electric circuits i dr. hassan ahmad

Electric Circuits I

Dr. Eng.

Hassan M. Ahmad [email protected], [email protected]

كــلــيـــة هندسة الحاسوب والمعلوماتية

Computer and Informatics Engineering

Faculty

Dr. H

assan A

hmad

2

Chapter 8 Second-Order Circuits

8.1 Examples of 2nd order RCL circuit

8.2 Finding Initial and Final Values

8.3 The source-free series RLC circuit

8.4 The source-free parallel RLC circuit

8.5 Step response of a series RLC circuit

8.6 Step response of a parallel RLC

9/23/2018 Dr. Eng. Hassan Ahmad Dr. H

assan A

hmad

8.1 Examples of 2nd order RCL circuit

A second-order circuit is characterized by a second-

order differential equation. It consists of resistors and

the equivalent of two energy storage elements.

Typical examples of second-order circuits:

(a) series RLC circuit,

(b) parallel RLC circuit,

(c) RL circuit,

(d) RC circuit.

3 9/23/2018 Dr. Eng. Hassan Ahmad Dr. H

assan A

hmad

8.2 Finding Initial and Final Values

The major problem in analysis of the second-order circuits is finding the

initial and final conditions on circuit variables: v(0), i(0), dv(0)∕dt, di(0)∕dt,

i(∞), and v(∞).

There are two key points to keep in mind in determining the initial conditions.

• First—we must carefully handle the polarity of voltage v(t) across the

capacitor and the direction of the current i(t) through the inductor.

• Second, keep in mind that the capacitor voltage is always continuous so

that

and the inductor current is always continuous so that

where denotes the time just before a switching event and is the

time just after the switching event, assuming that the switching event takes

place at t = 0.

4 9/23/2018 Dr. Eng. Hassan Ahmad

(0 ) (0 )v v

(0 ) (0 )i i

0t 0t

Dr. H

assan A

hmad

Example 8.1. The switch in Fig. has been closed for a long time. It is open

at t = 0. Find: i(0+), v(0+), di(0+)∕dt, dv(0+)∕dt, i(∞), v(∞).

Solution:

a) If the switch is closed a long time before

t = 0, it means that the circuit has reached

dc steady state at t = 0.

At dc steady state, the inductor acts like

a short circuit, while the capacitor acts

like an open circuit, so we have the circuit in Fig.(a) at t = 0−.

Thus,

As the inductor current and the capacitor

voltage cannot change abruptly,

9/23/2018 Dr. Eng. Hassan Ahmad 5

12(0 ) 2A, (0 ) 2 (0 ) 4V

4 2i v i

(0 ) 2A,(0 ) (0 ) (0 ) 4Vi vi v

Dr. H

assan A

hmad

b) At t = 0+, the switch is open; Fig.(b).

The same current flows through both the

inductor and capacitor.

Hence,

Similarly,

Applying KVL to the loop:


(0 ) (0 ) 2A

(0 ) 220 V /

0.

(s

1

0 )

C

C CC

i i

i idv dv

dt

dvi C

dt dt C C

LL

di di vv L

dt dt L

12 4 (0 ) (0 ) (0 ) 0

(0 ) 12 8 4 0

(0 ) 00 A / s

0.25

(0 )

L

L

Ldi

t

i v v

v

v

Ld

Dr. H

assan A

hmad

c) For t > 0, the circuit undergoes transience.

But as t → ∞, the circuit reaches steady

state again.

The inductor acts like a short circuit and

the capacitor like an open circuit, so that

the circuit in Fig.(b) becomes that shown

in Fig. (c), from which we have


( ) 0A, ( ) 12Vi v

Dr. H

assan A

hmad

8.3 The Source-Free Series RLC Circuits

The solution of the source-free series RLC circuit is

called as the natural response of the circuit.

The circuit is excited by the energy initially stored in

the capacitor (V0) and inductor (I0).

• Thus, at t = 0,

• Applying KVL around the loop:

• By differentiate with respect to t, we get

• This is a second-order differential equation.


0

0 0

1(0) , (0)v idt V i I

C

01( ) 0

diRi L i d

dt C

2

20

d i R di i

dt L dt LC

Dr. H

assan A

hmad

The initial value i is:

For the first-order circuit, the solution for i is of

exponential form:

where A and s are constants to be determined.

