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.
• Know how to analyze circuits with sequential switching.
Characteristics of R, L, C
• Resistors resist current flow.
• Inductors resist change of current.
• Capacitors resist change of voltage.
L and C are “dynamic” elements with respect to 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.
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
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
Transient Response
• RL Circuit
• RC Circuit
• RLC Circuit
First-order differential equation
Chapter 7
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
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.
Substitute i (t ) = Ae st in the equation
R st
Ase + Ae = 0
L
R
( s + ) Ae st = 0
L
st
s=−
R
L
i (t ) = Ae
R
− t
L
di (t ) R
+ i (t ) = 0
dt
L
Solving Source Free RL Circuits - continued
Initial condition: i (0) = I 0
from
i (t ) = Ae
R
− t
L
I 0 = Ae 0
I0 = A
Therefore
i (t ) = I 0 e
R
− t
L
Solving Source Free RL Circuits
Method 2: Direct integration
t
di (t ) R
+ i (t ) = 0
dt
L
di (t )
R
= − i (t )
dt
L
di (t )
R
= − dt
i (t )
L
i (t )
∫
I0
t
di (t )
R
= ∫ − dt
i (t ) 0 L
i (t )
ln i (t ) I
0
R
=− t
L 0
R
ln i (t ) − ln I 0 = − (t − 0)
L
i (t ) = I 0 e
R
− t
L
Time Constant
The ratio L/R is called the “time constant” and is denoted
by the symbol τ.
L
τ=
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
− t
L
= I 0e
e −1 = 0.368
−
t
τ
Time Constant Graph
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 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)
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.
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).
iL = 4.5e
−10 6
t
4
mA
iL (5µs) = 1.289 mA
iSW = 9 – iL (5µs) = 9 – 1.289 = 7.711 mA
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.
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:
iL (t ) = I L 0e
R
− t
L
v C (t ) = VC 0 e
−
= I L 0e
t
RC
• Time constants:
L
τL =
R
τ C = RC
−
t
τ
= VC 0 e
−
t
τ
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)
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 (t ) = vC (0)e −125t
vC(0) = 192V
vC = 0.1vC(0) when t = 18.42mS
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
v(1.5ms) = 4.5V
v(3ms) = 1.0V
i(0+) = 0.1A
i(1.5ms) = 0A
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
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.
Step Functions
v(t)
v(t)
1V
1V
0V
Unit Step function
0V
t
t
Example
I = 1A
I = 2A
Suppose the voltage source changes abruptly from 1V to 2V.
Does the current change abruptly as well?
Voltage
Example - continued
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
Complete Solution
• Solve by assuming i(t) is of the form
k1 + k 2 e
−
t
τ
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
R
R
− t
− t
V
i (t ) = 1 − e L  + i (0)e L
R

Solution – Direct Integration
For t > 0 (letting V = u)
di (t )
Ri(t ) + L
=V
dt
di(t )
L
= V − Ri (t )
dt
Ldi (t )
= dt
V − Ri(t )
Ldi (t )
∫ V − Ri(t ) = ∫ dt
L
− ln(V − Ri(t )) = t + c1
R
L
− ln(V − Ri (t )) = t + c1
R
R
R
ln(V − Ri(t )) = − t − c1
L
L
V − Ri(t ) = e
V − Ri(t ) = e
R
− t
L
R
− t
L
Ri(t ) = V − c2e
⋅e
R
− c1
L
⋅ c2
R
− t
L
R
V c2 − L t
i (t ) = − e
R R
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
e
Therefore, we have i (t ) = −
R
R
− t
− t
V
L
i (t ) = 1 − e  + i (0)e L
R

R
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
k1 + k 2 e
−
t
τ
• Find the time constant τ (may use Thevenin’s if necessary).
• Solve for k1, 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.
Driven RL Circuits
V0
− Rt / L
i( t ) =
1
−
e
u(t)
(
)
R
Example: RL Circuit with Step Input
Find i(t)
i(t)=25+25(1-e-t/2)u(t) A
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
Example
The switch is in the position shown for a long time before t = 0.
Find i(t).
i (t ) = k1 + k 2 e
−
t
τ
Time constant τ = 1 sec
Example - continued
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
t
τ
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
Example - continued
Find k1, k2 using i(0) = 0, i(∞) = 0.25
At t = 0, i(0) = 0 A
0 = k1 + k 2
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
iL (t ) = k1 + k 2 e
−
t
τ
Example Graph
v1 (t )
v1 (t ) = 0
v1 (t ) = 0.25 − 0.25e −2t
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(∞) = 5A
(
)
iL (t ) = 5 1 − e −40t − 15e −40t amps
RC Circuit Response
Chapter 7 Summary
• Showed how to determine the natural response of both RL
and RC circuits.
• Showed hot to determine the step response of both RL and
RC circuits.
• Showed how to analyze circuits with sequential switching.