0 exp(−t/τ)
M. B. Patil, IIT Bombay
RC circuits: charging and discharging transients
t0 V
i
Cv
R
Vs
Vs
V0
(A)Let v(t) = A exp(−t/τ) + B, t > 0
(1)
(2)
Conditions on v(t):
v(0−) = Vs(0−) = 0 V
v(0+) ≃ v(0−) = 0 V
Note that we need the condition at 0+ (and not at 0−)
because Eq. (A) applies only for t > 0.
As t → ∞ , i → 0 → v(∞) = Vs(∞) = V0
Imposing (1) and (2) on Eq. (A), we get
i.e., B = V0 ,A = −V0
t = 0+: 0 = A+ B ,
t → ∞: V0 = B .
v(t) = V0 [1− exp(−t/τ)]
0 V
i
t
Cv
RVs
Vs
V0
(A)Let v(t) = A exp(−t/τ) + B, t > 0
(1)
(2)
Conditions on v(t):
Note that we need the condition at 0+ (and not at 0−)
because Eq. (A) applies only for t > 0.
v(0−) = Vs(0−) = V0
v(0+) ≃ v(0−) = V0
As t → ∞ , i → 0 → v(∞) = Vs(∞) = 0 V
Imposing (1) and (2) on Eq. (A), we get
t = 0+: V0 = A+ B ,
i.e., A = V0 ,B = 0
t → ∞: 0 = B .
v(t) = V0 exp(−t/τ)
M. B. Patil, IIT Bombay
RC circuits: charging and discharging transients
t0 V
i
Cv
R
Vs
Vs
V0
(A)Let v(t) = A exp(−t/τ) + B, t > 0
(1)
(2)
Conditions on v(t):
v(0−) = Vs(0−) = 0 V
v(0+) ≃ v(0−) = 0 V
Note that we need the condition at 0+ (and not at 0−)
because Eq. (A) applies only for t > 0.
As t → ∞ , i → 0 → v(∞) = Vs(∞) = V0
Imposing (1) and (2) on Eq. (A), we get
i.e., B = V0 ,A = −V0
t = 0+: 0 = A+ B ,
t → ∞: V0 = B .
v(t) = V0 [1− exp(−t/τ)]
0 V
i
t
Cv
RVs
Vs
V0
(A)Let v(t) = A exp(−t/τ) + B, t > 0
(1)
(2)
Conditions on v(t):
Note that we need the condition at 0+ (and not at 0−)
because Eq. (A) applies only for t > 0.
v(0−) = Vs(0−) = V0
v(0+) ≃ v(0−) = V0
As t → ∞ , i → 0 → v(∞) = Vs(∞) = 0 V
Imposing (1) and (2) on Eq. (A), we get
t = 0+: V0 = A+ B ,
i.e., A = V0 ,B = 0
t → ∞: 0 = B .
v(t) = V0 exp(−t/τ)
M. B. Patil, IIT Bombay
RC circuits: charging and discharging transients
t0 V
i
Cv
R
Vs
Vs
V0
(A)Let v(t) = A exp(−t/τ) + B, t > 0
(1)
(2)
Conditions on v(t):
v(0−) = Vs(0−) = 0 V
v(0+) ≃ v(0−) = 0 V
Note that we need the condition at 0+ (and not at 0−)
because Eq. (A) applies only for t > 0.
As t → ∞ , i → 0 → v(∞) = Vs(∞) = V0
Imposing (1) and (2) on Eq. (A), we get
i.e., B = V0 ,A = −V0
t = 0+: 0 = A+ B ,
t → ∞: V0 = B .
v(t) = V0 [1− exp(−t/τ)]
0 V
i
t
Cv
RVs
Vs
V0
(A)Let v(t) = A exp(−t/τ) + B, t > 0
(1)
(2)
Conditions on v(t):
Note that we need the condition at 0+ (and not at 0−)
because Eq. (A) applies only for t > 0.
v(0−) = Vs(0−) = V0
v(0+) ≃ v(0−) = V0
As t → ∞ , i → 0 → v(∞) = Vs(∞) = 0 V
Imposing (1) and (2) on Eq. (A), we get
t = 0+: V0 = A+ B ,
i.e., A = V0 ,B = 0
t → ∞: 0 = B .
v(t) = V0 exp(−t/τ)
M. B. Patil, IIT Bombay
RC circuits: charging and discharging transients
t0 V
i
Cv
R
Vs
Vs
V0
(A)Let v(t) = A exp(−t/τ) + B, t > 0
(1)
(2)
Conditions on v(t):
v(0−) = Vs(0−) = 0 V
v(0+) ≃ v(0−) = 0 V
Note that we need the condition at 0+ (and not at 0−)
because Eq. (A) applies only for t > 0.
As t → ∞ , i → 0 → v(∞) = Vs(∞) = V0
Imposing (1) and (2) on Eq. (A), we get
i.e., B = V0 ,A = −V0
t = 0+: 0 = A+ B ,
t → ∞: V0 = B .
v(t) = V0 [1− exp(−t/τ)]
0 V
i
t
Cv
RVs
Vs
V0
(A)Let v(t) = A exp(−t/τ) + B, t > 0
(1)
(2)
Conditions on v(t):
Note that we need the condition at 0+ (and not at 0−)
because Eq. (A) applies only for t > 0.
v(0−) = Vs(0−) = V0
v(0+) ≃ v(0−) = V0
As t → ∞ , i → 0 → v(∞) = Vs(∞) = 0 V
Imposing (1) and (2) on Eq. (A), we get
t = 0+: V0 = A+ B ,
i.e., A = V0 ,B = 0
t → ∞: 0 = B .
v(t) = V0 exp(−t/τ)
M. B. Patil, IIT Bombay
RC circuits: charging and discharging transients
t0 V
i
Cv
R
Vs
Vs
V0
(A)Let v(t) = A exp(−t/τ) + B, t > 0
(1)
(2)
Conditions on v(t):
v(0−) = Vs(0−) = 0 V
v(0+) ≃ v(0−) = 0 V
Note that we need the condition at 0+ (and not at 0−)
because Eq. (A) applies only for t > 0.
As t → ∞ , i → 0 → v(∞) = Vs(∞) = V0
Imposing (1) and (2) on Eq. (A), we get
i.e., B = V0 ,A = −V0
t = 0+: 0 = A+ B ,
t → ∞: V0 = B .
v(t) = V0 [1− exp(−t/τ)]
0 V
i
t
Cv
RVs
Vs
V0
(A)Let v(t) = A exp(−t/τ) + B, t > 0
(1)
(2)
Conditions on v(t):
Note that we need the condition at 0+ (and not at 0−)
because Eq. (A) applies only for t > 0.
v(0−) = Vs(0−) = V0
v(0+) ≃ v(0−) = V0
As t → ∞ , i → 0 → v(∞) = Vs(∞) = 0 V
Imposing (1) and (2) on Eq. (A), we get
t = 0+: V0 = A+ B ,
i.e., A = V0 ,B = 0
t → ∞: 0 = B .
v(t) = V0 exp(−t/τ)
M. B. Patil, IIT Bombay
RC circuits: charging and discharging transients
t0 V
i
Cv
R
Vs
Vs
V0
(A)Let v(t) = A exp(−t/τ) + B, t > 0
(1)
(2)
Conditions on v(t):
v(0−) = Vs(0−) = 0 V
v(0+) ≃ v(0−) = 0 V
Note that we need the condition at 0+ (and not at 0−)
because Eq. (A) applies only for t > 0.
As t → ∞ , i → 0 → v(∞) = Vs(∞) = V0
Imposing (1) and (2) on Eq. (A), we get
i.e., B = V0 ,A = −V0
t = 0+: 0 = A+ B ,
t → ∞: V0 = B .
v(t) = V0 [1− exp(−t/τ)]
0 V
i
t
Cv
RVs
Vs
V0
(A)Let v(t) = A exp(−t/τ) + B, t > 0
(1)
(2)
Conditions on v(t):
Note that we need the condition at 0+ (and not at 0−)
because Eq. (A) applies only for t > 0.
v(0−) = Vs(0−) = V0
v(0+) ≃ v(0−) = V0
As t → ∞ , i → 0 → v(∞) = Vs(∞) = 0 V
Imposing (1) and (2) on Eq. (A), we get
t = 0+: V0 = A+ B ,
i.e., A = V0 ,B = 0
t → ∞: 0 = B .
v(t) = V0 exp(−t/τ)
M. B. Patil, IIT Bombay
RC circuits: charging and discharging transients
t0 V
i
Cv
RVs
Vs
V0
Compute i(t), t > 0 .
(A) i(t) = Cd
dtV0 [1− exp(−t/τ)]
=CV0
τexp(−t/τ) =
V0
Rexp(−t/τ)
Using these conditions, we obtain
(B) Let i(t) = A′ exp(−t/τ) + B′ , t > 0 .
t = 0+: v = 0 , Vs = V0 ⇒ i(0+) = V0/R .
t → ∞: i(t) = 0 .
A′ =V0
R, B′ = 0 ⇒ i(t) =
V0
Rexp(−t/τ)
0 V
i
t
Cv
RVs
Vs
V0
Compute i(t), t > 0 .
(A) i(t) = Cd
dtV0 [exp(−t/τ)]
= −CV0
τexp(−t/τ) = −V0
Rexp(−t/τ)
Using these conditions, we obtain
(B) Let i(t) = A′ exp(−t/τ) + B′ , t > 0 .
t → ∞: i(t) = 0 .
t = 0+: v = V0 , Vs = 0 ⇒ i(0+) = −V0/R .
A′ = −V0
R, B′ = 0 ⇒ i(t) = −V0
Rexp(−t/τ)
M. B. Patil, IIT Bombay
RC circuits: charging and discharging transients
t0 V
i
Cv
RVs
Vs
V0
Compute i(t), t > 0 .
(A) i(t) = Cd
dtV0 [1− exp(−t/τ)]
=CV0
τexp(−t/τ) =
V0
Rexp(−t/τ)
Using these conditions, we obtain
(B) Let i(t) = A′ exp(−t/τ) + B′ , t > 0 .
t = 0+: v = 0 , Vs = V0 ⇒ i(0+) = V0/R .
t → ∞: i(t) = 0 .
A′ =V0
R, B′ = 0 ⇒ i(t) =
V0
Rexp(−t/τ)
0 V
i
t
Cv
RVs
Vs
V0
Compute i(t), t > 0 .
(A) i(t) = Cd
dtV0 [exp(−t/τ)]
= −CV0
τexp(−t/τ) = −V0
Rexp(−t/τ)
Using these conditions, we obtain
(B) Let i(t) = A′ exp(−t/τ) + B′ , t > 0 .
t → ∞: i(t) = 0 .
t = 0+: v = V0 , Vs = 0 ⇒ i(0+) = −V0/R .
A′ = −V0
R, B′ = 0 ⇒ i(t) = −V0
Rexp(−t/τ)
M. B. Patil, IIT Bombay
RC circuits: charging and discharging transients
t0 V
i
Cv
RVs
Vs
V0
Compute i(t), t > 0 .
(A) i(t) = Cd
dtV0 [1− exp(−t/τ)]
=CV0
τexp(−t/τ) =
V0
Rexp(−t/τ)
Using these conditions, we obtain
(B) Let i(t) = A′ exp(−t/τ) + B′ , t > 0 .
t = 0+: v = 0 , Vs = V0 ⇒ i(0+) = V0/R .
t → ∞: i(t) = 0 .
A′ =V0
R, B′ = 0 ⇒ i(t) =
V0
Rexp(−t/τ)
0 V
i
t
Cv
RVs
Vs
V0
Compute i(t), t > 0 .
(A) i(t) = Cd
dtV0 [exp(−t/τ)]
= −CV0
τexp(−t/τ) = −V0
Rexp(−t/τ)
Using these conditions, we obtain
(B) Let i(t) = A′ exp(−t/τ) + B′ , t > 0 .
t → ∞: i(t) = 0 .
t = 0+: v = V0 , Vs = 0 ⇒ i(0+) = −V0/R .
A′ = −V0
R, B′ = 0 ⇒ i(t) = −V0
Rexp(−t/τ)
M. B. Patil, IIT Bombay
RC circuits: charging and discharging transients
t0 V
i
Cv
RVs
Vs
V0
Compute i(t), t > 0 .
(A) i(t) = Cd
dtV0 [1− exp(−t/τ)]
=CV0
τexp(−t/τ) =
V0
Rexp(−t/τ)
Using these conditions, we obtain
(B) Let i(t) = A′ exp(−t/τ) + B′ , t > 0 .
t = 0+: v = 0 , Vs = V0 ⇒ i(0+) = V0/R .
t → ∞: i(t) = 0 .
A′ =V0
R, B′ = 0 ⇒ i(t) =
V0
Rexp(−t/τ)
0 V
i
t
Cv
RVs
Vs
V0
Compute i(t), t > 0 .