Thus,

Since is the assumed solution, so only

with characteristic roots

or


0 00

(0 (0) 1( )

)(0) 0

diRI V

diRi L V

d dt Lt

sti Ae

22

2

2 1

0 0

0

st st s

st

td i R di i AR AAs e se e

dt L dt

R

LC L L

Ae s s

C

L LC

sti Ae 2 10

Rs s

L LC

0

1,

2

R

L LC

2

1,2

1

2 2

R Rs

L L LC

2 2

1,2 0s 2 2

02 0s s

Dr. H

assan A

hmad

The two values of s indicate that there are two possible solutions for i,

where the constants A1 and A2 are determined from the initial values i(0) and

di(0)∕dt.

There are three types of solutions:

1. Overdamped Case ( متضائل فوق ),

– A typical overdamped response in Fig.(a).

2. Critically Damped Case ( حرج بشكل متضائل )

A typical critically damped response Fig.(b).


1 2 1 2

1 1 2 2 1 2, ( )s t s t s t s t

i Ae i A e i t Ae A e

1 22

0 1 2, i.e. 4 ( )s t s t

C L R i t Ae A e

2

0 1 2

1 2 1 2 3

, i.e. 42

( ) ( )t t t t

RC L R s s

L

i t Ae A e A A e A e

( )

1 max

ti t te

t

Dr. H

assan A

hmad

3. Underdamped Case ( متضائلة تحت )

• ω0 is often called the undamped natural frequency,

• ωd is called the damped natural frequency.

The natural response is

For simply,

The response has a time constant of 1 ∕α and a period of T = 2π∕ωd.


2

2 2

1,2 0

0 , i

(

e 4

)

. .

ds j

C L R

2 2

01, dj

1 2

1 1 2 2 1 2

( ) ( cos sin )

, ( )

t

d di t e B t B t

B A A B j A A

1 2( ) ( cos sin )t

d di t e A t A t

Dr. H

assan A

hmad

Example 8.2. In Fig., R = 40 Ω, L = 4 H, and C = 1∕4 F.

Calculate the characteristic roots of the circuit. Is the natural response

overdamped, underdamped, or critically damped?

Solution,

Since α > ω0, we conclude that the response is overdamped.


0

2 2

1,2 0

1 2

40 1 15, 1

2 2(4) 14

4

5 25 1

0.101, 9.899

R

L LC

s

s s

Dr. H

assan A

hmad

Example 8.3. Find i(t) in the circuit of Fig. Assume that the circuit has

reached steady state at t = 0–.

Solution,

For t < 0, the switch is closed.

capacitor open circuit,

inductor a shunted circuit.

The equivalent circuit, Fig.(a).

Thus, at t = 0,

• the initial current through the inductor is

• the initial voltage across the capacitor


10(0) 1A

4 6i

(0) 6 (0) 6Vv i

Dr. H

assan A

hmad

For t > 0, the switch is opened.

The voltage source is disconnected, Fig.(b).

(source free series RLC circuit),

where the 3-Ω and 6-Ω resistors are in series

when the switch is opened:

The roots are calculated as follows:


3 6 9R

0

2 2

1,2 0

9 1 19, 10

12 1 12( )2 2 50

9 81 10

9, 4.

0

9

3

4

9

.359

5

d

d

R

L L

j j

C

s

Dr. H

assan A

hmad

Hence, the response is underdamped (α < ω0); that is,

At t = 0,

But,

Note: , because the polarity of v in Fig. (b) is opposite that in Fig. for

source-free series RLC circuit (slid 8).

Thus,

At t = 0,

Finally,


9

1 2 1 2( ) ( cos sin ) ( cos4.359 sin 4.359 )t t

d di t e A t A t e A t A t

1(0) 1A=i A

0 0

0

1 1( ) [ (0) (0)] 2[9(1) 6] 6A/ s

t

diRI V Ri v

dt L L

0 (0)V v

9

1 2

9 9

1 2 1 2

( cos4.359 sin 4.359 )

9 ( cos4.359 sin 4.359 ) (4.359)( sin 4.359 cos4.359 )

t

t t

di de A t A t

dt dt

e A t A t e A t A t

1 2 1 2

0

6 9( 0) 4.359( 0 ), for 1 0.6882t

diA A A A

dt

9( ) (cos4.359 0.6882sin 4.359 ) Ati t e t t Dr. H

assan A

hmad

8.4 Source-Free Parallel RLC Circuits

Parallel RLC circuits find many practical applications, notably in

communications networks and filter designs.

