Muhammad Irfan Yousuf (Peon of Holy Prophet (P.B.U.H)) 2000-E-41 118
cos t:A + [1/CR1]B = 0 A = -B/CR1
sin t:-B + [1/CR1]A = I0/C-B + [1/CR1][-B/CR1] = I0/C
B = I0/C[- - 1/C2R12]
A = -[I0/C[- - 1/C2R12]]/CR1
Complete solutionV(t) = Ke-(1/CR1)t + [-[I0/C[- - 1/C2R1
2]]/CR1]sin t + [I0/C[- - 1/C2R12]]cos t
At t = 0+V(0+) = Ke-(1/CR1)0+ + [-[I0/C[- - 1/C2R1
2]]/CR1]sin (0+) + [I0/C[- - 1/C2R12]]cos
(0+)I0sin t [R1 + R2] = K(1) + [I0/C[- - 1/C2R1
2]]I0sin t [R1 + R2] = K + [I0/C[- - 1/C2R1
2]]
K = I0sin t [R1 + R2] - [I0/C[- - 1/C2R12]]
V(t) = [I0sin t [R1 + R2] - [I0/C[- - 1/C2R12]]]e-(1/CR1)t + [-[I0/C[- - 1/C2R1
2]]/CR1]sin t + [I0/C[- - 1/C2R1
2]]cos t
Q#6.29: Consider a series RLC network which is excited by a voltage source.1. Determine the characteristic equation.2. Locus of the roots of the equation.3. Plot the roots of the equation.
Solution:
R LC
V(t)i(t)
For t 0 According to KVL di 1L + idt + Ri = V(t) dt CDifferentiating with respect to ‘t’ d2i i diL + + R = 0
Muhammad Irfan Yousuf (Peon of Holy Prophet (P.B.U.H)) 2000-E-41 119
dt2 C dtDividing both sides by ‘L’d2i i Rdi + + = 0 (i)dt2 LC Ldt The characteristic equation can be found by substituting the trial solution i = est or by the equivalent of substituting s2 for (d2i/dt2), and s for (di/dt); thus
1 Rs2 + + s = 0 LC L
2)
= 0 jjn
= 1
-jn
= 0
1 Rs2 + + s = 0 LC LCharacteristic equation:as2 + bs + c = 0Here
a 1
bR
L
c1
LC -b b2 – 4acs1, s2 = 2a
Muhammad Irfan Yousuf (Peon of Holy Prophet (P.B.U.H)) 2000-E-41 120
R R 2 1- - 4(1)
L L LC
s1, s2 = 2(1)
R R 2 1- - 4(1)
L L LC
s1, s2 = 2 2
R R 2 1- - 4(1)
L L LC
s1, s2 = 2 4
R R 2 1
= - - 4(1) 2L 2L 4LC
R R 2 1 = - - 2L 2L LC radical term (ii)
Hint: 4 = 2To convert equation (i) to a standard form, we define the value of resistance that causes the radical (pertaining to the root) term in the above equation as the critical resistance, Rcr. This value is found by solving the equation
2 R 1
- = 0 2L LC R = Rcr
Muhammad Irfan Yousuf (Peon of Holy Prophet (P.B.U.H)) 2000-E-41 121
2 Rcr 1
- = 0 2L LC 2 Rcr 1
= 2L LC Taking square root of both the sides
2 Rcr 1
= 2L LC
Rcr 1
= 2L LC
Using cross multiplication
LRcr = 2
C
Hint: 1 = 1
R =
Rcr
R C =
2 L
Muhammad Irfan Yousuf (Peon of Holy Prophet (P.B.U.H)) 2000-E-41 122
1n =
LC
R2n =
L
1n
2 = LC
Substituting the corresponding values in equation (i) we gets2 + 2ns + n
2 = 0roots of the characteristic equation are
Characteristic equation:as2 + bs + c = 0Here
a 1b 2n
c n2
-b b2 – 4acs1, s2 = 2a
-2n (2n)2 – 4(1)(n2)
s1, s2 = 2(1)
-2n 42n2 – 4n
2
s1, s2 = 2 2
Simplifying we get
s1, s2 = -n n2 – 1
when = 0
s1, s2 = -(0)n n(0)2 – 1
s1, s2 = n–1
Muhammad Irfan Yousuf (Peon of Holy Prophet (P.B.U.H)) 2000-E-41 123
s1, s2 = jn
Hint: –1 = j
3)
R 500 L 1 HC 1 10-6 F
Substituting the corresponding values in equation (ii)
