Chapter 7

Engr228
Circuit Analysis

Dr Curtis Nelson

Chapter 7 Objectives

• Be able to determine the natural response of both RL and

RC circuits;
• Be able to determine the step response of both RL and RC
circuits.

Engr228 - Chapter 7, Nilsson 9E 1

 Characteristics of R, L, C

• Resistors resist current flow;

• Inductors resist change of current;
• Capacitors resist change of voltage.

L and C are considered “dynamic” elements because

their V-I characteristics are a function of time.

Review: DC Characteristics of an Inductor

diL (t )
vL (t ) = L
dt

When iL(t) is constant, diL(t) = 0 thus, vL(t) = 0.

In other words, the inductor can be replaced with a short circuit.

Engr228 - Chapter 7, Nilsson 9E 2

 Review: DC Characteristics of a Capacitor

dvC (t )
iC (t ) = C
dt

When vC(t) is constant, dvC(t) = 0 thus, iC(t) = 0.

In other words, the capacitor can be replaced with an open circuit.

Types of Responses

• Transient Response, natural response, homogeneous

solution (e.g. oscillation, temporary position change)
– Fades over time
– Resists change
• Forced Response, steady-state response, particular solution
(e.g. permanent position change)
– Follows the input
– Independent of time passed

Engr228 - Chapter 7, Nilsson 9E 3

 Pendulum Example

I am holding a ball with a rope attached. What is the movement

of the ball if I move my hand to another point?

Two movements:

1. Oscillation.

2. Forced position change.

Mechanical Analogue

Forced response Natural response

at a different time

Engr228 - Chapter 7, Nilsson 9E 4

 Transient Response

• RL Circuit

First-order differential equation

• RC Circuit Chapter 7

• RLC Circuit Second-order differential equation

Chapter 8

Source Free RL Circuits

Inductor L has energy stored

so that the initial current is I0.

Similar to a pendulum that is

at a height h (potential energy
is nonzero.)
height

Engr228 - Chapter 7, Nilsson 9E 5

 Source Free RL Circuits

di(t )
Ri(t ) + L =0
dt
di(t ) R
+ i (t ) = 0
dt L

There are 2 ways to solve first-order differential equations.

Solving Source Free RL Circuits

Method 1: Assume solution is of the form i(t ) = Ae st
where A and s are the parameters that we wish to solve for.

di (t ) R
Substitute i (t ) = Ae st in the equation + i (t ) = 0
dt L
R st
Ase st + Ae = 0
L
R
(s + ) Ae st = 0
L
R
s=−
L
R
− t
L
i (t ) = Ae

Engr228 - Chapter 7, Nilsson 9E 6

 Solving Source Free RL Circuits - continued

Initial condition: i (0) = I 0

R
− t
L
from i (t ) = Ae

I 0 = Ae 0
I0 = A
R
− t
L
Therefore i (t ) = I 0 e

Solving Source Free RL Circuits

Method 2: Direct integration
t
di(t ) R i (t ) R
+ i (t ) = 0 ln i (t ) I =− t
dt L
0
L 0
di(t ) R R
= − i (t ) ln i (t ) − ln I 0 = − (t − 0)
dt L L
R
di(t ) R − t
= − dt i (t ) = I 0 e L
i (t ) L
i (t ) t
di (t ) R
∫
I0
= ∫ − dt
i (t ) 0 L

Engr228 - Chapter 7, Nilsson 9E 7

 Time Constant

The ratio L/R is called the “time constant” and is denoted

by the symbol τ (tau).
L
τ= Units: seconds
R

One time constant is defined as the amount of time

required for the output to go from 100% to 36.8%.

R t
− t −
L τ
i (t ) = I 0 e = I 0e

e −1 = 0.368

Time Constant Graph

Engr228 - Chapter 7, Nilsson 9E 8

 1st Order Response Observations

• The voltage on a capacitor or the current through an inductor is

the same prior to and after a switch at t = 0 seconds because
these quantities cannot change instantaneously.
• Resistor voltage (or current) prior to the switch v(0-) can be
different from the voltage (or current) after the switch v(0+).
• All voltages and all currents in an RC or RL circuit follow the
same natural response e-t/τ.

General RL Circuits

The time constant of a single-inductor circuit will be τ = L/Req

where Req is the resistance seen by the inductor.

Example: Req=R3+R4+R1R2 / (R1+R2)

Engr228 - Chapter 7, Nilsson 9E 9

 Example: RL with a Switch

Find the voltage v(t) at t = 200 ms.

v(t) = -12.99 volts at t = 200 ms

Textbook Problem 8.8 Hayt 7E

After being in the configuration shown for hours, the switch

in the circuit below is closed at t = 0 seconds. At t = 5µs,
calculate iL and iSW.