Consider the parallel RLC circuit shown in Fig.

• Assume initial inductor current I0 and initial capacitor voltage V0,

• Thus, applying KCL at the top node gives

• Taking the derivative with respect to t

and dividing by C results in

where s is first derivative, s2 is second derivative.


0

0 0

1(0) ( ) , (0)i I v t dt v V

L

01( ) 0

v dvv d C

R L dt

2

2

2 1 10

1 10

d v dvv

dt RC dt LCs s

RC LC

Dr. H

assan A

hmad

The roots of characteristic equation are

or

There are three possible solutions,

1. Overdamped Case.

The response is

2. Critically Damped Case.

The response is


1

2

2

,2

1 1 11

2

1

20 s

RC RCL Cs

Ls

RC C

2 2

1,2 0s 0

1 1,

2RC LC

1 2

2

2 2

1 2 1,

0

2 0( )

, i.e. 4

, s t s t

v t Ae A e s

L R C

2

2 2

0

1 1

, . . 4

(real, )( ) ( ) tv t A A t e s s

i e L R C

Dr. H

assan A

hmad

3. Underdamped Case.

The response is

• The constants A1 and A2 in each case can be determined from the initial

conditions. We need v(0) and dv(0)∕dt.

at the top node


2 2

1 2 0 1,2( ) cos sin , ,t

d d d dv t A t A t e s j

0

0 0

1(0) ( ) , (0)i I v t dt v V

L

0

0 0

00

( )(0

1 (0)(

o)

r

) 0

V RI

Vv dv dvv d C I C

R L dt R dt

dv

dt RC

2

0 , i.e. 4L R C

Dr. H

assan A

hmad

Example 8.4. In the parallel circuit of Fig., find v(t) for t > 0, assuming

v(0) = 5 V, i(0) = 0, L = 1 H, and C = 10 mF. Consider these cases: R = 1.923 Ω,

R = 5 Ω, and R = 6.25 Ω.

Solution:

CASE 1: R = 1.923 Ω,

Since α > ω0 in this case, the response is overdamped. So,

the corresponding response is

We now apply the initial conditions to get A1 and A2 :


3

03

1 126

2 2 1.923 10 10

1 110

1 10 10

RC

LC

2 2

1,2 0 1 2 2, 50s s s 2 50

1 2( ) t tv t Ae A e

1 2(0) 5 (1)v A A

0

3

0 (0) (0) (0) 5 0260

1.923 10

)0)

10

(( dv v RiV RIdv

dt RC dt RC

Dr. H

assan A

hmad

But

At t = 0,

From Eqs. (1) and (2), we obtain A1 = −0.2083 and A2 = 5.208.

Finally,

CASE 2: R = 5 Ω,

Since α > ω0 in this case, the response is damped. So,

To get A1 and A2 , we apply the initial conditions:

But,

At t = 0,

Finally,


2 50 2 50

1 2 1 2( ) 2 50t t t tdvv t Ae A e Ae A e

dt

1 2260 2 50 (2)A A

2 50( ) 0.2083 5.208t tv t e e

03 3

1 1 1 110, 10

2 2 5 10 10 1 10 10RC LC

2 2 10

1,2 0 1 2 1 2 1 2 10, ( ) ( ) ( ) ( )t ts s s v t A A t e v t A A t e

1 3

(0) (0) (0) 5 0(0) , 100

5 10 105

dv v Riv

d RCA

t

10

1 2 2( 10 10 ) tdvA A t A e

dt

21 2 50100 10A A A

10( ) (5 50 ) Vtv t t e Dr. H

assan A

hmad

CASE 3: R = 6.25 Ω,

Since α < ω0 in this case, the response is underdamped. So,

To get A1 and A2 , we apply the initial conditions:

But,

At t = 0,

Finally,


03 3

1 1 1 110, 10

2 2 5 10 10 1 10 10RC LC

2 2

1,2 0 1,2 8 6 6d ds s j j

31

(0) (0) (0) 5 0(0) , 80

6.25 10 15

0

dv v Ri

dAv

t RC

8

1 2

8

1 2 1 2

cos6 sin 6

( 8 cos6 8 sin 6 6 sin 6 6 cos6 )

t

t

dv dA t A t e

dt dt

A t A t A t A t e

1 2 2 6.6 780 8 6 6A A A

8( ) 5cos6 6.667sin 6 tv t t t e

8

1 2( ) cos6 sin 6 tv t A t A t e

Dr. H

assan A

hmad

Notice that by increasing the value of R, the degree of damping decreases and

the responses differ ( تتباين = تختلف ). Figure plots the three cases of Example 8.4.