(A) i(t) = Cd
dtV0 [exp(−t/τ)]
= −CV0
τexp(−t/τ) = −V0
Rexp(−t/τ)
Using these conditions, we obtain
(B) Let i(t) = A′ exp(−t/τ) + B′ , t > 0 .
t → ∞: i(t) = 0 .
t = 0+: v = V0 , Vs = 0 ⇒ i(0+) = −V0/R .
A′ = −V0
R, B′ = 0 ⇒ i(t) = −V0
Rexp(−t/τ)
M. B. Patil, IIT Bombay
RC circuits: charging and discharging transients
t0 V
i
Cv
RVs
Vs
V0
Compute i(t), t > 0 .
(A) i(t) = Cd
dtV0 [1− exp(−t/τ)]
=CV0
τexp(−t/τ) =
V0
Rexp(−t/τ)
Using these conditions, we obtain
(B) Let i(t) = A′ exp(−t/τ) + B′ , t > 0 .
t = 0+: v = 0 , Vs = V0 ⇒ i(0+) = V0/R .
t → ∞: i(t) = 0 .
A′ =V0
R, B′ = 0 ⇒ i(t) =
V0
Rexp(−t/τ)
0 V
i
t
Cv
RVs
Vs
V0
Compute i(t), t > 0 .
(A) i(t) = Cd
dtV0 [exp(−t/τ)]
= −CV0
τexp(−t/τ) = −V0
Rexp(−t/τ)
Using these conditions, we obtain
(B) Let i(t) = A′ exp(−t/τ) + B′ , t > 0 .
t → ∞: i(t) = 0 .
t = 0+: v = V0 , Vs = 0 ⇒ i(0+) = −V0/R .
A′ = −V0
R, B′ = 0 ⇒ i(t) = −V0
Rexp(−t/τ)
M. B. Patil, IIT Bombay
RC circuits: charging and discharging transients
t0 V
i
Cv
RVs
Vs
V0
Compute i(t), t > 0 .
(A) i(t) = Cd
dtV0 [1− exp(−t/τ)]
=CV0
τexp(−t/τ) =
V0
Rexp(−t/τ)
Using these conditions, we obtain
(B) Let i(t) = A′ exp(−t/τ) + B′ , t > 0 .
t = 0+: v = 0 , Vs = V0 ⇒ i(0+) = V0/R .
t → ∞: i(t) = 0 .
A′ =V0
R, B′ = 0 ⇒ i(t) =
V0
Rexp(−t/τ)
0 V
i
t
Cv
RVs
Vs
V0
Compute i(t), t > 0 .
(A) i(t) = Cd
dtV0 [exp(−t/τ)]
= −CV0
τexp(−t/τ) = −V0
Rexp(−t/τ)
Using these conditions, we obtain
(B) Let i(t) = A′ exp(−t/τ) + B′ , t > 0 .
t → ∞: i(t) = 0 .
t = 0+: v = V0 , Vs = 0 ⇒ i(0+) = −V0/R .
A′ = −V0
R, B′ = 0 ⇒ i(t) = −V0
Rexp(−t/τ)
M. B. Patil, IIT Bombay
RC circuits: charging and discharging transients
t0 V
i
C= 1µFv
R=1 kVs
Vs5 V
v(t) = V0 [1− exp(−t/τ)]
i(t) =V0
Rexp(−t/τ)
0
5
v (
Vo
lts) v
Vs
5
0
i (m
A)
time (msec)
−2 0 2 4 6 8
t0 V
i
C1µF
v
R=1 kVs
Vs5 V
v(t) = V0 exp(−t/τ)
i(t) = −V0
Rexp(−t/τ)
v (
Vo
lts)
5
0
v
Vs
i (m
A)
0
−5
time (msec)
−2 0 2 4 6 8
M. B. Patil, IIT Bombay
RC circuits: charging and discharging transients
t0 V
i
C= 1µFv
R=1 kVs
Vs5 V
v(t) = V0 [1− exp(−t/τ)]
i(t) =V0
Rexp(−t/τ)
0
5
v (
Vo
lts) v
Vs
5
0
i (m
A)
time (msec)
−2 0 2 4 6 8
t0 V
i
C1µF
v
R=1 kVs
Vs5 V
v(t) = V0 exp(−t/τ)
i(t) = −V0
Rexp(−t/τ)
v (
Vo
lts)
5
0
v
Vs
i (m
A)
0
−5
time (msec)
−2 0 2 4 6 8
M. B. Patil, IIT Bombay
RC circuits: charging and discharging transients
t0 V
i
C= 1µFv
R=1 kVs
Vs5 V
v(t) = V0 [1− exp(−t/τ)]
i(t) =V0
Rexp(−t/τ)
0
5
v (
Vo
lts) v
Vs
5
0
i (m
A)
time (msec)
−2 0 2 4 6 8
t0 V
i
C1µF
v
R=1 kVs
Vs5 V
v(t) = V0 exp(−t/τ)
i(t) = −V0
Rexp(−t/τ)
v (
Vo
lts)
5
0
v
Vs
i (m
A)
0
−5
time (msec)
−2 0 2 4 6 8
M. B. Patil, IIT Bombay
RC circuits: charging and discharging transients
t0 V
i
C= 1µFv
R=1 kVs
Vs5 V
v(t) = V0 [1− exp(−t/τ)]
i(t) =V0
Rexp(−t/τ)
0
5
v (
Vo
lts) v
Vs
5
0
i (m
A)
time (msec)
−2 0 2 4 6 8
t0 V
i
C1µF
v
R=1 kVs
Vs5 V
v(t) = V0 exp(−t/τ)
i(t) = −V0
Rexp(−t/τ)
v (
Vo
lts)
5
0
v
Vs
i (m
A)
0
−5
time (msec)
−2 0 2 4 6 8
M. B. Patil, IIT Bombay
RC circuits: charging and discharging transients
t0 V
i
C= 1µFv
R=1 kVs
Vs5 V
v(t) = V0 [1− exp(−t/τ)]
i(t) =V0
Rexp(−t/τ)
0
5
v (
Vo
lts) v
Vs
5
0
i (m
A)
time (msec)
−2 0 2 4 6 8
t0 V
i
C1µF
v
R=1 kVs
Vs5 V
v(t) = V0 exp(−t/τ)
i(t) = −V0
Rexp(−t/τ)
v (
Vo
lts)
5
0
v
Vs
i (m
A)
0
−5
time (msec)
−2 0 2 4 6 8
M. B. Patil, IIT Bombay
RC circuits: charging and discharging transients
t0 V
i
C= 1µFv
R=1 kVs
Vs5 V
v(t) = V0 [1− exp(−t/τ)]
i(t) =V0
Rexp(−t/τ)
0
5
v (
Vo
lts) v
Vs
5
0
i (m
A)
time (msec)
−2 0 2 4 6 8
t0 V
i
C1µF
v
R=1 kVs
Vs5 V
v(t) = V0 exp(−t/τ)
i(t) = −V0
Rexp(−t/τ)
v (
Vo
lts)
5
0
v
Vs
i (m
A)
0
−5
time (msec)
−2 0 2 4 6 8
M. B. Patil, IIT Bombay
RC circuits: charging and discharging transients
0 V 0 V
0
5
v (
Volts)
v (
Volts)
5
0
i i
t t
time (msec) time (msec)−1 0 1 2 3 4 5 6 −1 0 1 2 3 4 5 6
C= 1µF C= 1µFv v
R RVs Vs
Vs Vs5 V 5 V
R = 1 kΩ
R = 100 Ω
R = 1 kΩ
R = 100 Ω
v(t) = V0 exp(−t/τ)v(t) = V0 [1− exp(−t/τ)]
M. B. Patil, IIT Bombay
Analysis of RC/RL circuits with a piece-wise constant source
* Identify intervals in which the source voltages/currents are constant.For example,
(1) t < t1
(3) t > t2
(2) t1 < t < t2
t1 t2
Vs
0 t
* For any current or voltage x(t), write general expressions such as,x(t) = A1 exp(−t/τ) + B1 , t < t1 ,x(t) = A2 exp(−t/τ) + B2 , t1 < t < t2 ,x(t) = A3 exp(−t/τ) + B3 , t > t2 .
* Work out suitable conditions on x(t) at specific time points using
(a) If the source voltage/current has not changed for a “long” time(long compared to τ), all derivatives are zero.
⇒ iC = CdVc
dt= 0 , and VL = L
diL
dt= 0 .
(b) When a source voltage (or current) changes, say, at t = t0 ,Vc (t) or iL(t) cannot change abruptly, i.e.,
Vc (t+0 ) = Vc (t−0 ) , and iL(t+
0 ) = iL(t−0 ) .
* Compute A1, B1, · · · using the conditions on x(t).
M. B. Patil, IIT Bombay
Analysis of RC/RL circuits with a piece-wise constant source
* Identify intervals in which the source voltages/currents are constant.For example,
(1) t < t1
(3) t > t2
(2) t1 < t < t2
t1 t2
Vs
0 t
* For any current or voltage x(t), write general expressions such as,x(t) = A1 exp(−t/τ) + B1 , t < t1 ,x(t) = A2 exp(−t/τ) + B2 , t1 < t < t2 ,x(t) = A3 exp(−t/τ) + B3 , t > t2 .
* Work out suitable conditions on x(t) at specific time points using
(a) If the source voltage/current has not changed for a “long” time(long compared to τ), all derivatives are zero.
⇒ iC = CdVc
dt= 0 , and VL = L
diL
dt= 0 .
(b) When a source voltage (or current) changes, say, at t = t0 ,Vc (t) or iL(t) cannot change abruptly, i.e.,
Vc (t+0 ) = Vc (t−0 ) , and iL(t+
0 ) = iL(t−0 ) .
* Compute A1, B1, · · · using the conditions on x(t).
M. B. Patil, IIT Bombay
Analysis of RC/RL circuits with a piece-wise constant source
* Identify intervals in which the source voltages/currents are constant.For example,
(1) t < t1
(3) t > t2
(2) t1 < t < t2
t1 t2
Vs
0 t
* For any current or voltage x(t), write general expressions such as,x(t) = A1 exp(−t/τ) + B1 , t < t1 ,x(t) = A2 exp(−t/τ) + B2 , t1 < t < t2 ,x(t) = A3 exp(−t/τ) + B3 , t > t2 .
* Work out suitable conditions on x(t) at specific time points using
(a) If the source voltage/current has not changed for a “long” time(long compared to τ), all derivatives are zero.
⇒ iC = CdVc
dt= 0 , and VL = L
diL
dt= 0 .
(b) When a source voltage (or current) changes, say, at t = t0 ,Vc (t) or iL(t) cannot change abruptly, i.e.,
Vc (t+0 ) = Vc (t−0 ) , and iL(t+
0 ) = iL(t−0 ) .
* Compute A1, B1, · · · using the conditions on x(t).
M. B. Patil, IIT Bombay
Analysis of RC/RL circuits with a piece-wise constant source
* Identify intervals in which the source voltages/currents are constant.For example,
(1) t < t1
(3) t > t2
(2) t1 < t < t2
t1 t2
Vs
0 t
* For any current or voltage x(t), write general expressions such as,x(t) = A1 exp(−t/τ) + B1 , t < t1 ,x(t) = A2 exp(−t/τ) + B2 , t1 < t < t2 ,x(t) = A3 exp(−t/τ) + B3 , t > t2 .
* Work out suitable conditions on x(t) at specific time points using
(a) If the source voltage/current has not changed for a “long” time(long compared to τ), all derivatives are zero.
⇒ iC = CdVc
dt= 0 , and VL = L
diL
dt= 0 .
(b) When a source voltage (or current) changes, say, at t = t0 ,Vc (t) or iL(t) cannot change abruptly, i.e.,
Vc (t+0 ) = Vc (t−0 ) , and iL(t+
0 ) = iL(t−0 ) .
* Compute A1, B1, · · · using the conditions on x(t).
M. B. Patil, IIT Bombay
Analysis of RC/RL circuits with a piece-wise constant source
* Identify intervals in which the source voltages/currents are constant.For example,
(1) t < t1
(3) t > t2
(2) t1 < t < t2
t1 t2
Vs
0 t
* For any current or voltage x(t), write general expressions such as,x(t) = A1 exp(−t/τ) + B1 , t < t1 ,x(t) = A2 exp(−t/τ) + B2 , t1 < t < t2 ,x(t) = A3 exp(−t/τ) + B3 , t > t2 .
* Work out suitable conditions on x(t) at specific time points using
(a) If the source voltage/current has not changed for a “long” time(long compared to τ), all derivatives are zero.
⇒ iC = CdVc
dt= 0 , and VL = L
diL
dt= 0 .
(b) When a source voltage (or current) changes, say, at t = t0 ,Vc (t) or iL(t) cannot change abruptly, i.e.,
Vc (t+0 ) = Vc (t−0 ) , and iL(t+
0 ) = iL(t−0 ) .
* Compute A1, B1, · · · using the conditions on x(t).
M. B. Patil, IIT Bombay
Analysis of RC/RL circuits with a piece-wise constant source
* Identify intervals in which the source voltages/currents are constant.For example,
(1) t < t1
(3) t > t2
(2) t1 < t < t2
t1 t2
Vs
0 t
* For any current or voltage x(t), write general expressions such as,x(t) = A1 exp(−t/τ) + B1 , t < t1 ,x(t) = A2 exp(−t/τ) + B2 , t1 < t < t2 ,x(t) = A3 exp(−t/τ) + B3 , t > t2 .
