Survival Skills for
Circuit Analysis
What you need to know from ECE 109
Phyllis R. Nelson
prnelson@csupomona.edu
Professor, Department of Electrical and Computer Engineering
California State Polytechnic University, Pomona
P. R. Nelson
Fall 2010
WhatToKnow —- – p. 1/46
Basic Circuit
Concepts
All circuits can be analyzed by many
methods.
P. R. Nelson
Fall 2010
WhatToKnow —- – p. 2/46
Basic Circuit
Concepts
All circuits can be analyzed by many
methods.
Some methods have specific advantages for
calculating a particular result in a specific
circuit.
P. R. Nelson
Fall 2010
WhatToKnow —- – p. 2/46
Basic Circuit
Concepts
All circuits can be analyzed by many
methods.
Some methods have specific advantages for
calculating a particular result in a specific
circuit.
Some methods yield useful intuition about
circuit behavior.
P. R. Nelson
Fall 2010
WhatToKnow —- – p. 2/46
Basic Circuit
Concepts
All circuits can be analyzed by many
methods.
Some methods have specific advantages for
calculating a particular result in a specific
circuit.
Some methods yield useful intuition about
circuit behavior.
No single analysis method will be the
best for all possible circuits!
P. R. Nelson
Fall 2010
WhatToKnow —- – p. 2/46
Kirchhoff’s Laws
These two methods are the most general tools of
circuit analysis.
Sign errors in writing Kirchhoff’s laws are the
most common error I observe in student work in
higher-level classes.
Sign errors are not trivial errors!
P. R. Nelson
Fall 2010
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Kirchhoff’s current law
is equivalent to stating that electrical charge
cannot be stored, created, or destroyed locally in
a conductor.
The algebraic sum of all currents
entering a node is zero.
Alternate statements:
The algebraic sum of the currents leaving a
node is zero.
The total current entering a node is equal to
the total current leaving the node.
P. R. Nelson
Fall 2010
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Kirchhoff’s voltage law
is derived by considering the forces on an
electrical charge that travels around a closed
loop in a circuit and has the same starting and
ending velocity. The total force on the charge
must be zero, so the total change in electrical
potential energy (voltage) around the loop must
be zero.
The algebraic sum of the voltage
drops around any closed loop in a
circuit is zero.
P. R. Nelson
Fall 2010
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Solving K*L equations
Kirchhoff’s laws applied directly result in
X
Ii = 0 (KCL)
i
X
Vi = 0 (KVL)
i
P. R. Nelson
Fall 2010
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Solving K*L . . .
Solving KCL means finding the node voltages.
Solving KVL means finding the branch currents.
⇒ need relationships between current
and voltage for each circuit element.
These relationships describe the operation of the
circuit elements.
P. R. Nelson
Fall 2010
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Ohm’s law
+ V −
I
V = IR
Ohm’s law is an example of a current-voltage
(I-V ) relation.
To use K*L, you must know the I-V relation
for every circuit element.
I-V relations include definition of the
relationship between the direction of the
current and the sign of the voltage.
P. R. Nelson
Fall 2010
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Series resistors,
voltage divider
I
+
R1
VT
R2
+V − +V −
1
2
−
Ri
Rn
+V −
i
+V −
n
VT = I Req
Prove that these equations are true:
Req =
n
X
j=1
Rj
Ri
Vi = Pn
j=1 Rj
P. R. Nelson
VT
Fall 2010
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VT =
n
X
Vj =
j=1
n
X
I Rj = I
n
X
Rj = I Req
j=1
j=1
⇒ Req =
n
X
Rj
j=1
Vi = I Ri =
VT
Req
Ri =
Ri
Pn
j=1 Rj
P. R. Nelson
Fall 2010
!
VT
WhatToKnow —- – p. 10/46
Parallel resistors,
current divider
I
+
V
−
R1
I1 R2
I2
V = I Req
Prove that these equations are true:
−1
Req
=
2
X
i=1
−1
Ri
R1 R2
Req =
R1 + R2
Ix =
Ry
Rx + Ry
I
Which can be extended to more resistors?