9/23/2018 Dr. Eng. Hassan Ahmad 22 Dr. H

assan A

hmad

8.5 Step-Response of Series RLC Circuits

• The step response is obtained by the sudden application of a dc voltage.

• Applying KVL around the loop for t > 0,

The solution to this Eq. has two components: the transient response vt(t) and the

steady-state response vss(t); that is,

The transient response vt(t) is the component of the total response that dies out with

time.

Therefore, the transient response for three cases:


2

2

,s

sVd v R dv v

d

di dvL Ri v V i C

d

t L dt LC L

t dt

C

( ) ( ) ( )t ssv t v t v t

1 2

1 2

1 2

1 2

( ) (Ovedamped)

( ) ( ) (Criticallydamped)

( ) cos sin (Underdamped)

s t s t

t

t

d d

v t Ae A e

v t A A t e

v t A t A t e

Dr. H

assan A

hmad

• The steady-state response is the final value of v(t).

• Thus, the complete solutions are:

• The values of the constants A1 and A2 are obtained from the initial conditions:

v(0) and dv(0)∕dt.


1 2

1 2

1 2

1 2

( ) (Ovedamped)



s t s t

s

t

s

t

s d d

v t V Ae A e

v t V A A t e

v t V A t A t e

( ) ( )ss sv t v V

Dr. H

assan A

hmad

Example 8.5. For the circuit in Fig., find v(t) and i(t) for t > 0. Consider these

cases: R = 5 Ω, R = 4 Ω, and R = 1 Ω.

Solution:

CASE 1: R = 5 Ω,

For t < 0, the switch is closed for a long time.

capacitor open circuit, inductor short circuit.

For t > 0, the switch is opened, 1- Ω resistor disconnected.

The characteristic roots are determined as follows:

Since α > ω0 , we have the overdamped natural response. The total response is

therefore


24(0) 4A, (0) 1 (0) 4V

5 1i v i

2 2

0 1,2 0

5 1 12.5, 2 1, 4

2 2 1 1 0.25

Rs

L LC

4 4

1 2 1 2( ) ( ) ( ) 24 ( )t t t t

ssv t v Ae A e v t Ae A e

Dr. H

assan A

hmad

To find A1 and A2 using the initial conditions:

But,

At t = 0,

From Eqs. (1) and (2):

Finally,

Since the inductor and capacitor are in series for t > 0, the inductor current is the

same as the capacitor current. Hence,


1 2 1 2 20 (1(0) 4 4 )2v A A A A

(0) (0) 4 4(0) 4 16

0.25

dv dvi C

dt dt C

4 4

1 2 1 224 ( ) 4t t t tdv dAe A e Ae A e

dt dt

1 21 2)

4 ( )(

60d

Av

dtA

1 264 3, 4 3A A

44( ) 24 ( 16 ) V

3

t tv t e e

44( ) ( ) (4 ) A

3

t tdvi t C i t e e

dt

Dr. H

assan A

hmad

CASE 2: R = 4 Ω,



So, have the critically damped natural response. The total response is therefore

At t = 0,


Finally,


24(0) 4.8A (0) 1 (0) 4,

1V

4.8v ii

0 1 2

4 1 12, 2 2

2 2 1 1 0.25

Rs s

L LC

2

1 12 1( ) 24 ( ) , (0) 4.8 2 19.24 (1)tv t A A t e v A A

4.8(0) (0) 4.8

(0) 19.2dv dv

i Cdt dt C

2 2

1 2 1 2 224 ( ) ( 2 2 )t tdv dA A t e A tA A e

dt dt

1 219.2 2(

)0)

(2A Adv

dt

1 219.2, 19.2A A

2( ) 24 19.2(1 ) Vtv t t e 2( ) ( ) (4.8 9.6 ) At tdv

i t C i t e edt

Dr. H

assan A

hmad

CASE 3: R = 1 Ω,



Since α < ω0 in this case, the response is underdamped. So,

The total response is:


