Chapter 6: First Order Circuits

1. The source-free RC circuits

2. The source-free RL circuits

3. Singularity Functions

4. Step Response of an RC Circuit

5. Step Response of an RL Circuit

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Learning Outcomes...

At the end of this topic, students should be able to: • Describe the RC and RL circuit and its connection with

the time constant• Formulate and solve the differential equation to

describe the circuit’s behaviour• Explain the natural and forced responses of the first

order RC and RL circuits• Determine the natural and forced responses of the first

order RC and RL circuits

Source-free RC Circuit

• Consider a circuit below:

Apply Kirchhoff’s laws to purely resistive circuit results in algebraic

equations.

Apply the laws to RC and RL circuits produces differential equations.

• A first-order circuit is characterized by a first-order differential equation.

Ohms law Capacitor law

0 =+dt

dvC

R

v0 =+ CR ii

By KCL

Source-free RC Circuit

Solving this homogeneous equation:

Therefore,

DCR

tv

dtRC

dvv

RC

v

dt

dvdt

dvC

R

v

+−=

−=

−=

=+

∫∫

ln

11

0

D

t

DCR

t

eeeev τ==−

CR=τTime constant

Initial condition, v(0)

Source-Free RC Circuit (cont.)

• The natural response of a circuit refers to the behavior (in terms of voltages and currents) of the circuit itself, with no external sources of excitation.

• The time constant τ of a circuit is the time required for the response to decay by a factor of 1/e or 36.8% of its initial value.

• v decays faster for small τ and slower for large τ.

CR=τTime constantDecays more slowly

Decays faster

Source-Free RC Circuit (cont.)

The key to working with a source-free RC circuit is finding:

1. The initial voltage v(0) = V0 across the capacitor.

2. The time constant τ = RC.

τ−=

/

0)(

teVtv CR=τwhere

Example 1

Refer to the circuit below, determine vC, vx, and io for t ≥ 0.

Assume that vC(0) = 30 V.

• Please refer to lecture or textbook for more detail elaboration.

Answer: vC = 30e–0.25t V ; vx = 10e

–0.25t ; io = –2.5e–0.25t A

Example 2

The switch in circuit below is opened at t = 0, find v(t) for t ≥ 0.

• Please refer to lecture or textbook for more detail elaboration.

Answer: V(t) = 8e–2t V

Example 3

For the circuit shown below, find v when v(0) = 10 V.

• Please refer to lecture or textbook for more detail elaboration.

Answer: v(t) = 10e–5000t V

Source-Free RL Circuit

• A first-order RL circuit consists of a inductor L (or its equivalent) and a resistor (or its equivalent)

0 =+ RL vvBy KVL

0 =+ iRdt

diL

Inductors law Ohms law

dtL

R

i

di−=

LtReIti

/

0 )(−

=

Source-Free RL Circuit (cont.)

• The time constant τ of a circuit is the time required for the response to decay by a factor of 1/e or 36.8% of its initial value.

• i(t) decays faster for small τ and slower for large τ.

• The general form is very similar to a RC source-free circuit.

τ−=

/

0)(t

eIti

R

L=τ

A general form representing a RL

where

Source-Free RL Circuit (cont.)

τ−=

/

0)(t

eItiR

L=τ

A RL source-free circuit

where τ/

0)(t

eVtv−

= RC=τ

A RC source-free circuit

where

Comparison between a RL and RC circuit

Source-Free RL Circuit (cont.)

The key to working with a source-free RL circuit is finding:

1. The initial voltage i(0) = I0 through the inductor.

2. The time constant τ = L/R.

τ/

0)(

teIti

−=

R

L=τwhere

Example 1

Determine the current i(t) for t ≥ 0s for the circuit shown below. Assume that the switch was closed a long time.

Answer: i= 2/7 e-100t

Example 2

Determine the inductor current, iL(t) for t ≥0s for the circuit shown below. Assume that switch was at position 1 for a long time.

Answer: iL= 10 e-76.9t mA

Example 3

Find i and vx in the circuit. Assume that i(0) = 5 A.

• Please refer to lecture or textbook for more detail elaboration.

Answer: i(t) = 5e–53t A

Example 4

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

• Please refer to lecture or textbook for more detail elaboration.

Answer: i(t) = 2e–2t A

Unit-Step Function

• The unit step function u(t) is 0 for negative values of t

and 1 for positive values of t.

>

<=

0,1

0,0)(

t

ttu

>

<=−

o

o

ott

ttttu

,1

,0)(

−>

−<=+

o

o

ott

ttttu

,1

,0)(

Unit-Step Function

1. voltage source.

2. for current source:

Represent an abrupt change for:

• Initial condition: v(0-) = v(0+) = V0

• Applying KCL,

or

• Where u(t) is the unit-step function

Step-Response of a RC Circuit

• The step response of a circuit is its behavior when the excitation is the step function, which may be a voltage or a current source.

0)(

=−

+R

tuVv

dt

dvC s

)(tuRC

Vv

dt

dv s−−=

Step-Response of a RC Circuit

Solving this equation:

Therefore,

( )

( )

CR

t

VV

Vv

dtRC

dvVv

tuRCdt

dv

Vv

tuRC

Vv

dt

dv

S

S

tv

VS

S

S

−=−

−

−=−

−=−

−−=

∫∫

0

0

ln

11

11

0

( ) CR

t

SS eVVVv−

−+= 0

CR=τTime constant

Initial condition, v(0)

CR

t

S

S eVV

Vv −

=−

−⇒

0

Step-Response of a RC Circuit (cont.)