P. R. Nelson
Fall 2010
WhatToKnow —- – p. 11/46
V
V
I = I1 + I2 =
+
=V
R1 R2
2
X
Ri−1
i=1
!
−1
=
⇒ Req
V
=
Req
2
X
Ri−1
i=1
1
1
R1 R2
Req =
+
=
R1 R2
R1 + R2
P. R. Nelson
Fall 2010
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Let x = 1 and y = 2. Then
R1 R2
R1 +R2
V
I Req
R2
I1 =
=
=
I=
I
R1
R1
R1
R1 + R2
If x = 2 and y = 1 then
R1
I
I2 =
R1 + R2
P. R. Nelson
Fall 2010
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For n resistors in parallel,
n
X
n
X
V
=V
I=
Ij =
R
j
j=1
j=1
n
X
!
V
=
Req
−1
Req
n
X
Rj−1
j=1
⇒
=
−1
Rj
j=1
This is the only formula for parallel resistors that
extends easily to more than two resistors.
P. R. Nelson
Fall 2010
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Illegal circuits
The following configurations are not allowed:
a short-circuited voltage source
an open-circuited current source
Why?
P. R. Nelson
Fall 2010
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KCL Example
R1
+
Vx −
R2
R3
Iy
R4
Find VR2 and VR4 .
P. R. Nelson
Fall 2010
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KCL Example
R1
+
Vx −
R2
R3
Iy
R4
Find VR2 and VR4 .
Hint: Since KCL is applied at nodes, how many
KCL equations will be required?
P. R. Nelson
Fall 2010
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KCL Example - 1
R1
+
Vx −
R2
R3
Iy
R4
Find VR2 and VR4 .
P. R. Nelson
Fall 2010
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KCL Example - 1
R1
+
Vx −
R2
R3
Iy
R4
Find VR2 and VR4 .
1. Since KCL is applied at nodes, count the
nodes.
P. R. Nelson
Fall 2010
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KCL Example - 1
R1
+
Vx −
R2
R3
Iy
R4
Find VR2 and VR4 .
1. Since KCL is applied at nodes, count the
nodes. (4)
P. R. Nelson
Fall 2010
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KCL Example - 2, 3
R1
+
Vx −
R2
R3
Iy
R4
Find VR2 and VR4 .
P. R. Nelson
Fall 2010
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KCL Example - 2, 3
R1
+
Vx −
R2
R3
Iy
R4
Find VR2 and VR4 .
2. Choose one node as ground. Choose for
convenience!
3. Label node voltages.
P. R. Nelson
Fall 2010
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R1 V2 R3 V4
+
Vx −
R2
Iy
R4
P. R. Nelson
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R1 V2 R3 V4
+
Vx −
R2
Iy
R4
Both VR2 and VR4 are connected to ground.
The voltage source is connected to ground
⇒ one less KCL equation.
P. R. Nelson
Fall 2010
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KCL Example - 4
R1 V2 R3 V4
+
Vx −
R2
Iy
R4
2 unknown node voltages ⇒ 2 KCL equations.
4. Define currents for all elements connected to
nodes “2” and “4”.
