Solving Networks with Multiple Sources
Using Superposition Method
VS1
+
R1
R2
ϕ1
R3
Zeroing a voltage source
ϕ2
+
R4
A
A
B
B
IS2
VS1
VS = 0
-
ϕ0
Consider the circuit solved earlier using a nodal analysis.
The solution was:
ϕ1 = 9.09 V; ϕ2 = 10.91 V
If the voltage of the voltage source VS = 0, then the voltage VAB = 0.
The potentials of the point A and point B are equal.
Compare to the wire connecting points A and B:
The superposition technique solves the circuit by
TURNING ON ONE source at a time and ZEROING all the others.
Zeroing a current source source
No matter what current is flowing through the wire, VAB = 0.
The voltage source with ZERO voltage is equivalent to a wire (shortcircuit) connecting its terminals.
Step 1. Consider the circuit with the current source Is zeroed
VS1
A
R1
R2
ϕ1
ϕ2
A
+
R3
R4
+
IS2
VS1
IS = 0
-
ϕ0
B
B
The circuit can be solved by series – parallel technique.
If the current of the current source IS = 0, then
the current between the point A and point B is zero.
R24 = R2 +R4 = 12 Ohm
No matter what the voltage VAB is, the current between A and B is zero.
Req1 = R1 + R324 = 1 + 3.53 = 4.53 Ohm
The current source with ZERO current is equivalent to a break
(an open circuit) between its terminals.
Total partial current IT1 = Vs1/Req1 = 10/4.53 = 2.2 A
R324 = R3*R24/(R3+R24) = 5*12/17 = 3.53 Ohm
The partial potential ϕ11 = VS1 – IT1*R1 = 10 V – 2.2A*1 Ohm = 7.8 V
1
Step 2. Consider the circuit with the voltage source Vs zeroed
VS1
R1
+
R2
ϕ1
R3
ϕ2
R4
+
Step 2. Consider the circuit with the voltage source Vs zeroed
(cont.)
R1
R2
VS1
ϕ1
ϕ2
+
IS2
VS1
R3
R4
+
IS2
VS1
-
-
-
-
ϕ0
ϕ0
Total partial current IT2 = IS2;
The circuit can be solved by series – parallel technique.
The partial potential ϕ22 = IS2 *Req2 = 2 A*2.2 Ohm = 4.4 V;
R13 = R1 *R3/(R1+R3) = 1*5/6 = 0.83 Ohm
The partial current through R4, IR42 = ϕ22/R4 = 4.4 V/10 Ohm = 0.44 A
R213 = R2 + R13 = 2 + 0.83 = 2.83 Ohm;
The partial current through R2: IR22 = IS2 – IR42 = 2 A – 0.44 A = 1.56 A
Req2 = R4*R213/(R4+R213) =10*2.83/12.83 = 2.2 Ohm;
Total partial current IT2 = IS2;
The partial potential ϕ22 = IS2 *Req2 = 2 A*2.2 Ohm = 4.4 V;
Voltage across R2: V22 = IR22*R2 = 1.56*2 = 3.12 V;
Partial potential ϕ12 = ϕ22 – V22 = 4.4 – 3.12 =1.28 V;
Total potential ϕ1 = ϕ11 + ϕ12 = 7.8 V +1.28 V = 9.08 V
Linearity Property
For any circuit, if the voltage (or current) of one the voltage (or
current) source changes k times, all the currents and all the voltages
in the circuit due to this source change k-times
Example 1
Use linearity to find the actual value of Io in the circuit shown below.
Assume I0 = 1 A;
V1 = I0*(3 Ohm+5 Ohm) = 8 V;
I1 = 8V/4 Ohm = 2 A
I2 = I1 + I0 = 3 A;
V2 – V1 = I2* 2 Ohm = 6 V; V2 = V1 +6 V = 14 V;
I3 = V2/7 Ohm = 14/7 = 2 A;
I4 = I2 + I3 = 3 A+ 2 A = 5 A
The actual current I4 = IS = 15 A, is 3 times higher.
Hence the assumed I0 value needs to be corrected by a factor of 3, hence
the actual value is I0 = 3 A
2
Example 2
Solve the circuit using source transformation
VS1
R1
+
R2
ϕ1
Transform the current source into a voltage source
ϕ2
R3
+
+
R4
VTH
IS2
a
RTH
-
VS1
b
-
ϕ0
Consider the circuit solved earlier using a nodal analysis.
Equivalent Thevenin voltage, VTH = VOC = 2 A * 10 Ohm = 20 V.
The solution was:
Equivalent series resistance is the resistance at the current source terminals with
the current source set to zero, RTH = 10 Ohm.
ϕ1 = 9.09 V; ϕ2 = 10.91 V
Modified circuit after current source transformation
VS1
R1
R2
ϕ1
Example 3
Replace the circuit attached to a-b with the voltage source
a
+
1
+
RTH
R3
VS1
10 Ω
-
20 V
IV
S2 TH
-
b
His circuit has a single unknown potential ϕ1, for which the KCL equation is:
ϕ1 (1/R1 +1/R3 + 1/R2TH) = VS1/R1 +VTH/R2TH
ϕ1 *1.28 = 11.67; ϕ1 = 9.09 V
IS2
-
ϕ0
Where R2TH = R2+RTH = 12 Ω
VS1
First, find the open-circuit voltage between terminals a and b:
VTH = VAB (open-circuit)
Applying nodal analysis to the node “1”:
G11=1/5+1/20 = 0.25; (there is no current through the 4-Ohm
resistor when the circuit is open);
Is=25/5+3 = 8 A
V1oc=G^(-1)*Is = (1/0.25)*8 = 32;
VTH = Vab(open circuit) = 32 V
3
Example (cont.)
Example (cont.)
1
1
+
VTH
-
VS1
VS1
a
RTH
IS2
IS2
Second, find the resistance between a and b terminals when all the
sources are set to zero.
Rab = 5//20 +4 = 4 + 4 = 8 Ohm
The circuit on the left can be replaced with the Thevenin equivalent voltage
source having the equivalent open-circuit voltage VTH = 32 V and
equivalent resistance RTH = 8 Ohm
RTH = 8 Ohm
If the load resistance RL connected to the terminals a and b needs to be
optimized for the maximum power dissipated in it, then RLopt = RTH.
In the above circuit, the optimal load resistance,
RLopt = 8 Ohm
b
Example 4:
convert voltage source into current source
+
VS
9V
RS 3 Ω
IN
RN
-
1) find the output resistance with the voltage source zeroed:
Rout = RS = 3 Ω
2) find the short-circuit current between the terminals:
ISC = VS/RS = 9 V/ 3 Ω = 3 Α
Equivalent (Norton) current source parameters:
IN = ISC = VS/RS = 3A
RN = ROUT = RS = 3 Ω
4