Engr228 - Chapter 7, Nilsson 9E 10

 Textbook Problem 8.8 - continued

Step 1 – Find the initial current through the inductor.

iL(0) = 9V/2kΩ = 4.5 mA

Step 2 – Close the switch and calculate iL(t).

−10 6
t
4
iL = 4.5e mA
iL (5µs) = 1.289 mA
iSW = 9 – iL (5µs) = 9 – 1.289 = 7.711 mA

Source-Free RC Circuits

As you might expect, source-free RC circuits are an analogue

of source-free RL circuits. The derivation for the capacitor
voltage is now a node equation rather than a loop equation.

Engr228 - Chapter 7, Nilsson 9E 11

 Source-Free RC Circuits
• To be consistent with the direction of
assigned voltage, we write
v(t ) dv(t )
+C =0
R dt
dv(t ) v(t )
+ =0
dt RC

• Comparing the last equation to that of the RL circuit, it

becomes obvious that the form of the solution is the same
with the time constant τ = RC rather than L/R

di (t ) R
+ i (t ) = 0
dt L

Comparison Between Source-free RL and RC

• Transient response equations:

R t
− t −
L τ
iL (t ) = I L 0 e = I L 0e
t t
− −
RC τ
v C (t ) = VC 0 e = VC 0 e

• Time constants:

L
τL =
R
τ C = RC

Engr228 - Chapter 7, Nilsson 9E 12

 RC Natural Response

The time constant is τ = RC

for a first-order RC circuit.

General RC Circuits
The time constant of a single-capacitor circuit will be τ = ReqC
where Req is the resistance seen by the capacitor.

Example: Req=R2+R1R3 / (R1+R3)

Engr228 - Chapter 7, Nilsson 9E 13

 The Source Free RC Circuit

Find the voltage v(t) at t = 200µS.

v(t) = 321mV at t = 200µS

Textbook Problem 8.22 Hayt 7E

(a) Find vC(t) for all time in the circuit below.

(b) At what time is vC = 0.1vC(0)?

vC(0) = 192V vC (t ) = vC (0)e −125t

vC = 0.1vC(0) when t = 18.42mS

Engr228 - Chapter 7, Nilsson 9E 14

 Textbook Problem 8.25 Hayt 7E

Determine the value of the current labeled i and the voltage

labeled v at t = 0+, t = 1.5mS, and t = 3.0 mS.

v(0+) = 20V i(0+) = 0.1A

v(1.5ms) = 4.5V i(1.5ms) = 0A
v(3ms) = 1.0V i(3ms) = 0A

Example: L and R Circuit

Find i1(t) and iL(t) for t > 0.

Answer: τ = 20 µs; i1=-0.24e-t/τ, iL=0.36e-t/τ for t > 0

Engr228 - Chapter 7, Nilsson 9E 15

 Driven RL and RC Circuits

• Many RL and RC circuits are driven by a DC or an AC source.

The complete response of a driven RL or RC source is the sum
of a transient response and a forced response:

i (t ) = tran (t ) + forced (t )

The Unit Step Function u(t)

• A Step Function is often used to drive circuits.

• The forcing function of 1u(t) represents a function that has
zero value up until t = 0, and then a value of 1 forever after.

Engr228 - Chapter 7, Nilsson 9E 16

 Step Functions

v(t) v(t)

1V 1V

0V 0V
t t
Unit Step function

Example

I = 1A I = 2A

Suppose the voltage source changes abruptly from 1V to 2V.

Does the current change abruptly as well?

Engr228 - Chapter 7, Nilsson 9E 17

 Example - continued
Voltage

2V

1V
time
Current

2A

1A
time

Transient Response + Forced Response

Complete Response

• The differential equation for the circuit above now becomes

di (t )
Ri(t ) + L = vS (t )
dt
• The transient part of the complete solution is determined by
setting the forcing function vS(t) = 0
di ( t )
Ri ( t ) + L =0
dt

