network analysis by van valkenburg solution CHAP#6

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


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


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


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


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


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


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


R 3000 L 1 HC 1 10-6 F


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


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


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)

- +

+

-


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


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


-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


-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


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


-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


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


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


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)


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

+

-


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


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.


THE END

network analysis by van valkenburg solution CHAP#6

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