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Linear Circuit Analysis
Linear Circuits
Superposition
Linear Circuits
• Most circuits we will study are linear
• Linear circuits contain linear elements – those
that have a linear relationship between their
voltage and their current
– Resistors
– Voltage and Current Sources
– Dependent sources that depend on a voltage or
current (but not if they depend on a product of
current and voltage or some current or voltage to
a power different than one)
Superposition
• Linear circuits may be analyzed by looking at
the voltages and currents produced by each
independent source when acting alone and
then adding them together
• When one source acts on a circuit, the other
independent sources must be set to zero
– A voltage source set to zero is equivalent to a
short circuit
– A current source set to zero is equivalent to an
open circuit
Sources when set to Zero
• Voltage Sources
+
Is
Ξ
Vs=0
-
• Current Sources
Short
Circuit
Ξ
Open
Circuit
Superposition
• Superposition allows us to reduce the
complexity of a circuit
• Superposition requires that more circuits be
analyzed (more simpler circuits)
• Superposition allows us to analyze circuits that
contain sources of different types
– DC, AC, different frequency AC sources, triangular
waves, square waves, etc.
Circuit Analysis using Linearity
• In a linear circuit with one independent
source, if you double the value of the source,
then every current and voltage in the circuit
doubles
• Therefore, you can guess at any voltage or
current in the circuit and see what the source
would have to be in order to get that current
and then apply a scale factor
Linear Circuit Analysis
• When a circuit has one independent source,
you can analyze it by choosing a current or
voltage somewhere in the circuit and then
determine what the source had to be in order
to get that value
• Then you can apply a scale factor to all the
assumed currents and voltages in the circuit to
get the actual values of the source and other
voltages and currents
Example
Let the 4 A current source act alone, by
zeroing out the voltage source (effectively
making it a short circuit.
Example
Short
Short
circuit
Circuit
Let the 4 A current source act alone, by
zeroing out the voltage source (effectively
making it a short circuit).
Example
Is
Short
Short
circuit
Circuit
Let the current source be Is, and choose a
convenient value for I4 of 10 A,
then VB = 25 volts and I3 = 6.25 A
Since I2=I4+I3, then I2 = 16.25 A
Example
Is
If I2 = 16.25 A, and VA-VB = I2∙2 Ω
Since VB=25 v, then VA=57.5 v
Then I1=VA/14Ω= 4.11 A
So Is = I1+ I2 = 20.36 A
Short
Short
circuit
Circuit
Example
Is
Short
Short
circuit
Circuit
But Is = 4 A, so if we multiply
everything by a scale factor of 4/20.36
we will get the correct voltages and currents.
So I4 = 10 A ∙ (4 / 20.36) = 1.96 A
Linear Circuit Analysis
• When a circuit has one independent source,
you can analyze it by choosing a current or
voltage somewhere in the circuit and then
determine what the source had to be in order
to get that value
• Then you can apply a scale factor to all the
assumed currents and voltages in the circuit to
get the actual values of the source and other
voltages and currents
Superposition Example
Is
Short
Short
circuit
Circuit
•We found if I4=10 A, Is=20.36 A, VB = 25 volts,
I3 = 6.25 A, I2 = 16.25 A, VA=57.5 v & I1= 4.11 A
•Knowing that Is = 4 A, we can multiply each
voltage and current by 4/20.36, so Is=4 A,
I4=1.96 A, VB = 4.91 v, I3 = 1.23 A, I2 = 3.19 A,
VA= 11.30 v & I1= .81 A
Example
Open
Circuit
To finish the analysis by superposition,
let the 34 v source act alone, by zeroing
out the current source (effectively
making it an open circuit.
Example
Open
Circuit
Vs
•Pretend that the source is Vs and choose
a convenient value for such as I1 = -I2 = 1 A.
•So VA = 14 volts and VB = 16 volts and
I3 = 4 A and I4 = I2 - I3 = -5 A
Example
Open
Circuit
Vs
•If I4 = -5 A, then VB-Vs = I4∙2.5 Ω = -12.5 v
But VB= 16 v, so Vs = 28.5 volts
•Vs is really 34 v, so our scale factor is
34/28.5, so I4 = -5 A∙(34/28.5) = -5.96 A
Example
•If I4 = 1.96 A when the 4 A source was acting
alone and if I4 = -5.96 A when the 34 volt
source was acting alone,
•Then I4 = 1.96 - 5.96 = -4.00 A when both
sources are working together
Summary
• Linear circuits allow us to analyze the
response to each independent source and add
them together (superposition)
– Especially valuable for different types of sources
• Linear circuit analysis (where a current or
voltage is assumed in order to get a scale
factor to get the true voltages and currents)
can be used for any circuit with a single
independent source