P. R. Nelson
Fall 2010
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KCL Example - Eqn’s
R1 V2 R3 V4
+
Vx −
I1
R2
I3
I2 R4
I4
Iy
KCL at 2:
P. R. Nelson
Fall 2010
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KCL Example - Eqn’s
R1 V2 R3 V4
+
Vx −
I1
R2
I3
I2 R4
I4
Iy
KCL at 2:
I1 − I2 − I3 = 0
P. R. Nelson
Fall 2010
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KCL Example - Eqn’s
R1 V2 R3 V4
+
Vx −
I1
R2
I3
I2 R4
I4
Iy
KCL at 2:
I1 − I2 − I3 = 0
KCL at 4:
P. R. Nelson
Fall 2010
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KCL Example - Eqn’s
R1 V2 R3 V4
+
Vx −
I1
R2
I3
I2 R4
I4
Iy
KCL at 2:
I1 − I2 − I3 = 0
KCL at 4:
I3 − I4 − Iy = 0
P. R. Nelson
Fall 2010
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KCL Example - I-V’s
R1 V2 R3 V4
+
Vx −
I1
R2
I3
I2 R4
I4
Iy
Write unknown I’s in terms of node V’s:
P. R. Nelson
Fall 2010
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KCL Example - I-V’s
R1 V2 R3 V4
+
Vx −
I1
R2
I3
I2 R4
I4
Iy
Write unknown I’s in terms of node V’s:
Vx − V2
I1 =
R1
V2 − V4
I3 =
R3
V2
I2 =
R2
V4
I4 =
R4
P. R. Nelson
Fall 2010
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KCL Example Substitution
Vx − V2
V2
V2 − V4
−
−
= 0
R1
R2
R3
V4
V2 − V4
−
− Iy = 0
R3
R4
Check! 2 equations in 2 unknowns
This pair of equations can be solved, but it’s
messy unless resistor values are given.
P. R. Nelson
Fall 2010
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KVL Example
R1
+
VA −
R2
I3
R3
IB
R4
Find I3 .
P. R. Nelson
Fall 2010
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KVL Example
R1
+
VA −
R2
I3
R3
IB
R4
Find I3 .
Hint: Since KVL is applied around closed loops,
how many KVL equations will be required?
P. R. Nelson
Fall 2010
WhatToKnow —- – p. 24/46
KVL Example - 1
R1
+
VA −
R2
I3
R3
IB
R4
Find I3 .
P. R. Nelson
Fall 2010
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KVL Example - 1
R1
+
VA −
R2
I3
R3
IB
R4
Find I3 .
1. Count the number of “windowpanes” in the
circuit.
P. R. Nelson
Fall 2010
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KVL Example - 1
R1
+
VA −
R2
I3
R3
IB
R4
Find I3 .
1. Count the number of “windowpanes” in the
circuit. (3)
P. R. Nelson
Fall 2010
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KVL Example - 2, 3
R1
+
VA −
R2
I3
R3
IB
R4
Find I3 .
P. R. Nelson
Fall 2010
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KVL Example - 2, 3
R1
+
VA −
R2
I3
R3
IB
R4
Find I3 .
2. Choose one closed loop for each
windowpane. Choose for convenience!
3. Choose a direction for each loop.
P. R. Nelson
Fall 2010
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KVL Example
R1
+
VA −
R2
I3
R3
IB
R4
Only one loop passes through R3 , ensuring that
the unknown can be used as a “loop current.”
P. R. Nelson
Fall 2010
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KVL Example - 4, 5
R1
+
VA −
R2
I3
R3
IB
R4
4. Define loop current labels for each loop. Be
careful not to give different currents the same
label!
5. Define voltages for all branch elements.
P. R. Nelson
Fall 2010
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KVL Example - I & V
R1
+
VA −
I1
R2
R1
VA
R3
I3
IB
I4
R4
+
IB V4
−
R4
R3
+ V1 − + V3 −
+
R2 V2
−
−
P. R. Nelson
Fall 2010
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KVL Example - Eqn’s
KVL for loop 1:
P. R. Nelson
Fall 2010
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KVL Example - Eqn’s
KVL for loop 1:
−VA + V1 + V2 = 0
P. R. Nelson
Fall 2010
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KVL Example - Eqn’s
KVL for loop 1:
−VA + V1 + V2 = 0
KVL for loop 3:
P. R. Nelson
Fall 2010
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KVL Example - Eqn’s
KVL for loop 1:
−VA + V1 + V2 = 0
KVL for loop 3:
−V2 + V3 + V4 = 0
P. R. Nelson
Fall 2010
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KVL Example - Eqn’s
KVL for loop 1:
−VA + V1 + V2 = 0
KVL for loop 3:
−V2 + V3 + V4 = 0
KVL for loop 4: ??!?
