First Order RC and RL Transient

Circuits

Objectives

• To introduce the transients’ phenomena.

• To analyze step and natural responses of first order RC circuits.

• To analyze step and natural responses of first order RL circuits.

Transients

• In circuits with inductors and capacitors voltage and current cannot

change instantaneously.

• The application or removal of sources or circuit elements creates a

transient behavior.

• Transient is the process of going from one steady state to another

steady state following a sudden change in the circuit configuration.

• Sudden changes are mainly due to switching process or faults.

FIRST ORDER CIRCUITS

• Circuits that contain a single energy storing elements.

• Either a capacitor or an inductor.

SECOND ORDER CIRCUITS

• Circuits with two independent energy storing elements in any

combination

Transient Analysis

• The circuit is modeled in time domain using differential equations.

• The order of the differential equation equals the number of

independent energy storing elements in the circuits.

• Currents and Voltages of circuits with just one C or One L can be

obtained using first order differential equations.

First Order ODF

00)y(t YBAy

dt

dy

The first order ordinary differential equation in the form

Has a solution

A

1T ,Y

where,

)eY - ( Y(t)

F

T

t-

0

A

B

YyFF

Transient partSteady state

Initial Conditions

Initial conditions are the values of the capacitors voltage or the inductor

current at starting instant of the transient period.

• t = 0- is the instant just before switching.

• t = 0+ is the instant just after switching.

• In Capacitors VC(0-) = VC(0+). Where VC is the capacitor voltage.

• In inductors iL(0-) = iL(0+). Where iL is the Inductor current.

First Order RC circuit

0cTHRTH

VRiV

R

C

ci

dt

dvCi

VTH

RTH

C

+

vc_

THC

C

THvv

dt

dvCR

A. Step Response

The response of the circuit to sudden application of an energy supply.

t=0

TH

TH

C

TH

C vCR

vCRdt

dv 11

For t = 0+ to inf.

Since,

Then,

TH

TH

C

TH

C vCR

vCRdt

dv 11

o The following is the first order differential equation describing the

Change in the capacitor voltage during the transient period.

o The solution of this equation requires the initial conditions of the

capacitor voltage VC(0). Note that VC(0+) = VC(0-) = VCo

)eV -(V V(t) -

C0CC

t

FFCV

o The general solution of this equation is given by:

Where,

- VCF is the final value of VC

- is the circuit time constant

First Order RC circuit

Final Value of VC

• The final value of VC occurs when the capacitor is fully charge i.e.

the rate of change of Vc = 0

• Using the differential equation,

• Then, VCF = VTH

• VCF is usually refers to as VC(∞)

• VCF can be obtained by replacing the capacitor with an open circuit

since iCF = 0

TH

TH

CF

TH

vCR

vCR

110

VTH

RTH

+

vcF_

t = ∞

Time constant

The time constant is a measure of how fast is the charging process of

the Capacitor.

Mathematically it’s the time required to reach 63% of the final value.

CRTH

Example

6kv

i0tFor CO

kkkRTH 36||6

CFTHVVv 6

+_ 0t

k6

k6

k6Fm100

V12

)(tiO

0t(t),iFind O

Hence, if the capacitor voltage is known the problem is solved

+

- 0t

k6

k6

k6

k6V12

)(tiO

a b

TH

v

3.0101003000 6 CRTH

VVVV CCC 246)0()0( 0

mA e 0.6667-1(t)i

)e6 -2( 6(t)

3.0-

0

3.0-

C

t

t

V

4V+_

4V+_

1) Initial Conditions

2) Time Constant

3) Final Value

4) Diff. equation

0),( FIND tti

mAk

Vi 2

12

24

][32)2)(2(36)0( VkmAVvc

)0()0( cC vv

capacitor across voltageInitial :1 STEP

USE CIRCUIT IN STEADY STATE PRIOR TO THE SWITCHING

)( Determine :2 STEP C V

+

VCF

_

VV

mAi

CF 275.4236

5.48

036)(

constant Time :3 STEP

CRTH :circuit Capacitive

kkkRTH 5.16||2 FC m100

sF 15.0)10100)(105.1( 63

e 6

5

6

27

6

)()(

7)e2 -23( 27(t)

15.0-

15.0-

C

t

C

t

tVti

V

equation aldifferenti The :4 STEP

Example

B. Natural Response of RC circuits

• The response of the circuit due to the energy stored in the capacitor is known as

natural response (no sources in the circuit).

• For t >0 KVL equation around the circuit yields,

t=0

0

0

CC

C

vdt

dvRC

viR

e

)e0 -(V 0(t)

-

-

0CC

R

V

dt

dVC(t)i

V

t

CoCC

t

iC

Natural Response

Capacitor Voltage

CR :constant Time

In the circuit shown in the figure , the switch opens at t = 0.

Find the numerical expression for i(t).

Example

Solution

• Before t = 0, the circuit has reached steady state so that the capacitor

acts like an open circuit. The circuit is equivalent to that shown in Fig.

(a) after transforming the voltage source.

Initial Conditions

Time Constant

Final Value = 0

KCL yields

Solution

First Order R-L circuits

0)0()0( LL ii

The differential equation is

Solving the equation yields

At t = 0 the switch

closes,

The initial conditions

tLRsL

tLRSS

tLRLFLLF

eVdt

tdiLtv

eR

V

R

Vti

IIIti

)/(

)/(

)/(

)()(

)0()(

))0(()(

The final conditions

ILF = Vs/R

R-L Step Response

First Order R-L circuits step response

EQ

L =

R

Equivalent Resistance seen by an Inductor

• For the RL circuit in the previous example,

it was determined that = L/R.

• As with the RC circuit, the value of R should actually be the equivalent

(or Thevenin) resistance seen by the inductor.

• In general, a first-order RL circuit has the following time constant:

seen from the terminals of the inductor for t > 0EQwith independent sources killed

R = REQ

L =

R

LCircuit

Circuit

t > 0

independent

sources killed

REQ

Natural Response of R-L Circuit

• When the switch is closed (ON) – the inductor will store the

energy. As a result, the inductor is said to be charged.

• When the switch is opened (OFF), this will result in the

instantaneous change in the circuit. The inductor will supply

the energy stored to the resistor.

Using the Kirchhoff’s voltage law (KVL), to find the differential

equation of the loop:

By using differential and integral equations technique:

0 iRdt

diL

/)0()( teIti

The circuit when the switch is opened

where = L/R is a time constant.

Natural Response of R-L Circuit

/)0()( tL eRItv

The current generated against time shows the inductor loses its energy

exponentially

/)0()( teIti

Natural Response of R-L Circuit

Example

• Assuming that i(0) = 10 A, calculate i(t) and ix (t) in the circuit.

The equivalent resistance is the same as

the Thevenin resistance at the inductor

terminals.

Because of the dependent source, we

insert a voltage source with vo = 1 V

• Since there is no energy source in the circuit, the final value of the inductor current is

zero i.e. ILF = 0

• The time constant = L/R = 1.5 sec.

5.1/

/

10)(

)0()(

t

t

eti

eIti

The natural response equation yields

5.1/

/

3

10)(

)0()1

(5.0)(

tL

tL

etv

eIdt

diLtv

The inductor voltage

Ix (t)= VL(t)/2

Solution

28

The switch in the circuit shown below has been closed for a long time.

It opens at t = 0. Find i(t) for t > 0.

teti 102)(

Example

Answer:

29

For the circuit, find i(t) for t > 0.

Answer: i(t) = 2e–2t A