* Work out suitable conditions on x(t) at specific time points using
(a) If the source voltage/current has not changed for a “long” time(long compared to τ), all derivatives are zero.
⇒ iC = CdVc
dt= 0 , and VL = L
diL
dt= 0 .
(b) When a source voltage (or current) changes, say, at t = t0 ,Vc (t) or iL(t) cannot change abruptly, i.e.,
Vc (t+0 ) = Vc (t−0 ) , and iL(t+
0 ) = iL(t−0 ) .
* Compute A1, B1, · · · using the conditions on x(t).
M. B. Patil, IIT Bombay
RL circuit: example
i
v
t
10 V
t0 t1
R2
R1
Vs
Vs
R1= 10Ω
R2= 40Ω
L= 0.8H
t0=0
t1=0.1 s
Find i(t).
(1) t < t0
(2) t0 < t < t1
(3) t > t1
There are three intervals of constant Vs:
R2
R1
Vs
RTh seen by L is the same in all intervals:
τ = L/RTh
= 0.1 s
= 0.8 H/8Ω
RTh = R1 ‖ R2 = 8Ω
⇒ i(t−0 ) = 0 A ⇒ i(t+0 ) = 0 A .
At t = t−0 , v = 0 V, Vs = 0 V .
10 V
t
v(∞) = 0 V, i(∞) = 10 V/10 Ω = 1 A .
If Vs did not change at t = t1,
we would have
t1t0
Vs
i(t), t > 0 (See next slide).
Using i(t+0 ) and i(∞), we can obtain
M. B. Patil, IIT Bombay
RL circuit: example
i
v
t
10 V
t0 t1
R2
R1
Vs
Vs
R1= 10Ω
R2= 40Ω
L= 0.8H
t0=0
t1=0.1 s
Find i(t).
(1) t < t0
(2) t0 < t < t1
(3) t > t1
There are three intervals of constant Vs:
R2
R1
Vs
RTh seen by L is the same in all intervals:
τ = L/RTh
= 0.1 s
= 0.8 H/8Ω
RTh = R1 ‖ R2 = 8Ω
⇒ i(t−0 ) = 0 A ⇒ i(t+0 ) = 0 A .
At t = t−0 , v = 0 V, Vs = 0 V .
10 V
t
v(∞) = 0 V, i(∞) = 10 V/10 Ω = 1 A .
If Vs did not change at t = t1,
we would have
t1t0
Vs
i(t), t > 0 (See next slide).
Using i(t+0 ) and i(∞), we can obtain
M. B. Patil, IIT Bombay
RL circuit: example
i
v
t
10 V
t0 t1
R2
R1
Vs
Vs
R1= 10Ω
R2= 40Ω
L= 0.8H
t0=0
t1=0.1 s
Find i(t).
(1) t < t0
(2) t0 < t < t1
(3) t > t1
There are three intervals of constant Vs:
R2
R1
Vs
RTh seen by L is the same in all intervals:
τ = L/RTh
= 0.1 s
= 0.8 H/8Ω
RTh = R1 ‖ R2 = 8Ω
⇒ i(t−0 ) = 0 A ⇒ i(t+0 ) = 0 A .
At t = t−0 , v = 0 V, Vs = 0 V .
10 V
t
v(∞) = 0 V, i(∞) = 10 V/10 Ω = 1 A .
If Vs did not change at t = t1,
we would have
t1t0
Vs
i(t), t > 0 (See next slide).
Using i(t+0 ) and i(∞), we can obtain
M. B. Patil, IIT Bombay
RL circuit: example
i
v
t
10 V
t0 t1
R2
R1
Vs
Vs
R1= 10Ω
R2= 40Ω
L= 0.8H
t0=0
t1=0.1 s
Find i(t).
(1) t < t0
(2) t0 < t < t1
(3) t > t1
There are three intervals of constant Vs:
R2
R1
Vs
RTh seen by L is the same in all intervals:
τ = L/RTh
= 0.1 s
= 0.8 H/8Ω
RTh = R1 ‖ R2 = 8Ω
⇒ i(t−0 ) = 0 A ⇒ i(t+0 ) = 0 A .
At t = t−0 , v = 0 V, Vs = 0 V .
10 V
t
v(∞) = 0 V, i(∞) = 10 V/10 Ω = 1 A .
If Vs did not change at t = t1,
we would have
t1t0
Vs
i(t), t > 0 (See next slide).
Using i(t+0 ) and i(∞), we can obtain
M. B. Patil, IIT Bombay
RL circuit: example
i
v
t
10 V
t0 t1
R2
R1
Vs
Vs
R1= 10Ω
R2= 40Ω
L= 0.8H
t0=0
t1=0.1 s
Find i(t).
(1) t < t0
(2) t0 < t < t1
(3) t > t1
There are three intervals of constant Vs:
R2
R1
Vs
RTh seen by L is the same in all intervals:
τ = L/RTh
= 0.1 s
= 0.8 H/8Ω
RTh = R1 ‖ R2 = 8Ω
⇒ i(t−0 ) = 0 A ⇒ i(t+0 ) = 0 A .
At t = t−0 , v = 0 V, Vs = 0 V .
10 V
t
v(∞) = 0 V, i(∞) = 10 V/10 Ω = 1 A .
If Vs did not change at t = t1,
we would have
t1t0
Vs
i(t), t > 0 (See next slide).
Using i(t+0 ) and i(∞), we can obtain
M. B. Patil, IIT Bombay
RL circuit: example
i
v
t
10 V
t0 t1
R2
R1
Vs
Vs
R1= 10Ω
R2= 40Ω
L= 0.8H
t0=0
t1=0.1 s
Find i(t).
(1) t < t0
(2) t0 < t < t1
(3) t > t1
There are three intervals of constant Vs:
R2
R1
Vs
RTh seen by L is the same in all intervals:
τ = L/RTh
= 0.1 s
= 0.8 H/8Ω
RTh = R1 ‖ R2 = 8Ω
⇒ i(t−0 ) = 0 A ⇒ i(t+0 ) = 0 A .
At t = t−0 , v = 0 V, Vs = 0 V .
10 V
t
v(∞) = 0 V, i(∞) = 10 V/10 Ω = 1 A .
If Vs did not change at t = t1,
we would have
t1t0
Vs
i(t), t > 0 (See next slide).
Using i(t+0 ) and i(∞), we can obtain
M. B. Patil, IIT Bombay
RL circuit: example
i
v
t
10 V
0 0.2 0.4 0.6 0.8
i (A
mp)
1
0
time (sec)
t1t0
R2
R1
Vs
Vs
R1 = 10Ω
R2 = 40Ω
L = 0.8H
t0 = 0
t1 = 0.1 s
and we need to work out the
solution for t > t1 separately.
In reality, Vs changes at t = t1,
Consider t > t1.
For t0 < t < t1, i(t) = 1− exp(−t/τ) Amp.
i(t+1 ) = i(t−1 ) = 1− e−1 = 0.632 A (Note: t1/τ = 1).
i(∞) = 0 A.
Let i(t) = A exp(−t/τ) + B.
It is convenient to rewrite i(t) as
i(t) = A′ exp[−(t− t1)/τ ] + B.
Using i(t+1 ) and i(∞), we get
i(t) = 0.632 exp[−(t− t1)/τ ] A.
M. B. Patil, IIT Bombay
RL circuit: example
i
v
t
10 V
0 0.2 0.4 0.6 0.8
i (A
mp)
1
0
time (sec)
t1t0
R2
R1
Vs
Vs
R1 = 10Ω
R2 = 40Ω
L = 0.8H
t0 = 0
t1 = 0.1 s
and we need to work out the
solution for t > t1 separately.
In reality, Vs changes at t = t1,
Consider t > t1.
For t0 < t < t1, i(t) = 1− exp(−t/τ) Amp.
i(t+1 ) = i(t−1 ) = 1− e−1 = 0.632 A (Note: t1/τ = 1).
i(∞) = 0 A.
Let i(t) = A exp(−t/τ) + B.
It is convenient to rewrite i(t) as
i(t) = A′ exp[−(t− t1)/τ ] + B.
Using i(t+1 ) and i(∞), we get
i(t) = 0.632 exp[−(t− t1)/τ ] A.
M. B. Patil, IIT Bombay
RL circuit: example
i
v
t
10 V
0 0.2 0.4 0.6 0.8
i (A
mp)
1
0
time (sec)
t1t0
R2
R1
Vs
Vs
R1 = 10Ω
R2 = 40Ω
L = 0.8H
t0 = 0
t1 = 0.1 s
and we need to work out the
solution for t > t1 separately.
In reality, Vs changes at t = t1,
Consider t > t1.
For t0 < t < t1, i(t) = 1− exp(−t/τ) Amp.
i(t+1 ) = i(t−1 ) = 1− e−1 = 0.632 A (Note: t1/τ = 1).
i(∞) = 0 A.
Let i(t) = A exp(−t/τ) + B.
It is convenient to rewrite i(t) as
i(t) = A′ exp[−(t− t1)/τ ] + B.
Using i(t+1 ) and i(∞), we get
i(t) = 0.632 exp[−(t− t1)/τ ] A.
M. B. Patil, IIT Bombay
RL circuit: example
i
v
t
10 V
0 0.2 0.4 0.6 0.8
i (A
mp)
1
0
time (sec)
t1t0
R2
R1
Vs
Vs
R1 = 10Ω
R2 = 40Ω
L = 0.8H
t0 = 0
t1 = 0.1 s
and we need to work out the
solution for t > t1 separately.
In reality, Vs changes at t = t1,
Consider t > t1.
For t0 < t < t1, i(t) = 1− exp(−t/τ) Amp.
i(t+1 ) = i(t−1 ) = 1− e−1 = 0.632 A (Note: t1/τ = 1).
i(∞) = 0 A.
Let i(t) = A exp(−t/τ) + B.
It is convenient to rewrite i(t) as
i(t) = A′ exp[−(t− t1)/τ ] + B.
Using i(t+1 ) and i(∞), we get
i(t) = 0.632 exp[−(t− t1)/τ ] A.
M. B. Patil, IIT Bombay
RL circuit: example
i
v
t
10 V
0 0.2 0.4 0.6 0.8
i (A
mp)
1
0
time (sec)
t1t0
R2
R1
Vs
Vs
R1 = 10Ω
R2 = 40Ω
L = 0.8H
t0 = 0
t1 = 0.1 s
and we need to work out the
solution for t > t1 separately.
In reality, Vs changes at t = t1,
Consider t > t1.
For t0 < t < t1, i(t) = 1− exp(−t/τ) Amp.
i(t+1 ) = i(t−1 ) = 1− e−1 = 0.632 A (Note: t1/τ = 1).
i(∞) = 0 A.
Let i(t) = A exp(−t/τ) + B.
It is convenient to rewrite i(t) as
i(t) = A′ exp[−(t− t1)/τ ] + B.
Using i(t+1 ) and i(∞), we get
i(t) = 0.632 exp[−(t− t1)/τ ] A.
M. B. Patil, IIT Bombay
RL circuit: example
i
v
t
10 V
0 0.2 0.4 0.6 0.8
i (A
mp)
1
0
time (sec)
t1t0
R2
R1
Vs
Vs
R1 = 10Ω
R2 = 40Ω
L = 0.8H
t0 = 0
t1 = 0.1 s
and we need to work out the
solution for t > t1 separately.
In reality, Vs changes at t = t1,
Consider t > t1.
For t0 < t < t1, i(t) = 1− exp(−t/τ) Amp.
i(t+1 ) = i(t−1 ) = 1− e−1 = 0.632 A (Note: t1/τ = 1).
i(∞) = 0 A.
Let i(t) = A exp(−t/τ) + B.
It is convenient to rewrite i(t) as
i(t) = A′ exp[−(t− t1)/τ ] + B.
Using i(t+1 ) and i(∞), we get
i(t) = 0.632 exp[−(t− t1)/τ ] A.
M. B. Patil, IIT Bombay
RL circuit: example
i
v
t
10 V
0 0.2 0.4 0.6 0.8
i (A
mp)
1
0
time (sec)
t1t0
R2
R1
Vs
Vs
R1 = 10Ω
R2 = 40Ω
L = 0.8H
t0 = 0
t1 = 0.1 s
and we need to work out the
solution for t > t1 separately.
In reality, Vs changes at t = t1,
Consider t > t1.
For t0 < t < t1, i(t) = 1− exp(−t/τ) Amp.
i(t+1 ) = i(t−1 ) = 1− e−1 = 0.632 A (Note: t1/τ = 1).
i(∞) = 0 A.
Let i(t) = A exp(−t/τ) + B.
It is convenient to rewrite i(t) as
i(t) = A′ exp[−(t− t1)/τ ] + B.
Using i(t+1 ) and i(∞), we get
i(t) = 0.632 exp[−(t− t1)/τ ] A.