500 500 2 1
= - - 2(1) 2(1) (1)(10-6) (ii)
= -250 62500 - 1000000= -250 -937500= -250 937500-1
s1, s2 = -250 j968.246
R 1000 L 1 HC 1 10-6 F
Substituting the corresponding values in equation (ii)
1000 1000 2 1
= - - 2(1) 2(1) (1)(10-6) (ii)
= -500 250000 - 1000000= -500 -750000= -500 750000-1
s1, s2 = -500 j 866.025
Muhammad Irfan Yousuf (Peon of Holy Prophet (P.B.U.H)) 2000-E-41 124
R 3000 L 1 HC 1 10-6 F
Substituting the corresponding values in equation (ii)
3000 3000 2 1
= - - 2(1) 2(1) (1)(10-6) (ii)
= -1500 2250000 - 1000000= -1500 1250000= -1500 1118.034= (-1500 + 1118.034), (-1500 - 1118.034)
s1, s2 = -381.966, -2618.034
R 5000 L 1 HC 1 10-6 F
Substituting the corresponding values in equation (ii)
5000 5000 2 1
= - - 2(1) 2(1) (1)(10-6) (ii)
= -2500 6250000 - 1000000= -2500 5250000= -2500 2291.288= (-2500 + 2291.288), (-2500 - 2291.288)
s1, s2 = -208.712, -4791.288
Q#6.31: Analyze the network given in the figure on the loop basis, and determine the characteristic equation for the currents in the network as a function of k1. Find the values of k1 for which the roots of the characteristic equation are on the imaginary axis of the s plane. Find the range of values of k1 for which the roots of the characteristic equation have positive real parts. Solution:
1 H
Muhammad Irfan Yousuf (Peon of Holy Prophet (P.B.U.H)) 2000-E-41 125
i2
1 K1i1
1 1 1
V1(t)i1 i3
1 F
Loop i1:For t 0 According to KVLV1(t) = (i1)(1 ) + (i1 – i2)(1 ) + (i1 – i3)(1 ) + (i1 – i3)(XC)
1XC =
j2fc = 2fj2fc = jcj = s
1XC =
scc = 1 F
1XC = s(1 F)
1XC = s 1V1(t) = (i1)(1 ) + (i1 – i2)(1 ) + (i1 – i3)(1 ) + (i1 – i3)Simplifying s 1 1V1(t) = i1 + i1 – i2 + i1 – i3 + i1 - i3
s s
1 1V1(t) = (3 + )i1 – i2 – (1 + ) (i)
- +
+
-
Muhammad Irfan Yousuf (Peon of Holy Prophet (P.B.U.H)) 2000-E-41 126
s s
Loop i2:For t 0 According to KVL(i2 – i1)(1 ) + i2(XL) = 0XL = jLs = jXL = s(1 H)XL = s Substituting(i2 – i1)(1 ) + i2(s) = 0Simplifyingi2 – i1 + si2 = 0
(1 + s)i2 – i1 = 0 (ii)
Loop i3:For t 0 According to KVLSum of voltage rise = sum of voltage drop (a)Sum of voltage rise = k1i1
1Sum of voltage drop = (i3 – i1)(1 ) + (i3 – i1) + (i3)(1 )
sSubstituting in (a) 1(i3 – i1)(1 ) + (i3 – i1) + (i3)(1 ) = k1i1
s
Simplifying 1(i3 – i1)(1 ) + (i3 – i1) + (i3)(1 ) - k1i1 = 0 s
1 1i3 – i1 + i3 - i1 + i3 – k1i1 = 0 s s
1 1- + k1 + 1 i1 + 2 + i3 = 0 (iii) s s
Equations (i), (ii) & (iii) can be written in matrix form
Muhammad Irfan Yousuf (Peon of Holy Prophet (P.B.U.H)) 2000-E-41 127
1 1 3 + -1 - 1 + i1 V1
s s
-1 (1 + s) 0 i2
= 0
1 1 - 1 + k1 + 0 2 + i3 0 s s
A X B
Determinant of A =
1 1 1 1 3 + (1 + s) 2 + - (0)(0) - (-1) (-1) 2 + - 1 + k1 + s s s s
1 1(0) + (-) 1 + (-1)0 – (-) 1 + k1 + (1 + s)
s s