(0) 1 (0) 124

(0) 12A, 2V1 1

v ii

0

1 1 1 10.5, 2

2 2 1 1 0.25L LC

2 2

1,2 0 1,2 0.5 1.936 1.936d ds s j j

0.5

1 2( ) 24 cos1.936 sin1.936 tv t A t A t e

11 12 (1)(0) 12 24v A A

(0) (0) 12(0) 12 48

dv dvi C

dt dt C

Dr. H

assan A

hmad

But

At t = 0,


Finally,

The inductor current is:


0.5

1 2

0.5

1 2

0.5

1 2

24 cos1.936 sin1.936

( 1.936 sin1.936 1.936 cos1.936 )

0.5 ( cos1.936 sin1.936 )

t

t

t

dv dA t A t e

dt dt

A t A t e

e A t A t

2 148 ( 0 1.93(

6 ) 0.5( 0) (0

2))

Ad

dtA

v

1 212, 21.694A A

0.5( ) 24 21.694sin1.936 12cos1.936 Vtv t t t e

0.5( ) ( ) 3.1sin1.936 12cos1.936 Atdvi t C i t t t e

dt

Dr. H

assan A

hmad

Figure plots the responses for the three cases.

From this figure, we observe that the critically damped response approaches the step

input of 24 V the fastest.

9/23/2018 Dr. Eng. Hassan Ahmad 30 Dr. H

assan A

hmad

8.6 Step-Response of Parallel RLC Circuits

The step response is obtained by the sudden application of a dc current.

• Applying KCL at the top node for t > 0,

The complete solution consists of the transient response it(t) and the steady-state

response iss(t); that is,

The final value of the current through the inductor is the same as the source

current Is. Thus,

The constants A1 and A2 in each case can be determined from the initial

conditions for i and di∕dt.


s

v dvi C I

R dt

2

2

1with sIdi d i di i

v Ldt dt RC dt LC LC

( ) ( ) ( )t ssi t i t i t

1 2

1 2

1 2

1 2

( ) (Ovedamped)



s t s t

s

t

s

t

s d d

i t I Ae A e

i t I A A t e

i t I A t A t e

Dr. H

assan A

hmad

Example 8.6. In the circuit of Fig., find i(t) and iR(t) for t > 0.

Solution:

For t < 0, the switch is open, and the circuit is partitioned into two independent

subcircuits, Fig.(a).

• The 4-A current flows through the inductor, so that

• t < 0 30u(− t) = 30; t > 0 30u(− t) = 0, the voltage source is operative

for t < 0.


(0) 4Ai

Dr. H

assan A

hmad

• The capacitor acts like an open circuit,

and , Fig.(b).

• By VDR on Fig.(b)., the initial

capacitor voltage (t =0) is

For t > 0, the switch is closed, and we

have a parallel RLC circuit with a current

source, Fig.(c).

• t > 0 30u(− t) = 0, the voltage

source acts like a short-circuit, Fig.(d).

Thus, R = 20 ‖ 20 = 10 Ω.


(8mF) (20 )v v

20(30)

20 2(0) 15V

0v

03 3

2 2

1,2 0 1 2

1 1 1 16.25, 2.5

2 2 10 8 10 20 8 10

6.25 5.7282 11.978, 0.5218

RC LC

s s s

Dr. H

assan A

hmad

Since α > ω0 we have the overdamped case. Hence,

At t = 0,

But

so that at t = 0,

But

Thus,

From Eqs.(1) and (2):

The complete solution is as


1 2 11.978 0.5218

1 2 1 2( ) 4s t s t t t

si t I Ae A e Ae A e

1 2 2 1(0) 4 (1)4i A A A A

11.978 0.5218 11.978 0.5218

1 2 1 24 11.978 0.5218t t t tdi dAe A e Ae A e

dt dt

1 2

(0)11.978 0.5218

diA A

dt

( ) (0) (0)0.

15 15( ) (0)

275

015

di t di div t L v L

dt dt dt L

1 211.978 0.5218 0.75 (2)A A

2 10.0655, 0.0655A A

0.5218 11.978( ) 4 0.0655( ) At ti t e e

11.978 0.5218( ) 1( ) ( ) 0.785 0.0342 A

20 20

t t

R

v t dii t L i t e e

dt

Dr. H

assan A

hmad

35

9/23/2018 Dr. eng. Hassan Ahmad Dr. H

assan A

hmad

electric circuits i dr. hassan ahmad

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