M. B. Patil, IIT Bombay
RL circuit: example
i
v
t
10 V
0 0.2 0.4 0.6 0.8
i (A
mp)
1
0
time (sec)
i(t) = 0.632 exp[−(t− t1)/τ ] A.
t1t0
R2
R1
Vs
Vs
R1 = 10Ω
R2 = 40Ω
L = 0.8H
t0 = 0
t1 = 0.1 s
(SEQUEL file: ee101_rl1.sqproj)
0 0.2 0.4 0.6 0.8
0
1
i (A
mp)
time (sec)
Combining the solutions for t0 < t < t1 and t > t1,
we get
M. B. Patil, IIT Bombay
RL circuit: example
i
v
t
10 V
0 0.2 0.4 0.6 0.8
i (A
mp)
1
0
time (sec)
i(t) = 0.632 exp[−(t− t1)/τ ] A.
t1t0
R2
R1
Vs
Vs
R1 = 10Ω
R2 = 40Ω
L = 0.8H
t0 = 0
t1 = 0.1 s
(SEQUEL file: ee101_rl1.sqproj)
0 0.2 0.4 0.6 0.8
0
1
i (A
mp)
time (sec)
Combining the solutions for t0 < t < t1 and t > t1,
we get
M. B. Patil, IIT Bombay
RC circuit: The switch has been closed for a long time and opens at t = 0.
t=0i
5 k 1 k
6 V
R2R1
ic
R3=5k
vc5 µF
AND
i i
5 k1 k
5 k
5 k
5 k
6 V
ic
t < 0
ic
vc
t > 0
vc5µF 5µF
t = 0−: capacitor is an open circuit ⇒ i(0−) = 6 V/(5 k+ 1 k) = 1 mA.
vc(0−) = i(0−)R1 = 5V ⇒ vc(0
+) = 5V
⇒ i(0+) = 5 V/(5 k+ 5 k) = 0.5 mA.
Let i(t) = Aexp(-t/τ) + B for t > 0, with τ = 10 k× 5µF = 50 ms.
i(t) = 0.5 exp(-t/τ) mA.
Using i(0+) and i(∞) = 0 A, we get
0
1
time (sec)
i (mA)
0 0.5−0.5
0
0
5
time (sec)
time (sec)
0 0.5 0 0.5
ic (mA)vc (V)
(SEQUEL file: ee101_rc2.sqproj)
M. B. Patil, IIT Bombay
RC circuit: The switch has been closed for a long time and opens at t = 0.
t=0i
5 k 1 k
6 V
R2R1
ic
R3=5k
vc5 µF
AND
i i
5 k1 k
5 k
5 k
5 k
6 V
ic
t < 0
ic
vc
t > 0
vc5µF 5µF
t = 0−: capacitor is an open circuit ⇒ i(0−) = 6 V/(5 k+ 1 k) = 1 mA.
vc(0−) = i(0−)R1 = 5V ⇒ vc(0
+) = 5V
⇒ i(0+) = 5 V/(5 k+ 5 k) = 0.5 mA.
Let i(t) = Aexp(-t/τ) + B for t > 0, with τ = 10 k× 5µF = 50 ms.
i(t) = 0.5 exp(-t/τ) mA.
Using i(0+) and i(∞) = 0 A, we get
0
1
time (sec)
i (mA)
0 0.5−0.5
0
0
5
time (sec)
time (sec)
0 0.5 0 0.5
ic (mA)vc (V)
(SEQUEL file: ee101_rc2.sqproj)
M. B. Patil, IIT Bombay
RC circuit: The switch has been closed for a long time and opens at t = 0.
t=0i
5 k 1 k
6 V
R2R1
ic
R3=5k
vc5 µF
AND
i i
5 k1 k
5 k
5 k
5 k
6 V
ic
t < 0
ic
vc
t > 0
vc5µF 5µF
t = 0−: capacitor is an open circuit ⇒ i(0−) = 6 V/(5 k+ 1 k) = 1 mA.
vc(0−) = i(0−)R1 = 5V ⇒ vc(0
+) = 5V
⇒ i(0+) = 5 V/(5 k+ 5 k) = 0.5 mA.
Let i(t) = Aexp(-t/τ) + B for t > 0, with τ = 10 k× 5µF = 50 ms.
i(t) = 0.5 exp(-t/τ) mA.
Using i(0+) and i(∞) = 0 A, we get
0
1
time (sec)
i (mA)
0 0.5−0.5
0
0
5
time (sec)
time (sec)
0 0.5 0 0.5
ic (mA)vc (V)
(SEQUEL file: ee101_rc2.sqproj)
M. B. Patil, IIT Bombay
RC circuit: The switch has been closed for a long time and opens at t = 0.
t=0i
5 k 1 k
6 V
R2R1
ic
R3=5k
vc5 µF
AND
i i
5 k1 k
5 k
5 k
5 k
6 V
ic
t < 0
ic
vc
t > 0
vc5µF 5µF
t = 0−: capacitor is an open circuit ⇒ i(0−) = 6 V/(5 k+ 1 k) = 1 mA.
vc(0−) = i(0−)R1 = 5V ⇒ vc(0
+) = 5V
⇒ i(0+) = 5 V/(5 k+ 5 k) = 0.5 mA.
Let i(t) = Aexp(-t/τ) + B for t > 0, with τ = 10 k× 5µF = 50 ms.
i(t) = 0.5 exp(-t/τ) mA.
Using i(0+) and i(∞) = 0 A, we get
0
1
time (sec)
i (mA)
0 0.5−0.5
0
0
5
time (sec)
time (sec)
0 0.5 0 0.5
ic (mA)vc (V)
(SEQUEL file: ee101_rc2.sqproj)
M. B. Patil, IIT Bombay
RC circuit: The switch has been closed for a long time and opens at t = 0.
t=0i
5 k 1 k
6 V
R2R1
ic
R3=5k
vc5 µF
AND
i i
5 k1 k
5 k
5 k
5 k
6 V
ic
t < 0
ic
vc
t > 0
vc5µF 5µF
t = 0−: capacitor is an open circuit ⇒ i(0−) = 6 V/(5 k+ 1 k) = 1 mA.
vc(0−) = i(0−)R1 = 5V ⇒ vc(0
+) = 5V
⇒ i(0+) = 5 V/(5 k+ 5 k) = 0.5 mA.
Let i(t) = Aexp(-t/τ) + B for t > 0, with τ = 10 k× 5µF = 50 ms.
i(t) = 0.5 exp(-t/τ) mA.
Using i(0+) and i(∞) = 0 A, we get
0
1
time (sec)
i (mA)
0 0.5−0.5
0
0
5
time (sec)
time (sec)
0 0.5 0 0.5
ic (mA)vc (V)
(SEQUEL file: ee101_rc2.sqproj)
M. B. Patil, IIT Bombay
RC circuit: The switch has been closed for a long time and opens at t = 0.
t=0i
5 k 1 k
6 V
R2R1
ic
R3=5k
vc5 µF
AND
i i
5 k1 k
5 k
5 k
5 k
6 V
ic
t < 0
ic
vc
t > 0
vc5µF 5µF
t = 0−: capacitor is an open circuit ⇒ i(0−) = 6 V/(5 k+ 1 k) = 1 mA.
vc(0−) = i(0−)R1 = 5V ⇒ vc(0
+) = 5V
⇒ i(0+) = 5 V/(5 k+ 5 k) = 0.5 mA.
Let i(t) = Aexp(-t/τ) + B for t > 0, with τ = 10 k× 5µF = 50 ms.
i(t) = 0.5 exp(-t/τ) mA.
Using i(0+) and i(∞) = 0 A, we get
0
1
time (sec)
i (mA)
0 0.5−0.5
0
0
5
time (sec)
time (sec)
0 0.5 0 0.5
ic (mA)vc (V)
(SEQUEL file: ee101_rc2.sqproj)
M. B. Patil, IIT Bombay
RC circuit: The switch has been closed for a long time and opens at t = 0.
t=0i
5 k 1 k
6 V
R2R1
ic
R3=5k
vc5 µF
AND
i i
5 k1 k
5 k
5 k
5 k
6 V
ic
t < 0
ic
vc
t > 0
vc5µF 5µF
t = 0−: capacitor is an open circuit ⇒ i(0−) = 6 V/(5 k+ 1 k) = 1 mA.
vc(0−) = i(0−)R1 = 5V ⇒ vc(0
+) = 5V
⇒ i(0+) = 5 V/(5 k+ 5 k) = 0.5 mA.
Let i(t) = Aexp(-t/τ) + B for t > 0, with τ = 10 k× 5µF = 50 ms.
i(t) = 0.5 exp(-t/τ) mA.
Using i(0+) and i(∞) = 0 A, we get
0
1
time (sec)
i (mA)
0 0.5−0.5
0
0
5
time (sec)
time (sec)
0 0.5 0 0.5
ic (mA)vc (V)
(SEQUEL file: ee101_rc2.sqproj)
M. B. Patil, IIT Bombay
RC circuit: The switch has been closed for a long time and opens at t = 0.
t=0i
5 k 1 k
6 V
R2R1
ic
R3=5k
vc5 µF
AND
i i
5 k1 k
5 k
5 k
5 k
6 V
ic
t < 0
ic
vc
t > 0
vc5µF 5µF
t = 0−: capacitor is an open circuit ⇒ i(0−) = 6 V/(5 k+ 1 k) = 1 mA.
vc(0−) = i(0−)R1 = 5V ⇒ vc(0
+) = 5V
⇒ i(0+) = 5 V/(5 k+ 5 k) = 0.5 mA.
Let i(t) = Aexp(-t/τ) + B for t > 0, with τ = 10 k× 5µF = 50 ms.
i(t) = 0.5 exp(-t/τ) mA.
Using i(0+) and i(∞) = 0 A, we get
0
1
time (sec)
i (mA)
0 0.5
−0.5
0
0
5
time (sec)
time (sec)
0 0.5 0 0.5
ic (mA)vc (V)
(SEQUEL file: ee101_rc2.sqproj)
M. B. Patil, IIT Bombay
RC circuit: The switch has been closed for a long time and opens at t = 0.
t=0i
5 k 1 k
6 V
R2R1
ic
R3=5k
vc5 µF
AND
i i
5 k1 k
5 k
5 k
5 k
6 V
ic
t < 0
ic
vc
t > 0
vc5µF 5µF
t = 0−: capacitor is an open circuit ⇒ i(0−) = 6 V/(5 k+ 1 k) = 1 mA.
vc(0−) = i(0−)R1 = 5V ⇒ vc(0
+) = 5V
⇒ i(0+) = 5 V/(5 k+ 5 k) = 0.5 mA.
Let i(t) = Aexp(-t/τ) + B for t > 0, with τ = 10 k× 5µF = 50 ms.
i(t) = 0.5 exp(-t/τ) mA.
Using i(0+) and i(∞) = 0 A, we get
0
1
time (sec)
i (mA)
0 0.5−0.5
0
0
5
time (sec)
time (sec)
0 0.5 0 0.5
ic (mA)vc (V)
(SEQUEL file: ee101_rc2.sqproj)
M. B. Patil, IIT Bombay
RC circuit: The switch has been closed for a long time and opens at t = 0.
t=0i
5 k 1 k
6 V
R2R1
ic
R3=5k
vc5 µF
AND
i i
5 k1 k
5 k
5 k
5 k
6 V
ic
t < 0
ic
vc
t > 0
vc5µF 5µF
t = 0−: capacitor is an open circuit ⇒ i(0−) = 6 V/(5 k+ 1 k) = 1 mA.
vc(0−) = i(0−)R1 = 5V ⇒ vc(0
+) = 5V
⇒ i(0+) = 5 V/(5 k+ 5 k) = 0.5 mA.
Let i(t) = Aexp(-t/τ) + B for t > 0, with τ = 10 k× 5µF = 50 ms.
i(t) = 0.5 exp(-t/τ) mA.
Using i(0+) and i(∞) = 0 A, we get
0
1
time (sec)
i (mA)
0 0.5−0.5
0
0
5
time (sec)
time (sec)
0 0.5 0 0.5
ic (mA)vc (V)
(SEQUEL file: ee101_rc2.sqproj)
M. B. Patil, IIT Bombay
RC circuit: example
VS
R
i C(m
A)
VS,V
C(Volts)Vc
5 k
iC
C1µF
2
1
−1
0
10
0
0 2010 30 40 50
t (msec)
Find expressions for VC (t) and iC (t) insteady state (in terms of R, C , V0, T1,T2).
V0
0T1 T2
M. B. Patil, IIT Bombay
RC circuit: example
VS
R
i C(m
A)
VS,V
C(Volts)Vc
5 k
iC
C1µF
2
1
−1
0
10
0
0 2010 30 40 50
t (msec)
Find expressions for VC (t) and iC (t) insteady state (in terms of R, C , V0, T1,T2).
V0
0T1 T2
M. B. Patil, IIT Bombay
RC circuit: example
VS
R
C Vc
iC
0
I1
−I2
iC
VC
VS
V0
V2
V1
0
T1 T2
T1 T2
0 < t < T1 Let V(1)C (t) = Ae−t/τ + B
V(1)C (0) = V1, V
(1)C (∞) = V0
→ B = V0, A = V1 − V0.
