Chapter 6.2

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/C2R1

2]

A = -[I0/C[- - 1/C2R1

2]]/CR1

Complete solution

V(t) = Ke-(1/CR1)t

+ [-[I0/C[- - 1/C

2R1

2]]/CR1]sin t + [I0/C[- - 1/C

2R1

2]]cos t

At t = 0+

V(0+) = Ke-(1/CR1)0+

+ [-[I0/C[- - 1/C

2R1

2]]/CR1]sin (0+) + [I0/C[- - 1/C

2R1

2]]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/C2R1

2]]

V(t) = [I0sin t [R1 + R2] - [I0/C[- - 1/C2R1

2]]]e

-(1/CR1)t +

[-[I0/C[- -

1/C2R1

2]]/CR1]sin t + [I0/C[- - 1/C

2R1

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 L

C

V(t)

i(t)

For t 0

According to KVL

di 1

L + idt + Ri = V(t)

dt C


Differentiating with respect to ‘t’

d2i i di

L + + R = 0

dt2 C dt

Dividing 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 (d

2i/dt

2), and s for (di/dt); thus

1 R

s2 + + s = 0

LC L

2)

= 0 j

jn

= 1

-jn

= 0

1 R

s2 + + s = 0

LC L

Characteristic equation:

as2 + bs + c = 0

Here

a 1

b

R

L

c

1

LC


-b b2 – 4ac

s1, 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 = 2

To 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

L

Rcr = 2

C

Hint: 1 = 1

R

=

Rcr


R C

=

2 L

1

n =

LC

R

2n =

L

1

n2 =

LC

Substituting the corresponding values in equation (i) we get

s2 + 2ns + n

2 = 0

roots of the characteristic equation are


as2 + bs + c = 0

Here

a 1

b 2n

c n2

-b b2 – 4ac

s1, s2 =

2a

-2n (2n)2 – 4(1)(n

2)

s1, s2 =

2(1)

-2n 42n

2 – 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 H

C 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 H

C 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 H

C 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 H

C 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 KVL

V1(t) = (i1)(1 ) + (i1 – i2)(1 ) + (i1 – i3)(1 ) + (i1 – i3)(XC)

1

XC =

j2fc

= 2f

j2fc = jc

j = s

1

XC =

sc

c = 1 F

- +

+

-


1

XC =

s(1 F)

1

XC =

s 1

V1(t) = (i1)(1 ) + (i1 – i2)(1 ) + (i1 – i3)(1 ) + (i1 – i3)

Simplifying s

1 1

V1(t) = i1 + i1 – i2 + i1 – i3 + i1 - i3

s s

1 1

V1(t) = (3 + )i1 – i2 – (1 + ) (i) s s

Loop i2:

For t 0

According to KVL

(i2 – i1)(1 ) + i2(XL) = 0

XL = jL

s = j

XL = s(1 H)

XL = s

Substituting

(i2 – i1)(1 ) + i2(s) = 0

Simplifying

i2 – i1 + si2 = 0

(1 + s)i2 – i1 = 0 (ii)

Loop i3:

For t 0

According to KVL

Sum of voltage rise = sum of voltage drop (a)

Sum of voltage rise = k1i1

1

Sum of voltage drop = (i3 – i1)(1 ) + (i3 – i1) + (i3)(1 )

s

Substituting 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 1

i3 – 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


(5 – k1)s2 + (6 – 2k1)s + (2 – k1) = 0

When k1 = 0

(5 – 0)s2 + (6 – 2(0))s + (2 – 0) = 0

5s2 + 6s + 2 = 0

as2 + bs + c = 0

Here

a 5

b 6

c 2

-b b2 – 4ac

s1, s2 =

2a

-6 62 – 4(5)(2)

s1, s2 =

2(5)

-6 36 – 40

s1, s2 =

10

-6 -4

s1, s2 =

10

-6 -14

s1, s2 =

10

-6 j2

s1, s2 =

10

s1, 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) = 0

4s2 + 4s + 1 = 0

as2 + bs + c = 0

Here


a 4

b 4

c 1

-b b2 – 4ac

s1, s2 =

2a

-4 42 – 4(4)(1)

s1, s2 =

2(4)

