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 10E 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.
1V
Review: DC Characteristics of an Inductor i(t)
1 Ω
L 1V i(t)
1 Ω v
L
( t ) = L di
L dt
( t )
When i
L
(t) is constant, di
L
(t) = 0 thus, v
L
(t) = 0.
In other words, the inductor can be replaced with a short circuit.
Engr228 - Chapter 7, Nilsson 10E 2

1V
Review: DC Characteristics of a Capacitor i(t)
1 Ω
C
1V i(t)
1 Ω i
C
( t ) = C dv
C
( t ) dt
When v
C
(t) is constant, dv
C
(t) = 0 thus, i
C
(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 to zero 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 10E 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
Engr228 - Chapter 7, Nilsson 10E
Forced response
Natural response at a different time
4

• RL Circuit
• RC Circuit
• RLC Circuit
Transient Response
First-order differential equation
Chapter 7
Second-order differential equation
Chapter 8 height
-
+
R
Source Free RL Circuits i(t)
L
+
-
Inductor L has energy stored so that the initial current is I
0
.
Similar to a pendulum that is at a height h (potential energy is nonzero).
Engr228 - Chapter 7, Nilsson 10E 5

Source Free RL Circuits i(t)
-
+
R L
+
-
Ri ( t ) + di ( t ) dt
+
L di ( t dt
R
L i ( t )
)
=
= 0
0
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 constants that we wish to solve for.
Substitute i ( t ) = Ae st in the equation di ( t ) dt
+
R
L i ( t )
=
0
(
Ase st s +
+
R
L
Ae
R
L
) Ae st
= st
0
= 0 s = −
R
L i ( t ) = Ae
−
R
L t
Engr228 - Chapter 7, Nilsson 10E 6

Solving Source Free RL Circuits - continued
Initial condition: from i ( 0 )
=
I
0 i ( t ) = Ae
−
R
L t
Therefore
I
0
I
0
= Ae
= A
0 i ( t ) = I
0 e
−
R
L t
Solving Source Free RL Circuits
Method 2: Direct integration di ( t ) dt di ( t ) di i dt
( t
( t
)
)
+
=
=
R
L i ( t )
−
−
=
R i ( t )
L
R dt
L
0 i ( t ) di ( t )
I
0 i ( t ) t
=
0
−
R
L dt i ln i ( t ) ln i ( t ) i ( t )
I
0
− ln
= −
I
0
R
L t t
0
= −
R
L
( t − 0 )
( t ) = I
0 e
−
R
L t
Engr228 - Chapter 7, Nilsson 10E 7

Time Constant
The ratio L/R is called the time constant and is denoted by the symbol τ ( tau ).
τ =
L
R
Units: seconds
One time constant is defined as the amount of time required for the output to go from 100% to 36.8%.
i ( t ) = I
0 e
−
R
L t
= I
0 e
−
τ t e − 1
= 0 .
368
Time Constant Graph
Engr228 - Chapter 7, Nilsson 10E 8

1 st 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/R eq where R eq is the resistance seen by the inductor.
Example: R eq
=R
3
+R
4
+R
1
R
2
/ (R
1
+R
2
)
Engr228 - Chapter 7, Nilsson 10E 9

Example: RL with a Switch
Find the voltage v(t) at t = 200 ms.
v(t) = -12.99 volts at t = 200 ms
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 10E 10

Source-Free RC Circuits
• To be consistent with the direction of assigned voltage, we write v (
R t ) dv ( t ) dt
+ C
+ dv
RC
( dt v ( t ) t )
=
=
0
0
• 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 ) dt
+
R
L i ( t ) = 0
Comparison Between Source-free RL and RC
• Transient response equations: i
L
( t ) = I
L 0 e
−
R
L t v
C
( t ) = V
C t
0 e
−
RC
= I
L 0 e
−
τ t
= V
C 0 e
−
τ t
• Time constants:
τ
L
τ
C
L
=
=
R
RC
Engr228 - Chapter 7, Nilsson 10E 11

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 = R eq where R eq is the resistance seen by the capacitor.
C
Example: R eq
=R
2
+R
1
R
3
/ (R
1
+R
3
)
Engr228 - Chapter 7, Nilsson 10E 12