V(1)C (t) = −(V0 − V1)e−t/τ + V0 (1)
T1 < t < T2 Let V(2)C (t) = A′ e−t/τ + B′
V(2)C (T1) = V2, V
(2)C (∞) = 0
→ B′ = 0, A′ = V2 eT1/τ .
V(2)C (t) = V2e−(t−T1)/τ (2)
Now use the conditions:
V(1)C (T1) = V2, V
(2)C (T1 + T2) = V1.
V2 = −(V0 − V1)e−T1/τ + V0 (3)
V1 = V2e−(T1+T2−T1)/τ = V2e−T2/τ (4)
Rewrite with a ≡ e−T1/τ , b ≡ e−T2/τ .
V2 = −(V0 − V1)a + V0 (5)
V1 = b V2 (6)
Solve to get
V1 = b V01− a
1− ab, V2 = V0
1− a
1− ab
M. B. Patil, IIT Bombay
RC circuit: example
VS
R
C Vc
iC
0
I1
−I2
iC
VC
VS
V0
V2
V1
0
T1 T2
T1 T2
0 < t < T1 Let V(1)C (t) = Ae−t/τ + B
V(1)C (0) = V1, V
(1)C (∞) = V0
→ B = V0, A = V1 − V0.
V(1)C (t) = −(V0 − V1)e−t/τ + V0 (1)
T1 < t < T2 Let V(2)C (t) = A′ e−t/τ + B′
V(2)C (T1) = V2, V
(2)C (∞) = 0
→ B′ = 0, A′ = V2 eT1/τ .
V(2)C (t) = V2e−(t−T1)/τ (2)
Now use the conditions:
V(1)C (T1) = V2, V
(2)C (T1 + T2) = V1.
V2 = −(V0 − V1)e−T1/τ + V0 (3)
V1 = V2e−(T1+T2−T1)/τ = V2e−T2/τ (4)
Rewrite with a ≡ e−T1/τ , b ≡ e−T2/τ .
V2 = −(V0 − V1)a + V0 (5)
V1 = b V2 (6)
Solve to get
V1 = b V01− a
1− ab, V2 = V0
1− a
1− ab
M. B. Patil, IIT Bombay
RC circuit: example
VS
R
C Vc
iC
0
I1
−I2
iC
VC
VS
V0
V2
V1
0
T1 T2
T1 T2
0 < t < T1 Let V(1)C (t) = Ae−t/τ + B
V(1)C (0) = V1, V
(1)C (∞) = V0
→ B = V0, A = V1 − V0.
V(1)C (t) = −(V0 − V1)e−t/τ + V0 (1)
T1 < t < T2 Let V(2)C (t) = A′ e−t/τ + B′
V(2)C (T1) = V2, V
(2)C (∞) = 0
→ B′ = 0, A′ = V2 eT1/τ .
V(2)C (t) = V2e−(t−T1)/τ (2)
Now use the conditions:
V(1)C (T1) = V2, V
(2)C (T1 + T2) = V1.
V2 = −(V0 − V1)e−T1/τ + V0 (3)
V1 = V2e−(T1+T2−T1)/τ = V2e−T2/τ (4)
Rewrite with a ≡ e−T1/τ , b ≡ e−T2/τ .
V2 = −(V0 − V1)a + V0 (5)
V1 = b V2 (6)
Solve to get
V1 = b V01− a
1− ab, V2 = V0
1− a
1− ab
M. B. Patil, IIT Bombay
RC circuit: example
VS
R
C Vc
iC
0
I1
−I2
iC
VC
VS
V0
V2
V1
0
T1 T2
T1 T2
0 < t < T1 Let V(1)C (t) = Ae−t/τ + B
V(1)C (0) = V1, V
(1)C (∞) = V0
→ B = V0, A = V1 − V0.
V(1)C (t) = −(V0 − V1)e−t/τ + V0 (1)
T1 < t < T2 Let V(2)C (t) = A′ e−t/τ + B′
V(2)C (T1) = V2, V
(2)C (∞) = 0
→ B′ = 0, A′ = V2 eT1/τ .
V(2)C (t) = V2e−(t−T1)/τ (2)
Now use the conditions:
V(1)C (T1) = V2, V
(2)C (T1 + T2) = V1.
V2 = −(V0 − V1)e−T1/τ + V0 (3)
V1 = V2e−(T1+T2−T1)/τ = V2e−T2/τ (4)
Rewrite with a ≡ e−T1/τ , b ≡ e−T2/τ .
V2 = −(V0 − V1)a + V0 (5)
V1 = b V2 (6)
Solve to get
V1 = b V01− a
1− ab, V2 = V0
1− a
1− ab
M. B. Patil, IIT Bombay
RC circuit: example
VS
R
C Vc
iC
0
I1
−I2
iC
VC
VS
V0
V2
V1
0
T1 T2
T1 T2
0 < t < T1 Let V(1)C (t) = Ae−t/τ + B
V(1)C (0) = V1, V
(1)C (∞) = V0
→ B = V0, A = V1 − V0.
V(1)C (t) = −(V0 − V1)e−t/τ + V0 (1)
T1 < t < T2 Let V(2)C (t) = A′ e−t/τ + B′
V(2)C (T1) = V2, V
(2)C (∞) = 0
→ B′ = 0, A′ = V2 eT1/τ .
V(2)C (t) = V2e−(t−T1)/τ (2)
Now use the conditions:
V(1)C (T1) = V2, V
(2)C (T1 + T2) = V1.
V2 = −(V0 − V1)e−T1/τ + V0 (3)
V1 = V2e−(T1+T2−T1)/τ = V2e−T2/τ (4)
Rewrite with a ≡ e−T1/τ , b ≡ e−T2/τ .
V2 = −(V0 − V1)a + V0 (5)
V1 = b V2 (6)
Solve to get
V1 = b V01− a
1− ab, V2 = V0
1− a
1− ab
M. B. Patil, IIT Bombay
RC circuit: example
VS
R
C Vc
iC
0
I1
−I2
iC
VC
VS
V0
V2
V1
0
T1 T2
T1 T2
0 < t < T1 Let V(1)C (t) = Ae−t/τ + B
V(1)C (0) = V1, V
(1)C (∞) = V0
→ B = V0, A = V1 − V0.
V(1)C (t) = −(V0 − V1)e−t/τ + V0 (1)
T1 < t < T2 Let V(2)C (t) = A′ e−t/τ + B′
V(2)C (T1) = V2, V
(2)C (∞) = 0
→ B′ = 0, A′ = V2 eT1/τ .
V(2)C (t) = V2e−(t−T1)/τ (2)
Now use the conditions:
V(1)C (T1) = V2, V
(2)C (T1 + T2) = V1.
V2 = −(V0 − V1)e−T1/τ + V0 (3)
V1 = V2e−(T1+T2−T1)/τ = V2e−T2/τ (4)
Rewrite with a ≡ e−T1/τ , b ≡ e−T2/τ .
V2 = −(V0 − V1)a + V0 (5)
V1 = b V2 (6)
Solve to get
V1 = b V01− a
1− ab, V2 = V0
1− a
1− ab
M. B. Patil, IIT Bombay
RC circuit: example
VS
R
C Vc
iC
0
I1
−I2
iC
VC
VS
V0
V2
V1
0
T1 T2
T1 T2
0 < t < T1 Let V(1)C (t) = Ae−t/τ + B
V(1)C (0) = V1, V
(1)C (∞) = V0
→ B = V0, A = V1 − V0.
V(1)C (t) = −(V0 − V1)e−t/τ + V0 (1)
T1 < t < T2 Let V(2)C (t) = A′ e−t/τ + B′
V(2)C (T1) = V2, V
(2)C (∞) = 0
→ B′ = 0, A′ = V2 eT1/τ .
V(2)C (t) = V2e−(t−T1)/τ (2)
Now use the conditions:
V(1)C (T1) = V2, V
(2)C (T1 + T2) = V1.
V2 = −(V0 − V1)e−T1/τ + V0 (3)
V1 = V2e−(T1+T2−T1)/τ = V2e−T2/τ (4)
Rewrite with a ≡ e−T1/τ , b ≡ e−T2/τ .
V2 = −(V0 − V1)a + V0 (5)
V1 = b V2 (6)
Solve to get
V1 = b V01− a
1− ab, V2 = V0
1− a
1− ab
M. B. Patil, IIT Bombay
RC circuit: example
VS
R
C Vc
iC
0
I1
−I2
iC
VC
VS
V0
V2
V1
0
T1 T2
T1 T2
0 < t < T1 Let V(1)C (t) = Ae−t/τ + B
V(1)C (0) = V1, V
(1)C (∞) = V0
→ B = V0, A = V1 − V0.
V(1)C (t) = −(V0 − V1)e−t/τ + V0 (1)
T1 < t < T2 Let V(2)C (t) = A′ e−t/τ + B′
V(2)C (T1) = V2, V
(2)C (∞) = 0
→ B′ = 0, A′ = V2 eT1/τ .
V(2)C (t) = V2e−(t−T1)/τ (2)
Now use the conditions:
V(1)C (T1) = V2, V
(2)C (T1 + T2) = V1.
V2 = −(V0 − V1)e−T1/τ + V0 (3)
V1 = V2e−(T1+T2−T1)/τ = V2e−T2/τ (4)
Rewrite with a ≡ e−T1/τ , b ≡ e−T2/τ .
V2 = −(V0 − V1)a + V0 (5)
V1 = b V2 (6)
Solve to get
V1 = b V01− a
1− ab, V2 = V0
1− a
1− ab
M. B. Patil, IIT Bombay
RC circuit: example
VS
R
C Vc
iC
0
I1
−I2
iC
VC
VS
V0
V2
V1
0
T1 T2
T1 T2
0 < t < T1 Let V(1)C (t) = Ae−t/τ + B
V(1)C (0) = V1, V
(1)C (∞) = V0
→ B = V0, A = V1 − V0.
V(1)C (t) = −(V0 − V1)e−t/τ + V0 (1)
T1 < t < T2 Let V(2)C (t) = A′ e−t/τ + B′
V(2)C (T1) = V2, V
(2)C (∞) = 0
→ B′ = 0, A′ = V2 eT1/τ .
V(2)C (t) = V2e−(t−T1)/τ (2)
Now use the conditions:
V(1)C (T1) = V2, V
(2)C (T1 + T2) = V1.
V2 = −(V0 − V1)e−T1/τ + V0 (3)
V1 = V2e−(T1+T2−T1)/τ = V2e−T2/τ (4)
Rewrite with a ≡ e−T1/τ , b ≡ e−T2/τ .
V2 = −(V0 − V1)a + V0 (5)
V1 = b V2 (6)
Solve to get
V1 = b V01− a
1− ab, V2 = V0
1− a
1− ab
M. B. Patil, IIT Bombay
RC circuit: example
VS
R
C Vc
iC
0
I1
−I2
iC
VC
VS
V0
V2
V1
0
T1 T2
T1 T2
0 < t < T1 Let V(1)C (t) = Ae−t/τ + B
V(1)C (0) = V1, V
(1)C (∞) = V0
→ B = V0, A = V1 − V0.
V(1)C (t) = −(V0 − V1)e−t/τ + V0 (1)
T1 < t < T2 Let V(2)C (t) = A′ e−t/τ + B′
V(2)C (T1) = V2, V
(2)C (∞) = 0
→ B′ = 0, A′ = V2 eT1/τ .
V(2)C (t) = V2e−(t−T1)/τ (2)
Now use the conditions:
V(1)C (T1) = V2, V
(2)C (T1 + T2) = V1.
V2 = −(V0 − V1)e−T1/τ + V0 (3)
V1 = V2e−(T1+T2−T1)/τ = V2e−T2/τ (4)
Rewrite with a ≡ e−T1/τ , b ≡ e−T2/τ .
V2 = −(V0 − V1)a + V0 (5)
V1 = b V2 (6)
Solve to get
V1 = b V01− a
1− ab, V2 = V0
1− a
1− ab
M. B. Patil, IIT Bombay
RC circuit: example
VS
R
C Vc
iC
0
I1
−I2
iC
VC
VS
V0
V2
V1
0
T1 T2
T1 T2
0 < t < T1 Let V(1)C (t) = Ae−t/τ + B
V(1)C (0) = V1, V
(1)C (∞) = V0
→ B = V0, A = V1 − V0.
V(1)C (t) = −(V0 − V1)e−t/τ + V0 (1)
T1 < t < T2 Let V(2)C (t) = A′ e−t/τ + B′
V(2)C (T1) = V2, V
(2)C (∞) = 0
→ B′ = 0, A′ = V2 eT1/τ .
V(2)C (t) = V2e−(t−T1)/τ (2)
Now use the conditions:
V(1)C (T1) = V2, V
(2)C (T1 + T2) = V1.
V2 = −(V0 − V1)e−T1/τ + V0 (3)
V1 = V2e−(T1+T2−T1)/τ = V2e−T2/τ (4)
Rewrite with a ≡ e−T1/τ , b ≡ e−T2/τ .
