Chapter 5

Handy Circuit
Analysis
Techniques

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1 reproduction or display.
a linear circuit element has a linear voltage-
current relationship:
 if i(t) produces v(t), then Ki(t) produces Kv(t)
 if i1(t) produces v1(t) and i2(t) produces v2(t), then
i1(t) + i2(t) produces v1(t) + v2(t),
 resistors, sources are linear elements1
 a linear circuit is one with only linear
elements
Dependent sources need linear control equations to be linear elements.
1

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For the circuit shown, the solution can be expressed as:
0.7 0.2 v1  ia  Question: How much of
     v1 is due to source A,
0.2 1.2 v 2  ib  and how much is
because of source B?

We use the
superposition principle
to answer.

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If we define A as
0.7 0.2 
A   
0.2 1.2 
then

v1  1 ia  1 0  1 ia 

  A   A   A   Superposition:
v 2  ib  ib  0  the response is the sum
of experiments A and B.
Experiment B Experiment A
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In a linear network, the voltage across or the current through any element
may be calculated by adding algebraically all the individual voltages or
currents caused by the separate independent sources acting “alone”, i.e. with
 all other independent voltage sources replaced by short circuits and
 all other independent current sources replaced by open circuits.

Answer: i3(t) = 1.333 sin t V

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 Leave one source ON and turn all other
sources OFF:
 voltage sources: set v=0.
These become short circuits.
 current sources: set i=0.
These become open circuits.
Find the response from this source.
 Add the resulting responses
to find the total response.
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Use superposition to solve for the current ix

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First, turn the current source off:

3
ix   0.2
69

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Then, turn the voltage source off:
6
ix  (2)  0.8
69

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Finally, combine the results:

ix  ix  ix  0.2  0.8  1.0

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Determine the maximum positive current to
which the source Ix can be set before any
resistor exceeds its power rating.

Solution on next slide

Answer: Ix<42.49 mA

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Maximum current magnitude in 100Ω resistoris 0.25 /100 = 50m A
M axim um currentm agnitude in 64Ω resistoris 0.25 /64 = 62.5m A
Currentfrom voltage source alone is6/164 = 36.6m A flow ing clockw ise
64
Currentin 100Ω from Ix alone is Ix flow ing to the left.
164
64 64
Therefore 0.0366 − Ix < 0.05 or −0.05 < 0.0366 − Ix < 0.05
164 164
221.9m A > Ix > −34.33m A
100
Currentin 64Ω from Ix is Ix flow ing dow nw ard.
164
100 100
Therefore 0.0366 + Ix < 0.0625 or −0.0625 < 0.0366 + Ix < 0.0625
164 164
−0.1625 < Ix < 0.04247
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 When applying superposition to circuits with
dependent sources, these dependent sources
are never “turned off.”

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 current source off

voltage source off

ix = ix’+ix’’=2 + (−0.6) = 1.4 A

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 Idealvoltage sources: a first approximation
model for a battery.
 Why do real batteries have a current limit and
experience voltage drop as current increases?
 Two car battery models:

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For the car battery example:
VL = 12 – 0.01 IL

inc
r ea
se
R This line
L

represents all
possible RL

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The source has an internal resistance or output
resistance, which is modeled as Rs

short circuit current (when RL=0)

open circuit voltage (when RL=∞)

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The source has an internal parallel resistance
which is modeled as Rp

short circuit current (when RL=0)

open circuit voltage (when RL=∞)

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The sources are equivalent if
Rs=Rp and vs=isRs

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 The circuits (a) and (b) are
equivalent at the terminals.
 If given circuit (a), but
circuit (b) is more
convenient, switch them!
 This process is called
source transformation.

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We can find the current I in the circuit below
using source transformation, as shown.

I = (45-3)/(5+4.7+3) = 3.307 mA

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Thévenin’s theorem: a linear network can be
replaced by its Thévenin equivalent circuit, as
shown below:

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 We can repeatedly
apply source
transformation on
network A to find its
Thévenin equivalent
circuit.
 This method has
limitations- not all
circuits can be source
transformed.
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 Disconnect the load.
 Find the open circuit voltage voc
 Find the equivalent resistance Req of the
network with all independent sources turned
off.

Then:
VTH=voc and
RTH=Req
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Norton’s theorem: a linear network can be
replaced by its Norton equivalent circuit, as
shown below:

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 Replace the load with a short circuit.
 Find the short circuit current isc
 Find the equivalent resistance Req of the
network with all independent sources turned
off.

Then:
IN=isc and RN=Req
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 TheThévenin and Norton equivalents are
source transformations of each other!

RTH=RN =Req and vTH=iNReq

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Find the Thévenin and Norton equivalents for the network
faced by the 1-kΩ resistor.
Answer: next slide

the
load
resistor

this is the circuit we will simplify

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Thévenin Norton

Source
Transformation

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One method to find the Thévenin equivalent of
a circuit with a dependent source: find VTH and
IN and solve for RTH=VTH / IN
Example:

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Finding the ratio VTH / IN fails when both
quantities are zero!
Solution: apply a test source.

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 v test v test  (1.5i)
 1
2 3
i  1

Solve: vtest =0.6 V, and so RTH = 0.6 Ω



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What load resistor will allow the practical
source to deliver the maximum power to the
load?

Answer: RL=Rs

[Solve dpL/dRL=0.]

[Or: pL=i(vs-iRs), set dpL/di=0 to find imax=vs/2Rs. Hence RL=Rs]

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 The following resistors form a Δ:

 The following resistors form a Y:

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this Δ is equivalent to the Y if this Y is equivalent to the Δ if

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 How do we find the equivalent resistance of
the following network? Convert a Δ to a Y

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 use the Δ to Y
equations

use standard serial and parallel

combinations
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