Engr228 - Chapter 7, Nilsson 9E 18

 Complete Solution

• Solve by assuming i(t) is of the form

t
−
τ
k1 + k 2 e

di (t )
Ri (t ) + L = u (t )
dt

• As shown on the following pages through direct integration,

substituting the solution above into the circuit equation yields

V − t
R R
− t
i (t ) = 1 − e L  + i (0)e L
R 

Solution – Direct Integration

For t > 0 (letting V = u)
L
di(t ) − ln(V − Ri(t )) = t + c1
Ri(t ) + L =V R
dt R R
di (t ) ln(V − Ri(t )) = − t − c1
L = V − Ri(t ) L L
dt R
− t
R
− c1
L L
Ldi (t ) V − Ri(t ) = e ⋅e
= dt R
V − Ri(t ) − t
L
V − Ri(t ) = e ⋅ c2
Ldi(t )
∫ V − Ri(t ) = ∫ dt Ri(t ) = V − c2 e
R
− t
L

L V c2 − L t
R
− ln(V − Ri(t )) = t + c1 i (t ) = − e
R R R

Engr228 - Chapter 7, Nilsson 9E 19

 Direct Integration - continued
R
V c2 − L t
i (t ) = − e
R R
We can find c2 from the initial condition i(0)
V c2
i (0) = − ⋅1
R R
c2 = V − i (0) R
R
V V − i (0) R − L t
Therefore, we have i (t ) = − e
R R
V − t
R R
− t
i (t ) = 1 − e L  + i (0)e L
R 

Review: Solving First-Order Circuits

• Start by finding the initial current through the inductor L or

the initial voltage across the capacitor C;
• Assume that the desired response is of the form
t
−
τ
k1 + k 2 e

• Find the time constant τ (may use Thévenin if necessary);

• Solve for k1 and k2 using initial conditions and the status of
the circuit as time approaches infinity;
• After solving for the inductor current (or capacitor voltage),
find other values that the problem may ask for.

Engr228 - Chapter 7, Nilsson 9E 20

 Driven RL Circuits

V0
i( t ) =
R
(1 − e − Rt / L )u ( t )

Example: RL Circuit with Step Input

Find i(t)

i(t)=25+25(1-e-t/2)u(t) A

Engr228 - Chapter 7, Nilsson 9E 21

 Driven RC Circuits (Part 1 of 2)

Find Vc(t) and i(t)

vC(t)=20 + 80e-t/1.2 V and i(t)=0.1 + 0.4e−t/1.2 A

Driven RC Circuits (Part 2 of 2)

vC=20 + 80e-t/1.2 V

i=0.1 + 0.4e−t/1.2 A

Engr228 - Chapter 7, Nilsson 9E 22

 Example

The switch is in the position shown for a long time before t = 0.

Find i(t).

t
−
τ
i (t ) = k1 + k 2e

Time constant τ = 1 sec

Example - continued
t
−
τ
i (t ) = k1 + k 2 e

At t = 0, i(0) = 2 A 2 = k1 + k 2

At t = ∞, i(∞) = 1 A 1 = k1 + 0

Therefore, k1 = 1, k2 = 1

The answer is: i (t ) = 1 + e − t

Engr228 - Chapter 7, Nilsson 9E 23

 Example Graph

i (t )

2A

1A
i (t ) = 2 i (t ) = 1 + e − t

Example
L stores no energy at t = 0.
Find v1(t).

Find iL(t) first

Req = ( 2 || 2) + 1 = 2
L 1
τ= = = 0.5
Req 2

Engr228 - Chapter 7, Nilsson 9E 24

 Example - continued
t
−
τ
Find k1, k2 using i(0) = 0, i(∞) = 0.25 iL (t ) = k1 + k 2 e

At t = 0, i(0) = 0 A 0 = k1 + k2

At t = ∞, i(∞) = 0.25 A 0.25 = k1 + 0

Therefore k1= 0.25, k2 = -0.25

iL (t ) = 0.25 − 0.25e −2t

v1(t) = iL(t) R

v1 (t ) = 0.25 − 0.25e −2t

Example Graph

v1 (t )

v1 (t ) = 0 v1 (t ) = 0.25 − 0.25e −2t

Engr228 - Chapter 7, Nilsson 9E 25

 Textbook Problem 8.72 Hayt 7E
The switch has been closed for a very long time.
a) Find iL for t < 0
b) Find iL for t = 0+
c) Find iL(∞)
d) Find iL(t) for t > 0

iL(0-) = -15A
iL(0+) = -15A ( )
iL (t ) = 5 1 − e −40t − 15e −40t amps
iL(∞) = 5A

RC Circuit Response

Engr228 - Chapter 7, Nilsson 9E 26

 Chapter 7 Summary

• Showed how to determine the natural response of both RL

and RC circuits;
• Showed how to determine the step response of both RL
and RC circuits.

Engr228 - Chapter 7, Nilsson 9E 27