V2 = −(V0 − V1)a + V0 (5)
V1 = b V2 (6)
Solve to get
V1 = b V01− a
1− ab, V2 = V0
1− a
1− ab
M. B. Patil, IIT Bombay
RC circuit: example
VS
R
C Vc
iC
0
I1
−I2
iC
VC
VS
V0
V2
V1
0
T1 T2
T1 T2
0 < t < T1 Let V(1)C (t) = Ae−t/τ + B
V(1)C (0) = V1, V
(1)C (∞) = V0
→ B = V0, A = V1 − V0.
V(1)C (t) = −(V0 − V1)e−t/τ + V0 (1)
T1 < t < T2 Let V(2)C (t) = A′ e−t/τ + B′
V(2)C (T1) = V2, V
(2)C (∞) = 0
→ B′ = 0, A′ = V2 eT1/τ .
V(2)C (t) = V2e−(t−T1)/τ (2)
Now use the conditions:
V(1)C (T1) = V2, V
(2)C (T1 + T2) = V1.
V2 = −(V0 − V1)e−T1/τ + V0 (3)
V1 = V2e−(T1+T2−T1)/τ = V2e−T2/τ (4)
Rewrite with a ≡ e−T1/τ , b ≡ e−T2/τ .
V2 = −(V0 − V1)a + V0 (5)
V1 = b V2 (6)
Solve to get
V1 = b V01− a
1− ab, V2 = V0
1− a
1− ab
M. B. Patil, IIT Bombay
RC circuit: example
VS
R
C Vc
iC
0
I1
−I2
iC
VC
VS
V0
V2
V1
0
T1 T2
T1 T2
0 < t < T1 Let V(1)C (t) = Ae−t/τ + B
V(1)C (0) = V1, V
(1)C (∞) = V0
→ B = V0, A = V1 − V0.
V(1)C (t) = −(V0 − V1)e−t/τ + V0 (1)
T1 < t < T2 Let V(2)C (t) = A′ e−t/τ + B′
V(2)C (T1) = V2, V
(2)C (∞) = 0
→ B′ = 0, A′ = V2 eT1/τ .
V(2)C (t) = V2e−(t−T1)/τ (2)
Now use the conditions:
V(1)C (T1) = V2, V
(2)C (T1 + T2) = V1.
V2 = −(V0 − V1)e−T1/τ + V0 (3)
V1 = V2e−(T1+T2−T1)/τ = V2e−T2/τ (4)
Rewrite with a ≡ e−T1/τ , b ≡ e−T2/τ .
V2 = −(V0 − V1)a + V0 (5)
V1 = b V2 (6)
Solve to get
V1 = b V01− a
1− ab, V2 = V0
1− a
1− ab
M. B. Patil, IIT Bombay
RC circuit: example
VS
R
C Vc
iC
0
I1
−I2
iC
VC
VS
V0
V2
V1
0
T1 T2
T1 T2
0 < t < T1 Let V(1)C (t) = Ae−t/τ + B
V(1)C (0) = V1, V
(1)C (∞) = V0
→ B = V0, A = V1 − V0.
V(1)C (t) = −(V0 − V1)e−t/τ + V0 (1)
T1 < t < T2 Let V(2)C (t) = A′ e−t/τ + B′
V(2)C (T1) = V2, V
(2)C (∞) = 0
→ B′ = 0, A′ = V2 eT1/τ .
V(2)C (t) = V2e−(t−T1)/τ (2)
Now use the conditions:
V(1)C (T1) = V2, V
(2)C (T1 + T2) = V1.
V2 = −(V0 − V1)e−T1/τ + V0 (3)
V1 = V2e−(T1+T2−T1)/τ = V2e−T2/τ (4)
Rewrite with a ≡ e−T1/τ , b ≡ e−T2/τ .
V2 = −(V0 − V1)a + V0 (5)
V1 = b V2 (6)
Solve to get
V1 = b V01− a
1− ab, V2 = V0
1− a
1− ab
M. B. Patil, IIT Bombay
RC circuit: example
VS
R
C Vc
iC
0
I1
−I2
iC
VC
VS
V0
V2
V1
0
T1 T2
T1 T2
0 < t < T1 Let V(1)C (t) = Ae−t/τ + B
V(1)C (0) = V1, V
(1)C (∞) = V0
→ B = V0, A = V1 − V0.
V(1)C (t) = −(V0 − V1)e−t/τ + V0 (1)
T1 < t < T2 Let V(2)C (t) = A′ e−t/τ + B′
V(2)C (T1) = V2, V
(2)C (∞) = 0
→ B′ = 0, A′ = V2 eT1/τ .
V(2)C (t) = V2e−(t−T1)/τ (2)
Now use the conditions:
V(1)C (T1) = V2, V
(2)C (T1 + T2) = V1.
V2 = −(V0 − V1)e−T1/τ + V0 (3)
V1 = V2e−(T1+T2−T1)/τ = V2e−T2/τ (4)
Rewrite with a ≡ e−T1/τ , b ≡ e−T2/τ .
V2 = −(V0 − V1)a + V0 (5)
V1 = b V2 (6)
Solve to get
V1 = b V01− a
1− ab, V2 = V0
1− a
1− ab
M. B. Patil, IIT Bombay
RC circuit: example
VS
R
C Vc
iC
0
I1
−I2
iC
VC
VS
V0
V2
V1
0
T1 T2
T1 T2
0 < t < T1 Let V(1)C (t) = Ae−t/τ + B
V(1)C (0) = V1, V
(1)C (∞) = V0
→ B = V0, A = V1 − V0.
V(1)C (t) = −(V0 − V1)e−t/τ + V0 (1)
T1 < t < T2 Let V(2)C (t) = A′ e−t/τ + B′
V(2)C (T1) = V2, V
(2)C (∞) = 0
→ B′ = 0, A′ = V2 eT1/τ .
V(2)C (t) = V2e−(t−T1)/τ (2)
Now use the conditions:
V(1)C (T1) = V2, V
(2)C (T1 + T2) = V1.
V2 = −(V0 − V1)e−T1/τ + V0 (3)
V1 = V2e−(T1+T2−T1)/τ = V2e−T2/τ (4)
Rewrite with a ≡ e−T1/τ , b ≡ e−T2/τ .
V2 = −(V0 − V1)a + V0 (5)
V1 = b V2 (6)
Solve to get
V1 = b V01− a
1− ab, V2 = V0
1− a
1− ab
M. B. Patil, IIT Bombay
RC circuit: example
VS
R
C Vc
iC
0
I1
−I2
iC
VC
VS
V0
V2
V1
0
T1 T2
T1 T2
V1 = b V01− a
1− ab, V2 = V0
1− a
1− ab, with a = e−T1/τ , b = e−T2/τ .
V(1)C (t) = −(V0 − V1)e−t/τ + V0, V
(2)C (t) = V2e−(t−T1)/τ .
Current calculation:
Method 1:
iC (t) = CdVC
dt(home work)
Method 2:
Start from scratch!
STOP
M. B. Patil, IIT Bombay
RC circuit: example
VS
R
C Vc
iC
0
I1
−I2
iC
VC
VS
V0
V2
V1
0
T1 T2
T1 T2
V1 = b V01− a
1− ab, V2 = V0
1− a
1− ab, with a = e−T1/τ , b = e−T2/τ .
V(1)C (t) = −(V0 − V1)e−t/τ + V0, V
(2)C (t) = V2e−(t−T1)/τ .
Current calculation:
Method 1:
iC (t) = CdVC
dt(home work)
Method 2:
Start from scratch!
STOP
M. B. Patil, IIT Bombay
RC circuit: example
VS
R
C Vc
iC
0
I1
−I2
iC
VC
VS
V0
V2
V1
0
T1 T2
T1 T2
V1 = b V01− a
1− ab, V2 = V0
1− a
1− ab, with a = e−T1/τ , b = e−T2/τ .
V(1)C (t) = −(V0 − V1)e−t/τ + V0, V
(2)C (t) = V2e−(t−T1)/τ .
Current calculation:
Method 1:
iC (t) = CdVC
dt(home work)
Method 2:
Start from scratch!
STOP
M. B. Patil, IIT Bombay
RC circuit: example
VS
R
C Vc
iC
0
I1
−I2
iC
VC
VS
V0
V2
V1
0
T1 T2
T1 T2
V1 = b V01− a
1− ab, V2 = V0
1− a
1− ab, with a = e−T1/τ , b = e−T2/τ .
V(1)C (t) = −(V0 − V1)e−t/τ + V0, V
(2)C (t) = V2e−(t−T1)/τ .
Current calculation:
Method 1:
iC (t) = CdVC
dt(home work)
Method 2:
Start from scratch!
STOP
M. B. Patil, IIT Bombay
RC circuit: example
VS
R
C Vc
iC
0
I1
−I2
iC
VC
VS
V0
V2
V1
0
T1 T2
T1 T2
V1 = b V01− a
1− ab, V2 = V0
1− a
1− ab, with a = e−T1/τ , b = e−T2/τ .
V(1)C (t) = −(V0 − V1)e−t/τ + V0, V
(2)C (t) = V2e−(t−T1)/τ .
Current calculation:
Method 1:
iC (t) = CdVC
dt(home work)
Method 2:
Start from scratch!
STOP
M. B. Patil, IIT Bombay
RC circuit: example
VS
R
C Vc
iC
0
I1
−I2
iC
VC
VS
V0
V2
V1
0
T1 T2
T1 T2
V1 = b V01− a
1− ab, V2 = V0
1− a
1− ab, with a = e−T1/τ , b = e−T2/τ .
V(1)C (t) = −(V0 − V1)e−t/τ + V0, V
(2)C (t) = V2e−(t−T1)/τ .
Current calculation:
Method 1:
iC (t) = CdVC
dt(home work)
Method 2:
Start from scratch!
STOP
M. B. Patil, IIT Bombay
RC circuit: example
VS
R
C Vc
iC
0
I1
−I2
iC∆
VC
VS
V0
V2
V1
0T1 T2
T1 T2
∆
0 < t < T1 Let i(1)C (t) = Ae−t/τ + B
i(1)C (0) = I1, i
(1)C (∞) = 0
→ B = 0, A = I1.
i(1)C (t) = I1e−t/τ (1)
T1 < t < T2 Let i(2)C (t) = A′ e−t/τ + B′
i(2)C (T1) = −I2, i
(2)C (∞) = 0
→ B′ = 0, A′ = −I2 eT1/τ .
i(2)C (t) = −I2e−(t−T1)/τ (2)
Now use the conditions:
i(1)C (T1)− i
(2)C (T1) = ∆ = V0/R,
i(1)C (0)− i
(2)C (T1 + T2) = ∆ = V0/R.
I1e−T1/τ − (−I2) = ∆ (3)
I1 − (−I2e−(T1+T2−T1)/τ ) = ∆ (4)
a I1 + I2 = ∆ (5)
I1 + b I2 = ∆ (6)
Solve to get
I1 = ∆1− b
1− ab, I2 = ∆
1− a
1− ab
(a = e−T1/τ , b = e−T2/τ )
M. B. Patil, IIT Bombay
RC circuit: example
VS
R
C Vc
iC
0
I1
−I2
iC∆
VC
VS
V0
V2
V1
0T1 T2
T1 T2
∆
0 < t < T1 Let i(1)C (t) = Ae−t/τ + B
i(1)C (0) = I1, i
(1)C (∞) = 0
→ B = 0, A = I1.
i(1)C (t) = I1e−t/τ (1)
T1 < t < T2 Let i(2)C (t) = A′ e−t/τ + B′
i(2)C (T1) = −I2, i
(2)C (∞) = 0
→ B′ = 0, A′ = −I2 eT1/τ .
i(2)C (t) = −I2e−(t−T1)/τ (2)
Now use the conditions:
i(1)C (T1)− i
(2)C (T1) = ∆ = V0/R,
i(1)C (0)− i
(2)C (T1 + T2) = ∆ = V0/R.
I1e−T1/τ − (−I2) = ∆ (3)
I1 − (−I2e−(T1+T2−T1)/τ ) = ∆ (4)
a I1 + I2 = ∆ (5)
I1 + b I2 = ∆ (6)
Solve to get
I1 = ∆1− b
1− ab, I2 = ∆
1− a
1− ab
(a = e−T1/τ , b = e−T2/τ )
M. B. Patil, IIT Bombay
RC circuit: example
VS
R
C Vc
iC
0
I1
−I2
iC∆
VC
VS
V0
V2
V1
0T1 T2
T1 T2
∆
0 < t < T1 Let i(1)C (t) = Ae−t/τ + B
i(1)C (0) = I1, i
(1)C (∞) = 0
→ B = 0, A = I1.
i(1)C (t) = I1e−t/τ (1)
T1 < t < T2 Let i(2)C (t) = A′ e−t/τ + B′
i(2)C (T1) = −I2, i
(2)C (∞) = 0
→ B′ = 0, A′ = −I2 eT1/τ .
i(2)C (t) = −I2e−(t−T1)/τ (2)
Now use the conditions:
i(1)C (T1)− i
(2)C (T1) = ∆ = V0/R,
i(1)C (0)− i
(2)C (T1 + T2) = ∆ = V0/R.