P. R. Nelson
Fall 2010
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KVL Example - Eqn’s
KVL for loop 1:
−VA + V1 + V2 = 0
KVL for loop 3:
−V2 + V3 + V4 = 0
KVL for loop 4: ??!?
−V4 + V4 = 0 is a tautology . . .
P. R. Nelson
Fall 2010
WhatToKnow —- – p. 30/46
KVL Example - Eqn’s
KVL for loop 1:
−VA + V1 + V2 = 0
KVL for loop 3:
−V2 + V3 + V4 = 0
KVL for loop 4: ??!?
−V4 + V4 = 0 is a tautology . . .
. . . but the current source current is
IB = I4 − I3
P. R. Nelson
Fall 2010
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KVL Example - I-V’s
R1
VA
+ V1 −
+
R2
−
I1
R3
+
+ V3 −
IBV4
V2
I3
I4 −
−
R4
Write unknown V’s in terms of loop I’s:
P. R. Nelson
Fall 2010
WhatToKnow —- – p. 31/46
KVL Example - I-V’s
R1
VA
+ V1 −
+
R2
−
I1
R3
+
+ V3 −
IBV4
V2
I3
I4 −
−
R4
Write unknown V’s in terms of loop I’s:
V1 = I1 R1
V3 = I3 R3
V2 = (I1 − I3 ) R2
V4 = I4 R4
P. R. Nelson
Fall 2010
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KVL Example Substitution
−VA + R1 I1 + (I1 − I3 ) R2 = 0
− (I1 − I3 ) R2 + I3 R3 + I4 R4 = 0
I4 − I3 = IB
Check! 3 equations in 3 unknowns
Collect terms, then solve this set of
equations by Kramer’s rule.
P. R. Nelson
Fall 2010
WhatToKnow —- – p. 32/46
KVL Example - Matrix

   
R1 + R2
−R2
0
I1
VA

   
R2 + R3 R4  I3  =  0 
 −R2
0
−1
1
I4
IB


R1 + R2
−R2
0


∆ =  −R2
R2 + R3 R4 
0
−1
1
= (R1 + R2 ) (R2 + R3 ) + (R1 + R2 ) R4 − R22
= R1 R2 + R1 R3 + R1 R4 + R2 R3 + R2 R4
P. R. Nelson
Fall 2010
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

R1 + R2 VA 0


∆ I3 =  −R2
0 R4 
0
IB 1
= R2 VA − (R1 + R2 ) R4 VA
I3
=
R2 VA − (R1 + R2 ) R4 IB
R1 R2 + R1 R3 + R1 R4 + R2 R3 + R2 R4
P. R. Nelson
Fall 2010
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Equivalent circuits
Equivalent circuits enable us to treat a portion of
a circuit like a “black box” in that we can give it
the simplest possible model.
There are two possible models that account for
both power sources and impedance:
a voltage source in series with a resistor
a current source in parallel with a resistor
Equivalent circuits can be used to
simplify a circuit before solving
for an unknown!
P. R. Nelson
Fall 2010
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Thèvenin equivalent
Req
Voc
+
−
I
+
V
−
V = Voc − Req I
Voc
V
I=
−
Req Req
If the terminals are connected to an open circuit,
I = 0 and V = Voc .
If the terminals are shorted, V = 0 and
I = Voc /Req = Isc .
P. R. Nelson
Fall 2010
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Norton equivalent
I
Isc
Req
+
V
−
V
I = Isc −
Req
V = Req Isc − Req I
If the terminals are shorted, V = 0 and I = Isc .
If the terminals are connected to an open circuit,
I = 0 and V = Req Isc .
P. R. Nelson
Fall 2010
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Graphical
representation
All possible combinations of pairs of voltage and
current values for either equivalent circuit model
can be represented as the line through the points
(V = 0, ISC ) and (VOC , 0).