• Integrating both sides and considering the initial conditions, the solution of the equation is:

>−+

<=

−0)(

0)(

/

0

0

teVVV

tVtv

t

ss

τ

Final value at t -> ∞

Initial value at t = 0

Source-free Response

Complete Response = Natural response + Forced Response(stored energy) (independent source)

= V0e–t/τ + Vs(1–e–t/τ)

Step-Response of a RC Circuit (cont.)

Three steps to find out the step response of an RC circuit:

1. The initial capacitor voltage v(0).

2. The final capacitor voltage v(∞) — DC voltage across C.

3. The time constant τ.

τ/ )]( )0( [ )( )(

tevvvtv

−∞−++∞=

Note: The above method is a short-cut method. You may also determine the solution by setting up the circuit formula directly using KCL, KVL , ohms law, capacitor and inductor VI laws.

Example 1

Find the capacitor voltage and current for t ≥ 0 s for the circuit shown below. At what time does the voltage v pass through zero?

Assume that the switch was in position 1 for a long time prior to t = 0-s.

Answer: v = 120 – 180 e-tV; t = 0.41 s

Solution

Determine the initial voltage, v at t ≤ 0s, i.e. v(0).

The switch has been in position 1 for a long time so the capacitor looks like an open circuit.

( )

V

v

60

8026

6

−=

−+

=

Therefore, the voltage across 6 Ω resistor is (use voltage divider rule)

-ve sign is due to the reference for v is opposite to that of 80 V voltage source.

Solution (cont.)

Determine the final value of voltage, v at t → ∞, i.e. v(∞).

After the switch has been in position 2 for a long time, the capacitor will look like an open circuit.

Hence the final value voltage, v is 120 V.

Solution (cont.)

Determine the time constant, τ at t → 0s.

s

RC

1

10110133

=

×××=

=−

τ

Time constant,

Solution (cont.)

We know that the total response, v(t) is given by

Therefore

To find t when v = 0 V,

( ) ( ) ( )[ ] τ

t

evvvv−

∞−+∞= 0

[ ]Ve

evt

t

−

−

−=

−−+=

180120

12060120 1

( ) st

e

ev

t

t

405.032ln3

2

180

120

0180120

=−=

==

=−=

−

−

Solution (cont.)

To find, i(t), we know that

Therefore

dt

dvCi =

( )

( )( )Ae

e

edt

di

t

t

t

−

−−

−−

=

−−××=

−×=

18.0

1180101

180120101

3

3

Find v(t) for t > 0 s in the circuit shown below. Assume the switch has been opened for a long time and is closed at t = 0 s.

Calculate v(t) at t = 0.5.

Example 2

• Please refer to lecture or textbook for more detail elaboration.

Answer: and v(0.5) = 0.5182V515)(2

−=− t

etv

Step-response of a RL Circuit

• The step response of a circuit is its behavior when the excitation is the step function, which may be a voltage or a current source.

• Initial current

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

• Final inductor currenti(∞) = Vs/R

• Time constant τ = L/R

)()()( tueR

VI

R

Vti

t

s

o

s τ−

−+=

Step-Response of a RL Circuit (cont.)

Three steps to find out the step response of an RL circuit:

1. The initial inductor current i(0) at t = 0+.

2. The final inductor current i(∞).

3. The time constant τ.

Note: The above method is a short-cut method. You may also determine the solution by setting up the circuit formula directly using KCL, KVL , ohms law, capacitor and inductor VI laws.

τ/ )]( )0( [ )( )(

teiiiti

−∞−++∞=

The switch in the circuit shown below has been closed for a long time. It opens at t = 0 s.

Find i(t) for t > 0 s.

Example 1

• Please refer to lecture or textbook for more detail elaboration.

Answer: teti

102)(

−+=

Example 2

The current and voltage at the terminals of the inductor in the RL circuit shown below are

Specify the numerical values of VS, R, IO, and L.

[ ]stVev

stAeit

t

0;100

0;4440

40

≥−=

≥+=−

−

Summary

Source-free response of RL and RC circuits

The response depends on two quantities:

1. The circuit time constant, τ

2. The value of inductor current or capacitor voltage at t = 0 s.

vC(t)=vC(0) e-t/RCRC

iL(t)=iL(0) e-Rt/LRL

Source-free ResponseCircuit

Summary (cont.)

RL and RC circuits excited by a constant source

The response depends on two quantities:

1. The circuit time constant, τ

2. The value of inductor current or capacitor voltage at t = 0 s.

3. The value of inductor current or capacitor voltage at t → ∞.

vC(t)= vC(∞) + (vC(0)- vC(∞)) e-t/RCRC

iL(t)= iL(∞)+(iL(0)- iL(∞)) e-t/τRL

ResponseCircuit

References

• Alexander Sadiku, Fundamentals of Electric Circuits, 4th

edition, McGraw-Hill, 2009

• Richard C. Dorf and James A. Svoboda, Introduction to Electric Circuits, 3rd edition, John Wiley, 1996

• Thomas L. Floyd and David M. Buchla, Electric Circuits Fundamentals, 8th edition, Pearson, 2010

• James W. Nilsson & Susan A. Riedel, Electric Circuits, 9th

edition, Pearson-Prentice Hall, 2011