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 v
C
(t) for all time in the circuit below.
(b) At what time is v
C
= 0.1
v
C
(0)?
v
C
(0) = 192V v
C
= 0.1v
C v
C
( t )
= v
C
( 0 ) e
(0) when t = 18.42mS
− 125 t
Engr228 - Chapter 7, Nilsson 10E 13

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 1 u(t) represents a function that has zero value up until t = 0, and then a value of 1 forever after.
Engr228 - Chapter 7, Nilsson 10E 14

1V v(t) t=0
Step Functions t=0
R 1V R v(t)
1V 1V
0V
Unit Step function t
0V t
Example
1V
I
1 Ω
L 2V
I
1 Ω
I = 2A
L
I = 1A
Suppose the voltage source changes abruptly from 1V to 2V.
Does the current change abruptly as well?
Engr228 - Chapter 7, Nilsson 10E 15

Voltage
1V
Current
1A
Example - continued
2V time
2A
Transient Response + Forced Response time
Complete Response
• The differential equation for the circuit above now becomes
Ri ( t )
+
L di ( t ) dt
= v
S
( t )
• The transient part of the complete solution is determined by setting the forcing function v
S
(t) = 0
Ri ( t ) + L di ( t ) dt
= 0
Engr228 - Chapter 7, Nilsson 10E 16

Complete Solution
• The complete solution is solved by assuming i(t) is of the form: i(t) k
1
+ k
2 e
−
τ t
AC
R
Ri ( t ) + L di ( t ) dt
= u ( t ) -
+ u(t)
• As shown on the following pages through direct integration, substituting the solution above into the circuit equation yields
L i ( t ) =
V
R
1 − e
−
R
L t
+ i ( 0 ) e
−
R
L t
Solution – Direct Integration
For t > 0 (letting V = u )
Ri ( t ) + L di ( t )
= V dt
L
V di ( t ) dt
Ldi ( t )
= V
=
− Ri ( t )
− dt
Ri ( t )
V
Ldi ( t )
− Ri ( t )
L
−
R ln( V
=
dt
− Ri ( t )) = t + c
1
−
L
R ln( V − Ri ( t )) = t + ln(
V
V
−
−
Ri (
Ri t )
( t )) =
= e
−
R
L t
−
R
L t −
⋅ e
−
R
L c
1 c
1
R
L c
1
V
Ri
− Ri ( t )
( t ) = V
= e
−
R
L t
⋅ c
2
− c
2 e
−
R
L t i ( t ) =
V
R
− c
2
R e
−
R
L t
Engr228 - Chapter 7, Nilsson 10E 17

Direct Integration - continued i ( t ) =
V
R
− c
2
R e
−
R
L t
We can find c
2 from the initial condition i(0) i c
(
2
0 ) =
= V
V
R
− c
R
2
− i ( 0 ) R
⋅ 1
Therefore, we have i ( t ) =
V
R
−
V − i ( 0 ) R
R e
−
R
L t i ( t ) =
V
R
1 − e
−
R
L t
+ i ( 0 ) e
−
R
L t
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 k
1
+ k
2 e
−
τ t
• Find the time constant τ ;
• Solve for k
1 and k
2 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 10E 18

Example
The switch is in the position shown for a long time before t = 0.
Find i(t).
t=0 i(t)
R=1 Ω
L=1H
1V 2V i ( t ) = k
1
+ k
2 e
−
τ t
Time constant τ = 1 sec
Example - continued
At t = 0, i(0) = 2 A i ( t )
= k
1
+ k
2 e
−
τ t
2
= k
1
+ k
2
At t = ∞, i( ∞ ) = 1 A 1 = k
1
+ 0
Therefore, k
1
= 1, k
2
= 1
The answer is: i ( t ) = 1 + e − t
Engr228 - Chapter 7, Nilsson 10E 19

2A i ( t ) = 2
Example Graph i ( t ) i ( t ) = 1 + e − t
1A
Example: RL Circuit with Step Input
Find i(t) i(t)=25+25(1-e -t/2 )u(t) A
Engr228 - Chapter 7, Nilsson 10E 20

Driven RC Circuits (Part 1 of 2)
Find V c
(t) and i(t) v
C
(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) v
C
=20 + 80e -t/1.2 V i=0.1 + 0.4e
− t/1.2
A
Engr228 - Chapter 7, Nilsson 10E 21

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 10E 22