I1e−T1/τ − (−I2) = ∆ (3)
I1 − (−I2e−(T1+T2−T1)/τ ) = ∆ (4)
a I1 + I2 = ∆ (5)
I1 + b I2 = ∆ (6)
Solve to get
I1 = ∆1− b
1− ab, I2 = ∆
1− a
1− ab
(a = e−T1/τ , b = e−T2/τ )
M. B. Patil, IIT Bombay
RC circuit: example
VS
R
C Vc
iC
0
I1
−I2
iC∆
VC
VS
V0
V2
V1
0T1 T2
T1 T2
∆
0 < t < T1 Let i(1)C (t) = Ae−t/τ + B
i(1)C (0) = I1, i
(1)C (∞) = 0
→ B = 0, A = I1.
i(1)C (t) = I1e−t/τ (1)
T1 < t < T2 Let i(2)C (t) = A′ e−t/τ + B′
i(2)C (T1) = −I2, i
(2)C (∞) = 0
→ B′ = 0, A′ = −I2 eT1/τ .
i(2)C (t) = −I2e−(t−T1)/τ (2)
Now use the conditions:
i(1)C (T1)− i
(2)C (T1) = ∆ = V0/R,
i(1)C (0)− i
(2)C (T1 + T2) = ∆ = V0/R.
I1e−T1/τ − (−I2) = ∆ (3)
I1 − (−I2e−(T1+T2−T1)/τ ) = ∆ (4)
a I1 + I2 = ∆ (5)
I1 + b I2 = ∆ (6)
Solve to get
I1 = ∆1− b
1− ab, I2 = ∆
1− a
1− ab
(a = e−T1/τ , b = e−T2/τ )
M. B. Patil, IIT Bombay
RC circuit: example
VS
R
C Vc
iC
0
I1
−I2
iC∆
VC
VS
V0
V2
V1
0T1 T2
T1 T2
∆
0 < t < T1 Let i(1)C (t) = Ae−t/τ + B
i(1)C (0) = I1, i
(1)C (∞) = 0
→ B = 0, A = I1.
i(1)C (t) = I1e−t/τ (1)
T1 < t < T2 Let i(2)C (t) = A′ e−t/τ + B′
i(2)C (T1) = −I2, i
(2)C (∞) = 0
→ B′ = 0, A′ = −I2 eT1/τ .
i(2)C (t) = −I2e−(t−T1)/τ (2)
Now use the conditions:
i(1)C (T1)− i
(2)C (T1) = ∆ = V0/R,
i(1)C (0)− i
(2)C (T1 + T2) = ∆ = V0/R.
I1e−T1/τ − (−I2) = ∆ (3)
I1 − (−I2e−(T1+T2−T1)/τ ) = ∆ (4)
a I1 + I2 = ∆ (5)
I1 + b I2 = ∆ (6)
Solve to get
I1 = ∆1− b
1− ab, I2 = ∆
1− a
1− ab
(a = e−T1/τ , b = e−T2/τ )
M. B. Patil, IIT Bombay
RC circuit: example
VS
R
C Vc
iC
0
I1
−I2
iC∆
VC
VS
V0
V2
V1
0T1 T2
T1 T2
∆
0 < t < T1 Let i(1)C (t) = Ae−t/τ + B
i(1)C (0) = I1, i
(1)C (∞) = 0
→ B = 0, A = I1.
i(1)C (t) = I1e−t/τ (1)
T1 < t < T2 Let i(2)C (t) = A′ e−t/τ + B′
i(2)C (T1) = −I2, i
(2)C (∞) = 0
→ B′ = 0, A′ = −I2 eT1/τ .
i(2)C (t) = −I2e−(t−T1)/τ (2)
Now use the conditions:
i(1)C (T1)− i
(2)C (T1) = ∆ = V0/R,
i(1)C (0)− i
(2)C (T1 + T2) = ∆ = V0/R.
I1e−T1/τ − (−I2) = ∆ (3)
I1 − (−I2e−(T1+T2−T1)/τ ) = ∆ (4)
a I1 + I2 = ∆ (5)
I1 + b I2 = ∆ (6)
Solve to get
I1 = ∆1− b
1− ab, I2 = ∆
1− a
1− ab
(a = e−T1/τ , b = e−T2/τ )
M. B. Patil, IIT Bombay
RC circuit: example
VS
R
C Vc
iC
0
I1
−I2
iC∆
VC
VS
V0
V2
V1
0T1 T2
T1 T2
∆
0 < t < T1 Let i(1)C (t) = Ae−t/τ + B
i(1)C (0) = I1, i
(1)C (∞) = 0
→ B = 0, A = I1.
i(1)C (t) = I1e−t/τ (1)
T1 < t < T2 Let i(2)C (t) = A′ e−t/τ + B′
i(2)C (T1) = −I2, i
(2)C (∞) = 0
→ B′ = 0, A′ = −I2 eT1/τ .
i(2)C (t) = −I2e−(t−T1)/τ (2)
Now use the conditions:
i(1)C (T1)− i
(2)C (T1) = ∆ = V0/R,
i(1)C (0)− i
(2)C (T1 + T2) = ∆ = V0/R.
I1e−T1/τ − (−I2) = ∆ (3)
I1 − (−I2e−(T1+T2−T1)/τ ) = ∆ (4)
a I1 + I2 = ∆ (5)
I1 + b I2 = ∆ (6)
Solve to get
I1 = ∆1− b
1− ab, I2 = ∆
1− a
1− ab
(a = e−T1/τ , b = e−T2/τ )
M. B. Patil, IIT Bombay
RC circuit: example
VS
R
C Vc
iC
0
I1
−I2
iC∆
VC
VS
V0
V2
V1
0T1 T2
T1 T2
∆
0 < t < T1 Let i(1)C (t) = Ae−t/τ + B
i(1)C (0) = I1, i
(1)C (∞) = 0
→ B = 0, A = I1.
i(1)C (t) = I1e−t/τ (1)
T1 < t < T2 Let i(2)C (t) = A′ e−t/τ + B′
i(2)C (T1) = −I2, i
(2)C (∞) = 0
→ B′ = 0, A′ = −I2 eT1/τ .
i(2)C (t) = −I2e−(t−T1)/τ (2)
Now use the conditions:
i(1)C (T1)− i
(2)C (T1) = ∆ = V0/R,
i(1)C (0)− i
(2)C (T1 + T2) = ∆ = V0/R.
I1e−T1/τ − (−I2) = ∆ (3)
I1 − (−I2e−(T1+T2−T1)/τ ) = ∆ (4)
a I1 + I2 = ∆ (5)
I1 + b I2 = ∆ (6)
Solve to get
I1 = ∆1− b
1− ab, I2 = ∆
1− a
1− ab
(a = e−T1/τ , b = e−T2/τ )
M. B. Patil, IIT Bombay
RC circuit: example
VS
R
C Vc
iC
0
I1
−I2
iC∆
VC
VS
V0
V2
V1
0T1 T2
T1 T2
∆
0 < t < T1 Let i(1)C (t) = Ae−t/τ + B
i(1)C (0) = I1, i
(1)C (∞) = 0
→ B = 0, A = I1.
i(1)C (t) = I1e−t/τ (1)
T1 < t < T2 Let i(2)C (t) = A′ e−t/τ + B′
i(2)C (T1) = −I2, i
(2)C (∞) = 0
→ B′ = 0, A′ = −I2 eT1/τ .
i(2)C (t) = −I2e−(t−T1)/τ (2)
Now use the conditions:
i(1)C (T1)− i
(2)C (T1) = ∆ = V0/R,
i(1)C (0)− i
(2)C (T1 + T2) = ∆ = V0/R.
I1e−T1/τ − (−I2) = ∆ (3)
I1 − (−I2e−(T1+T2−T1)/τ ) = ∆ (4)
a I1 + I2 = ∆ (5)
I1 + b I2 = ∆ (6)
Solve to get
I1 = ∆1− b
1− ab, I2 = ∆
1− a
1− ab
(a = e−T1/τ , b = e−T2/τ )
M. B. Patil, IIT Bombay
RC circuit: example
VS
R
C Vc
iC
0
I1
−I2
iC∆
VC
VS
V0
V2
V1
0T1 T2
T1 T2
∆
0 < t < T1 Let i(1)C (t) = Ae−t/τ + B
i(1)C (0) = I1, i
(1)C (∞) = 0
→ B = 0, A = I1.
i(1)C (t) = I1e−t/τ (1)
T1 < t < T2 Let i(2)C (t) = A′ e−t/τ + B′
i(2)C (T1) = −I2, i
(2)C (∞) = 0
→ B′ = 0, A′ = −I2 eT1/τ .
i(2)C (t) = −I2e−(t−T1)/τ (2)
Now use the conditions:
i(1)C (T1)− i
(2)C (T1) = ∆ = V0/R,
i(1)C (0)− i
(2)C (T1 + T2) = ∆ = V0/R.
I1e−T1/τ − (−I2) = ∆ (3)
I1 − (−I2e−(T1+T2−T1)/τ ) = ∆ (4)
a I1 + I2 = ∆ (5)
I1 + b I2 = ∆ (6)
Solve to get
I1 = ∆1− b
1− ab, I2 = ∆
1− a
1− ab
(a = e−T1/τ , b = e−T2/τ )
M. B. Patil, IIT Bombay
RC circuit: example
VS
R
C Vc
iC
0
I1
−I2
iC∆
VC
VS
V0
V2
V1
0T1 T2
T1 T2
∆
0 < t < T1 Let i(1)C (t) = Ae−t/τ + B
i(1)C (0) = I1, i
(1)C (∞) = 0
→ B = 0, A = I1.
i(1)C (t) = I1e−t/τ (1)
T1 < t < T2 Let i(2)C (t) = A′ e−t/τ + B′
i(2)C (T1) = −I2, i
(2)C (∞) = 0
→ B′ = 0, A′ = −I2 eT1/τ .
i(2)C (t) = −I2e−(t−T1)/τ (2)
Now use the conditions:
i(1)C (T1)− i
(2)C (T1) = ∆ = V0/R,
i(1)C (0)− i
(2)C (T1 + T2) = ∆ = V0/R.
I1e−T1/τ − (−I2) = ∆ (3)
I1 − (−I2e−(T1+T2−T1)/τ ) = ∆ (4)
a I1 + I2 = ∆ (5)
I1 + b I2 = ∆ (6)
Solve to get
I1 = ∆1− b
1− ab, I2 = ∆
1− a
1− ab
(a = e−T1/τ , b = e−T2/τ )
M. B. Patil, IIT Bombay
RC circuit: example
VS
R
C Vc
iC
0
I1
−I2
iC∆
VC
VS
V0
V2
V1
0T1 T2
T1 T2
∆
0 < t < T1 Let i(1)C (t) = Ae−t/τ + B
i(1)C (0) = I1, i
(1)C (∞) = 0
→ B = 0, A = I1.
i(1)C (t) = I1e−t/τ (1)
T1 < t < T2 Let i(2)C (t) = A′ e−t/τ + B′
i(2)C (T1) = −I2, i
(2)C (∞) = 0
→ B′ = 0, A′ = −I2 eT1/τ .
i(2)C (t) = −I2e−(t−T1)/τ (2)
Now use the conditions:
i(1)C (T1)− i
(2)C (T1) = ∆ = V0/R,
i(1)C (0)− i
(2)C (T1 + T2) = ∆ = V0/R.
I1e−T1/τ − (−I2) = ∆ (3)
I1 − (−I2e−(T1+T2−T1)/τ ) = ∆ (4)
a I1 + I2 = ∆ (5)
I1 + b I2 = ∆ (6)
Solve to get
I1 = ∆1− b
1− ab, I2 = ∆
1− a
1− ab
(a = e−T1/τ , b = e−T2/τ )
M. B. Patil, IIT Bombay
RC circuit: example
VS
R
C Vc
iC
0
I1
−I2
iC∆
VC
VS
V0
V2
V1
0T1 T2
T1 T2
∆
0 < t < T1 Let i(1)C (t) = Ae−t/τ + B
i(1)C (0) = I1, i
(1)C (∞) = 0
→ B = 0, A = I1.
i(1)C (t) = I1e−t/τ (1)
T1 < t < T2 Let i(2)C (t) = A′ e−t/τ + B′
i(2)C (T1) = −I2, i
(2)C (∞) = 0
→ B′ = 0, A′ = −I2 eT1/τ .
i(2)C (t) = −I2e−(t−T1)/τ (2)
Now use the conditions:
i(1)C (T1)− i
(2)C (T1) = ∆ = V0/R,
i(1)C (0)− i
(2)C (T1 + T2) = ∆ = V0/R.