I
ISC
V
VOC
P. R. Nelson
Fall 2010
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Superposition
A voltage or current in a linear circuit is the
superposition (sum) of the results for each
source alone.
Superposition can be used along with parallel
and series resistors, voltage and current dividers,
and Thèvenin and Norton equivalent circuits to
simplify the analysis of a circuit.
Superposition often gives solutions
in a mathematical form that enhances
intuition.
P. R. Nelson
Fall 2010
WhatToKnow —- – p. 39/46
Superposition
Example
R1
+
Vx −
R2
Iy
I2 R3
Find the current I2 using superposition.
Solve for the current I2 with
Iy = 0 so that Vx is the only source
Vx = 0 so that Iy is the only source
then add the results.
P. R. Nelson
Fall 2010
WhatToKnow —- – p. 40/46
Superposition
Example - V
Set Iy = 0 and solve for the current I2V .
R1
+
Vx −
R2
I2V R3
P. R. Nelson
Fall 2010
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Superposition
Example - V
Set Iy = 0 and solve for the current I2V .
R1
+
Vx −
I2V =
R2
I2V R3
R3
R1 R2 + R1 R3 + R2 R3
P. R. Nelson
Vx
Fall 2010
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Superposition
Example - I, Result
Set Vx = 0
and solve for
the current I2I .
R1
R2
Iy
I2I R3
P. R. Nelson
Fall 2010
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Superposition
Example - I, Result
R1
Set Vx = 0
and solve for
the current I2I .
I2I = −
R2
Iy
I2I R3
R1 R3
R1 R2 + R1 R3 + R2 R3
P. R. Nelson
Iy
Fall 2010
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Superposition
Example - I, Result
R1
Set Vx = 0
and solve for
the current I2I .
I2I = −
I2 =
R2
R1 R3
R1 R2 + R1 R3 + R2 R3
R3
R1 R2 + R1 R3 + R2 R3
Iy
I2I R3
Iy
(Vx − R1 Iy )
P. R. Nelson
Fall 2010
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Dependent Sources
Circuits containing dependent sources are solved
by the same methods used for other circuits.
There is one extra step, because the final result
cannot contain the dependent variable!
Not eliminating the dependent
variable from the final result is
another of the most common
errors I see in student work!
P. R. Nelson
Fall 2010
WhatToKnow —- – p. 43/46
Dependent Source
Example
R1
+
Vx −
R2
BV2
+−
+
+
V2 R3 V3
−
−
Find V3 .
P. R. Nelson
Fall 2010
WhatToKnow —- – p. 44/46
Dependent Source
Example
R1
+
Vx −
R2
BV2
+−
+
+
V2 R3 V3
−
−
Find V3 .
Hint: The answer may only contain
Vx , R1 , R2 , R3 , and B.
P. R. Nelson
Fall 2010
WhatToKnow —- – p. 44/46
Dependent Source
Example - Eqn’s
R1
+
Vx −
R2
BV2
+−
+
+
V2 R3 V3
−
−
Find V3 .
P. R. Nelson
Fall 2010
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Dependent Source
Example - Eqn’s
R1
+
Vx −
R2
BV2
+−
+
+
V2 R3 V3
−
−
Find V3 .
V2 − Vx V2
V3
+
+
=0
R1
R2 R3
P. R. Nelson
Fall 2010
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Dependent Source
Example - Eqn’s
R1
+
Vx −
R2
BV2
+−
+
+
V2 R3 V3
−
−
Find V3 .
V2 − Vx V2
V3
+
+
=0
R1
R2 R3
−V2 + B V2 + V3 = 0
P. R. Nelson
Fall 2010
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Dependent Source
Example - Result
R1
+
Vx −
R2
BV2
+−
+
+
V2 R3 V3
−
−
Find V3 .
V3 =
Vx
R1
h
R1 +R2
(1−B)R1 R2
+
1
R3
i
P. R. Nelson
Fall 2010
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