After simplifying Characteristic equation:(5 – k1)s2 + (6 – 2k1)s + (2 – k1) = 0When k1 = 0 (5 – 0)s2 + (6 – 2(0))s + (2 – 0) = 05s2 + 6s + 2 = 0as2 + bs + c = 0Here
a 5b 6c 2
-b b2 – 4acs1, s2 = 2a
Muhammad Irfan Yousuf (Peon of Holy Prophet (P.B.U.H)) 2000-E-41 128
-6 62 – 4(5)(2)s1, s2 = 2(5)
-6 36 – 40s1, s2 = 10
-6 -4s1, s2 = 10
-6 -14s1, s2 = 10
-6 j2s1, s2 = 10s1, s2 = -0.6 j0.2
s1, s2 = (-0.6 + j0.2), (-0.6 - j0.2)
When k1 = 1 (5 – 1)s2 + (6 – 2(1))s + (2 – 1) = 04s2 + 4s + 1 = 0as2 + bs + c = 0Here
a 4b 4c 1
-b b2 – 4acs1, s2 = 2a
-4 42 – 4(4)(1)s1, s2 = 2(4)
-4 16 – 16s1, s2 = 8
Muhammad Irfan Yousuf (Peon of Holy Prophet (P.B.U.H)) 2000-E-41 129
-4 0s1, s2 = 8 -4 0s1, s2 = 8
s1, s2 = -0.5, -0.5
When k1 = 2 (5 – 2)s2 + (6 – 2(2))s + (2 – 2) = 03s2 + 2s + 0 = 0as2 + bs + c = 0Here
a 3b 2c 0
-b b2 – 4acs1, s2 = 2a
-2 22 – 4(3)(0)s1, s2 = 2(3)
-2 4 – 0s1, s2 = 6 -2 4
s1, s2 = 6
-2 2
s1, s2 = 6
s1, s2 = 0, 0.667
When k1 = -1 (5 – (-1))s2 + (6 – 2(-1))s + (2 – (-1)) = 06s2 + 8s + 3 = 0as2 + bs + c = 0Here
a 6b 8
Muhammad Irfan Yousuf (Peon of Holy Prophet (P.B.U.H)) 2000-E-41 130
c 3 -b b2 – 4acs1, s2 = 2a
-8 82 – 4(6)(3)s1, s2 = 2(6)
-8 64 – 72s1, s2 = 12 -8 -8
s1, s2 = 6
-8 -18s1, s2 = 6 -8 j2.828s1, s2 = 6
s1, s2 = (-1.334 + j0.472), (-1.334 - j0.472)
Q#6.32: Show that equation 6-121 can be written in the form i = ke-nt cos (n1 - 2 t + )
Give the values for k and in terms of k5 and k6 of Eq. (6-121).
Solution:
Let k5 = kcos (i)k6 = -ksin (ii)k = (kcos)2 + (-ksin)2
k = k2cos2 + k2sin2k = k2(cos2 + sin2)k = k2(1)
k = k2 = k52 + k6
2
Dividing Eq. (i) by (ii)
kcos k5
= -cot = -ksin k6
Muhammad Irfan Yousuf (Peon of Holy Prophet (P.B.U.H)) 2000-E-41 131
-1 k5
= cot - k6
Using the trigonometric identitycos (x + y) = cos x cos y – sin x sin y
Q#6.33: A switch is closed at t = 0 connecting a battery of voltage V with a series RL circuit.
(a) Solution:sw
t = 0R L
Vi
For t 0 According to KVL
diV = iR + L dtDividing both sides by ‘L’ di R V + i = dt L LThis is a linear non-homogeneous equation of the first order and its solution is,
Thus RP = L VQ = L
Hence the solution of this equation
Muhammad Irfan Yousuf (Peon of Holy Prophet (P.B.U.H)) 2000-E-41 132
i = e-PtQePtdt + ke-Pt
Vi = e-(R/L)t e(R/L)tdt + ke-(R/L)t
L
Vi = e-(R/L)t e(R/L)tdt + ke-(R/L)t
L
e(R/L)t
e(R/L)tdt = d dt (R/L)t
L e(R/L)t
e(R/L)tdt =R
Substituting
V L e(R/L)t
i = e-(R/L)t + ke-(R/L)t L R
V i = + ke-(R/L)t Ri(0-) = i(0+) = 0Substituting i = 0 at t = 0
V 0 = + ke-(R/L)(0)
Re0 = 1 Vk = - RSubstituting
V -Vi = + e-(R/L)t R R
V
Muhammad Irfan Yousuf (Peon of Holy Prophet (P.B.U.H)) 2000-E-41 133
i = (1 - e-(R/L)t ) R
P = i2R
tWR = i2R dt 0
t V 2WR = (1 - e-(R/L)t )2Rdt 0 R(a - b)2 = a2 + b2 – 2ab
t V2
WR = (1 + e-2(R/L)t – 2(1)(e-(R/L)t))Rdt 0 R2
t V2
WR = (1 + e-2(R/L)t – 2e-(R/L)t)dt 0 R
V2 t t tWR = (1)dt + e-2(R/L)tdt + (-2e-(R/L)tdt) R 0 0 0Simplifying