I1e−T1/τ − (−I2) = ∆ (3)
I1 − (−I2e−(T1+T2−T1)/τ ) = ∆ (4)
a I1 + I2 = ∆ (5)
I1 + b I2 = ∆ (6)
Solve to get
I1 = ∆1− b
1− ab, I2 = ∆
1− a
1− ab
(a = e−T1/τ , b = e−T2/τ )
M. B. Patil, IIT Bombay
RC circuit: example
VS
R
C Vc
iC
0
I1
−I2
iC∆
VC
VS
V0
V2
V1
0T1 T2
T1 T2
∆
0 < t < T1 Let i(1)C (t) = Ae−t/τ + B
i(1)C (0) = I1, i
(1)C (∞) = 0
→ B = 0, A = I1.
i(1)C (t) = I1e−t/τ (1)
T1 < t < T2 Let i(2)C (t) = A′ e−t/τ + B′
i(2)C (T1) = −I2, i
(2)C (∞) = 0
→ B′ = 0, A′ = −I2 eT1/τ .
i(2)C (t) = −I2e−(t−T1)/τ (2)
Now use the conditions:
i(1)C (T1)− i
(2)C (T1) = ∆ = V0/R,
i(1)C (0)− i
(2)C (T1 + T2) = ∆ = V0/R.
I1e−T1/τ − (−I2) = ∆ (3)
I1 − (−I2e−(T1+T2−T1)/τ ) = ∆ (4)
a I1 + I2 = ∆ (5)
I1 + b I2 = ∆ (6)
Solve to get
I1 = ∆1− b
1− ab, I2 = ∆
1− a
1− ab
(a = e−T1/τ , b = e−T2/τ )
M. B. Patil, IIT Bombay
RC circuit: example
VS
R
C Vc
iC
0
I1
−I2
iC∆
VC
VS
V0
V2
V1
0T1 T2
T1 T2
∆
0 < t < T1 Let i(1)C (t) = Ae−t/τ + B
i(1)C (0) = I1, i
(1)C (∞) = 0
→ B = 0, A = I1.
i(1)C (t) = I1e−t/τ (1)
T1 < t < T2 Let i(2)C (t) = A′ e−t/τ + B′
i(2)C (T1) = −I2, i
(2)C (∞) = 0
→ B′ = 0, A′ = −I2 eT1/τ .
i(2)C (t) = −I2e−(t−T1)/τ (2)
Now use the conditions:
i(1)C (T1)− i
(2)C (T1) = ∆ = V0/R,
i(1)C (0)− i
(2)C (T1 + T2) = ∆ = V0/R.
I1e−T1/τ − (−I2) = ∆ (3)
I1 − (−I2e−(T1+T2−T1)/τ ) = ∆ (4)
a I1 + I2 = ∆ (5)
I1 + b I2 = ∆ (6)
Solve to get
I1 = ∆1− b
1− ab, I2 = ∆
1− a
1− ab
(a = e−T1/τ , b = e−T2/τ )
M. B. Patil, IIT Bombay
RC circuit: example
VS
R
C Vc
iC
0
I1
−I2
iC∆
VC
VS
V0
V2
V1
0T1 T2
T1 T2
∆
0 < t < T1 Let i(1)C (t) = Ae−t/τ + B
i(1)C (0) = I1, i
(1)C (∞) = 0
→ B = 0, A = I1.
i(1)C (t) = I1e−t/τ (1)
T1 < t < T2 Let i(2)C (t) = A′ e−t/τ + B′
i(2)C (T1) = −I2, i
(2)C (∞) = 0
→ B′ = 0, A′ = −I2 eT1/τ .
i(2)C (t) = −I2e−(t−T1)/τ (2)
Now use the conditions:
i(1)C (T1)− i
(2)C (T1) = ∆ = V0/R,
i(1)C (0)− i
(2)C (T1 + T2) = ∆ = V0/R.
I1e−T1/τ − (−I2) = ∆ (3)
I1 − (−I2e−(T1+T2−T1)/τ ) = ∆ (4)
a I1 + I2 = ∆ (5)
I1 + b I2 = ∆ (6)
Solve to get
I1 = ∆1− b
1− ab, I2 = ∆
1− a
1− ab
(a = e−T1/τ , b = e−T2/τ )M. B. Patil, IIT Bombay
RC circuit: example
Q
Q
insulator
conductor
conductor
Unit: Farad (F)
VS
R
VC
C Vc
iC
C =ǫA
tt
iC
iC
iC =dQ
dt= C
dVC
dt
0
I1
−I2
iC
VC
VS
V0
V2
V1
0T1 T2
T1 T2Charge conservation:
Periodic steady state: All quantities are periodic, i.e.,x(t0 + T ) = x(t0)
Capacitor charge: Q(t0 + T ) = Q(t0)
iC =dQ
dt→ Q =
∫iC dt.
Q(t0 + T ) = Q(t0)→ Q(t0 + T )− Q(t0) = 0
→∫ t0+T
t0
iC dt = 0.∫ T
0iC dt = 0→
∫ T1
0iC dt +
∫ T1+T2
T1
iC dt = 0
→∫ T1+T2
T1
iC dt = −∫ T1
0iC dt.
M. B. Patil, IIT Bombay
RC circuit: example
Q
Q
insulator
conductor
conductor
Unit: Farad (F)
VS
R
VC
C Vc
iC
C =ǫA
tt
iC
iC
iC =dQ
dt= C
dVC
dt
0
I1
−I2
iC
VC
VS
V0
V2
V1
0T1 T2
T1 T2Charge conservation:
Periodic steady state: All quantities are periodic, i.e.,x(t0 + T ) = x(t0)
Capacitor charge: Q(t0 + T ) = Q(t0)
iC =dQ
dt→ Q =
∫iC dt.
Q(t0 + T ) = Q(t0)→ Q(t0 + T )− Q(t0) = 0
→∫ t0+T
t0
iC dt = 0.∫ T
0iC dt = 0→
∫ T1
0iC dt +
∫ T1+T2
T1
iC dt = 0
→∫ T1+T2
T1
iC dt = −∫ T1
0iC dt.
M. B. Patil, IIT Bombay
RC circuit: example
Q
Q
insulator
conductor
conductor
Unit: Farad (F)
VS
R
VC
C Vc
iC
C =ǫA
tt
iC
iC
iC =dQ
dt= C
dVC
dt
0
I1
−I2
iC
VC
VS
V0
V2
V1
0T1 T2
T1 T2Charge conservation:
Periodic steady state: All quantities are periodic, i.e.,x(t0 + T ) = x(t0)
Capacitor charge: Q(t0 + T ) = Q(t0)
iC =dQ
dt→ Q =
∫iC dt.
Q(t0 + T ) = Q(t0)→ Q(t0 + T )− Q(t0) = 0
→∫ t0+T
t0
iC dt = 0.∫ T
0iC dt = 0→
∫ T1
0iC dt +
∫ T1+T2
T1
iC dt = 0
→∫ T1+T2
T1
iC dt = −∫ T1
0iC dt.
M. B. Patil, IIT Bombay
RC circuit: example
Q
Q
insulator
conductor
conductor
Unit: Farad (F)
VS
R
VC
C Vc
iC
C =ǫA
tt
iC
iC
iC =dQ
dt= C
dVC
dt
0
I1
−I2
iC
VC
VS
V0
V2
V1
0T1 T2
T1 T2Charge conservation:
Periodic steady state: All quantities are periodic, i.e.,x(t0 + T ) = x(t0)
Capacitor charge: Q(t0 + T ) = Q(t0)
iC =dQ
dt→ Q =
∫iC dt.
Q(t0 + T ) = Q(t0)→ Q(t0 + T )− Q(t0) = 0
→∫ t0+T
t0
iC dt = 0.∫ T
0iC dt = 0→
∫ T1
0iC dt +
∫ T1+T2
T1
iC dt = 0
→∫ T1+T2
T1
iC dt = −∫ T1
0iC dt.
M. B. Patil, IIT Bombay
RC circuit: example
Q
Q
insulator
conductor
conductor
Unit: Farad (F)
VS
R
VC
C Vc
iC
C =ǫA
tt
iC
iC
iC =dQ
dt= C
dVC
dt
0
I1
−I2
iC
VC
VS
V0
V2
V1
0T1 T2
T1 T2Charge conservation:
Periodic steady state: All quantities are periodic, i.e.,x(t0 + T ) = x(t0)
Capacitor charge: Q(t0 + T ) = Q(t0)
iC =dQ
dt→ Q =
∫iC dt.
Q(t0 + T ) = Q(t0)→ Q(t0 + T )− Q(t0) = 0
→∫ t0+T
t0
iC dt = 0.
∫ T
0iC dt = 0→
∫ T1
0iC dt +
∫ T1+T2
T1
iC dt = 0
→∫ T1+T2
T1
iC dt = −∫ T1
0iC dt.
M. B. Patil, IIT Bombay
RC circuit: example
Q
Q
insulator
conductor
conductor
Unit: Farad (F)
VS
R
VC
C Vc
iC
C =ǫA
tt
iC
iC
iC =dQ
dt= C
dVC
dt
0
I1
−I2
iC
VC
VS
V0
V2
V1
0T1 T2
T1 T2Charge conservation:
Periodic steady state: All quantities are periodic, i.e.,x(t0 + T ) = x(t0)
Capacitor charge: Q(t0 + T ) = Q(t0)
iC =dQ
dt→ Q =
∫iC dt.
Q(t0 + T ) = Q(t0)→ Q(t0 + T )− Q(t0) = 0
→∫ t0+T
t0
iC dt = 0.∫ T
0iC dt = 0→
∫ T1
0iC dt +
∫ T1+T2
T1
iC dt = 0
→∫ T1+T2
T1
iC dt = −∫ T1
0iC dt.
M. B. Patil, IIT Bombay
−10
10
0
0
2
4
6
8
10
0 1 2 3 4 5 6
t (msec)
VS
R
Vc
1 k
iC
C1µF
T1= 1.5msec, T2= 1.5msec.
V1= 1.8 V, V2= 8.2 V.
I1= 8.2mA, I2= 8.2mA.
VS
VC
iC (mA)
t (msec)
0 1 2 3 4 5 6
T1= 1msec, T2= 2msec.
V1= 0.9 V, V2= 6.7 V.
I1= 9.1mA, I2= 6.7mA.
t (msec)
0 1 2 3 4 5 6
T1= 2msec, T2= 1msec.
V1= 3.4 V, V2= 9.1 V.
I1= 6.7mA, I2 = 9.1mA.
SEQUEL file: ee101 rc1b.sqproj
M. B. Patil, IIT Bombay
−10
10
0
0
2
4
6
8
10
0 1 2 3 4 5 6
t (msec)
VS
R
Vc
1 k
iC
C1µF
T1= 1.5msec, T2= 1.5msec.
V1= 1.8 V, V2= 8.2 V.
I1= 8.2mA, I2= 8.2mA.
VS
VC
iC (mA)
t (msec)
0 1 2 3 4 5 6
T1= 1msec, T2= 2msec.
V1= 0.9 V, V2= 6.7 V.
I1= 9.1mA, I2= 6.7mA.
t (msec)
0 1 2 3 4 5 6
T1= 2msec, T2= 1msec.
V1= 3.4 V, V2= 9.1 V.
I1= 6.7mA, I2 = 9.1mA.
SEQUEL file: ee101 rc1b.sqproj
M. B. Patil, IIT Bombay
−10
10
0
0
2
4
6
8
10
0 1 2 3 4 5 6
t (msec)
VS
R
Vc
1 k
iC
C1µF
T1= 1.5msec, T2= 1.5msec.
V1= 1.8 V, V2= 8.2 V.
I1= 8.2mA, I2= 8.2mA.
VS
VC
iC (mA)
t (msec)
0 1 2 3 4 5 6
T1= 1msec, T2= 2msec.
V1= 0.9 V, V2= 6.7 V.
I1= 9.1mA, I2= 6.7mA.
t (msec)
0 1 2 3 4 5 6
T1= 2msec, T2= 1msec.
V1= 3.4 V, V2= 9.1 V.
I1= 6.7mA, I2 = 9.1mA.
SEQUEL file: ee101 rc1b.sqproj
M. B. Patil, IIT Bombay
−10
10
0
0
2
4
6
8
10
0 1 2 3 4 5 6
t (msec)
VS
R
Vc
1 k
iC
C1µF
T1= 1.5msec, T2= 1.5msec.
V1= 1.8 V, V2= 8.2 V.
I1= 8.2mA, I2= 8.2mA.
VS
VC
iC (mA)
t (msec)
0 1 2 3 4 5 6
T1= 1msec, T2= 2msec.
V1= 0.9 V, V2= 6.7 V.
I1= 9.1mA, I2= 6.7mA.
t (msec)
0 1 2 3 4 5 6
T1= 2msec, T2= 1msec.
V1= 3.4 V, V2= 9.1 V.
I1= 6.7mA, I2 = 9.1mA.
SEQUEL file: ee101 rc1b.sqproj
M. B. Patil, IIT Bombay
t (msec)
0 1 2 3 4 5 6
0
2
4
6
8
10
−100
100
0
VS
R
T1= 1msec, T2= 2msec.
V1≈ 0V, V2= 10V.
I1= 100mA, I2= 100mA.
Vc
0.1 k
iC
C1µF
iC (mA)
VS
VC
M. B. Patil, IIT Bombay