-4 16 – 16

s1, s2 =

8

-4 0

s1, s2 =

8

-4 0

s1, s2 =

8

s1, s2 = -0.5, -0.5

When k1 = 2

(5 – 2)s2 + (6 – 2(2))s + (2 – 2) = 0

3s2 + 2s + 0 = 0

as2 + bs + c = 0

Here

a 3

b 2

c 0

-b b2 – 4ac

s1, s2 =

2a

-2 22 – 4(3)(0)

s1, s2 =

2(3)

-2 4 – 0

s1, 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)) = 0

6s2 + 8s + 3 = 0

as2 + bs + c = 0

Here

a 6

b 8

c 3

-b b2 – 4ac

s1, s2 =

2a

-8 82 – 4(6)(3)

s1, s2 =

2(6)

-8 64 – 72

s1, s2 =

12

-8 -8

s1, s2 =

6

-8 -18

s1, s2 =

6

-8 j2.828

s1, 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 = k2cos

2 + k

2sin

2

k = k2(cos

2 + sin

2)

k = k2(1)

k = k2

= k5

2 + k6

2

Dividing Eq. (i) by (ii)

kcos k5

= -cot =

-ksin k6

-1 k5

= cot -

k6

Using the trigonometric identity

cos (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 = 0

R L

V

i

For t 0

According to KVL


di

V = iR + L

dt

Dividing both sides by ‘L’

di R V

+ i =

dt L L

This is a linear non-homogeneous equation of the first order and its solution is,

Thus

R

P =

L

V

Q =

L

Hence the solution of this equation

i = e-PtQe

Ptdt + ke

-Pt

V

i = e-(R/L)t

e(R/L)t

dt + ke-(R/L)t

L

V

i = e-(R/L)t

e(R/L)t

dt + ke-(R/L)t

L

e(R/L)t

e(R/L)t

dt =

d

dt (R/L)t

L e(R/L)t

e(R/L)t

dt =

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

R

i(0-) = i(0+) = 0

Substituting i = 0 at t = 0

V

0 = + ke-(R/L)(0)

R

e0

= 1

V

k = -

R

Substituting

V -V

i = + e-(R/L)t

R R

V

i = (1 - e-(R/L)t

)

R

P = i2R

t

WR = i2R dt

0

t V 2

WR = (1 - e-(R/L)t

)2Rdt

0 R

(a - b)2 = a

2 + b

2 – 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 t

WR = (1)dt + e-2(R/L)t

dt + (-2e-(R/L)t

dt)

R 0 0 0

Simplifying


V2

2L L 3L

WR = 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 3L

WR = (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 3L

WR = (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 1

100 sin t = L + iR + idt

dt C

Here

= n

di 1

100 sin nt = L + iR + idt … (i)

dt C

1

n =

LC

L = 1 H

C = 1 10-6

F

1

n =

(1 H)( 1 10-6

F)

After simplifying

n = 1000 rad/sec

Substituting in (i) we get

di 1

100 sin 1000t = L + iR + idt … (i)

dt C Differentiating both the sides & substituting the values of L & C we get

d2i di i

100 (1000) cos 1000t = (1) + (1000) +

dt2

dt 10-6

Simplifying we get

d2i di

100000cos 1000t = + (1000) + 1000000i

dt2

dt

The trial solution for the particular integral is

ip = 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 derivative

100000cos 1000t = -1000000A cos 1000t - B 1000000sin 1000t + 1000(-1000A sin 1000t

+ B 1000cos 1000t) + 1000000(A cos 1000t + B sin 1000t)

Simplifying


100000cos 1000t = -1000000A cos 1000t – 1000000B sin 1000t - 1000000A sin 1000t +

1000000B cos 1000t + 1000000A cos 1000t + 1000000B sin 1000t

Simplifying

Equating the coefficients

Cos:

100000 = 1000000B

100000

B =

1000000

B = 0.1

Sin:

0 = - 1000000B – 1000000A + 1000000B

0 = –1000000A

A = 0

ip = A cos 1000t + B sin 1000t

Substituting the values of A & B

ip = (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 resonance

XL = XC

In a series RLC circuit

Z = R + j(XL - XC)

Z = R + j(XC - XC)

Z = R


V

Im =

Z

100

Im =

1000

Im = 0.1 A

(2) = n1 - 2

Determine the values of n & substitute & simplify

Do yourself.

THE END

Documents

Chapter 6.2