V2 2L L 3LWR = t + e-(R/L)t - e-2(R/L)t -
R R 2R 2R
(b)Li2
WL = 2
LV2
WL = (1 - e-(R/L)t )2
2R2
(c)At t = 0
V2 2L L 3LWR = (0) + e-(R/L)(0) - e-2(R/L)(0) -
R R 2R 2R
V2 2L L 3L
Muhammad Irfan Yousuf (Peon of Holy Prophet (P.B.U.H)) 2000-E-41 134
WR = (0) + e0 - e0 -
R R 2R 2R
V2 2L L 3LWR = (1) - (1) -
R R 2R 2R
V2 WR = 0
R
WR = 0 joules
At t = 0 LV2
WL = (1 - e-(R/L)0)2
2R2
LV2
WL = (1 – e0)2
2R2
LV2
WL = (1 – 1)2
2R2
WL = 0 joules At t = LV2
WL = (1 - e-(R/L))2
2R2
LV2
WL = (1 – e-)2
2R2
LV2
WL = (1 – 0)2
2R2
LV2
WL = joules 2R2
(d)
Muhammad Irfan Yousuf (Peon of Holy Prophet (P.B.U.H)) 2000-E-41 135
In steady state total energy supply
W = WR + WL
V2 2L L 3L LV2
W = t + e-(R/L)t - e-2(R/L)t - + (1 – e-(R/L)t)2
R R 2R 2R 2R2
Q#6.34: In the series RLC circuit shown in the accompanying diagram, the frequency of the driving force voltage is (1) = n
(2) = n1 - 2
Solution:
1000 1 H
100 sin t i(t)1 F
For t 0 According to KVL di 1100 sin t = L + iR + idt
dt CHere = n
di 1100 sin nt = L + iR + idt … (i) dt C 1n =
LC
L = 1 HC = 1 10-6 F 1n =
(1 H)( 1 10-6 F)
After simplifying n = 1000 rad/sec
+
-
Muhammad Irfan Yousuf (Peon of Holy Prophet (P.B.U.H)) 2000-E-41 136
Substituting in (i) we get di 1100 sin 1000t = L + iR + idt … (i) dt CDifferentiating both the sides & substituting the values of L & C we get
d2i di i100 (1000) cos 1000t = (1) + (1000) + dt2 dt 10-6
Simplifying we get d2i di100000cos 1000t = + (1000) + 1000000i dt2 dt The trial solution for the particular integral isip = A cos 1000t + B sin 1000t d2ip dip
100000cos 1000t = + (1000) + 1000000ip
dt2 dt (ip) = -1000A sin 1000t + B 1000cos 1000t(ip) = -1000000A cos 1000t - B 1000000sin 1000t(ip) = Ist derivative(ip) = 2nd derivative100000cos 1000t = -1000000A cos 1000t - B 1000000sin 1000t + 1000(-1000A sin 1000t + B 1000cos 1000t) + 1000000(A cos 1000t + B sin 1000t)Simplifying100000cos 1000t = -1000000A cos 1000t – 1000000B sin 1000t - 1000000A sin 1000t + 1000000B cos 1000t + 1000000A cos 1000t + 1000000B sin 1000tSimplifyingEquating the coefficientsCos:100000 = 1000000B
100000B =
1000000
B = 0.1
Sin:
0 = - 1000000B – 1000000A + 1000000B0 = –1000000A
A = 0
ip = A cos 1000t + B sin 1000tSubstituting the values of A & Bip = (0) cos 1000t + (0.1) sin 1000t
Muhammad Irfan Yousuf (Peon of Holy Prophet (P.B.U.H)) 2000-E-41 137
ip = 0.1 sin 1000t
ejt – e-jt
sin t = 2j
Here = 1000
ej1000t – e-j1000t
sin 1000t = 2j
ej1000t – e-j1000t
ip = 0.1 Transient response 2j
In steady state At resonanceXL = XC
In a series RLC circuitZ = R + j(XL - XC)Z = R + j(XC - XC)Z = R
VIm =
Z
100Im =
1000
Im = 0.1 A
(2) = n1 - 2
Determine the values of n & substitute & simplifyDo yourself.
Muhammad Irfan Yousuf (Peon of Holy Prophet (P.B.U.H)) 2000-E-41 138
THE END
Top Related