Superposition Theorem
2
Superposition Theorem
The superposition principle states that the voltage across (or current through) an element in a linear circuits is the algebraic sum of the voltage across (or current through) that element due to each independent source acting alone.
Current Source open circuit(0 A)
Voltage Source short circuit (0 V)
3
Superposition Theorem
Step to apply:
1. Turn off all independent sources except one source. Find the output (voltage or current) due to that active source.
2. Repeat step 1 for each other independent sources.
3. Find the total contribution by adding algebraically all the contribution due to the independent source.
j
R1V
e +
-L N
i
j
+
-V1
R1
i1
L N
i2
-L N
+
V2
eR1
21;21 iiiVVV
Example
21 vvv
Vv
VvVv
10
82;21
determine the branches current using superposition theorem.
Solution The application of the superposition theorem is shown in
Figure 1, where it is used to calculate the branch current. We begin by calculating the branch current caused by the voltage source of 120 V. By substituting the ideal current with open circuit, we deactivate the current source, as shown in Figure 2.
120 V 3
6
12 A4
2
i1i2
i3i4
Figure 1
To calculate the branch current, the node voltage across the 3Ω resistor must be known. Therefore
120 V 3
6
4
2
i'1 i'2i'3 i'4
v1
Figure 2
42
v
3
v
6
120v 111
= 0
where v1 = 30 V
The equations for the current in each branch,
6
30120 = 15 A
i'2 = 3
30= 10 A
i'3 = i'
4 =
6
30= 5 A
In order to calculate the current cause by the current source, we deactivate the ideal voltage source with a short circuit, as shown
3
6
12 A4
2
i1"
i2"
i3"
i4"
i'1 =
To determine the branch current, solve the node voltages across the 3Ω dan 4Ω resistors as shown in Figure 4
The two node voltages are
3
6
12 A4
2
v3v4
+
-
+
-
2634333 vvvv
124
v
2
vv 434
= 0
= 0
By solving these equations, we obtain
v3 = -12 V v4 = -24 V
Now we can find the branches current,
To find the actual current of the circuit, add the currents due to both the current and voltage source,
Superposition - Problem
2k1k
2k12V
I0
2mA
4mA
– +
2mA Source Contribution
2k1k
2k
I’0
2mA
I’0 = -4/3 mA
4mA Source Contribution
2k1k
2k
I’’0
4mA
I’’0 = 0
12V Source Contribution
2k1k
2k12V
I’’’0
– +
I’’’0 = -4 mA
Example
find v using superposition
one independent source at a time, dependent source remains
KCL: i = i1 + i2
Ohm's law: i = v1 / 1 = v1
KVL: 5 = i (1 + 1) + i2(2)
KVL: 5 = i(1 + 1) + i1(2) + 2v1
10 = i(4) + (i1+i2)(2) + 2v1
10 = v1(4) + v1(2) + 2v1
v1 = 10/8 V
Consider the other independent source
KCL: i = i1 + i2
KVL: i(1 + 1) + i2(2) + 5 = 0i2(2) + 5 = i1(2) + 2v2Ohm's law: i(1) = v2
v2(2) + i2(2) +5 = 0 => i2 = -(5+2v2)/2i2(2) + 5 = i1(2) + 2v2-2v2 = (i - i2)(2) + 2v2
-2v2 = [v2 + (5+2v2)/2](2) + 2v2-4v2 = 2v2 + 5 +2v2
-8v2 = 5 => v2 = - 5/8 V
from superposition: v = -5/8 + 10/8 v = 5/8 V
Example
Note that the voltage source and the current source have two different frequencies. Thus, if we want to use phasors, the only way we've solved sinusoidal steady-state problems, we MUST use superposition to solve this problem. We will consider each source acting alone, and then find v0(t) by superposition.
sin cos 90t t Remember that
+
-AC
0.2F 1H 2cos10t30sin 5t+
-
v0(t)
Consider first the acting alone.Since, ,we have = 5 and
Example
30sin 5 30cos 5 90t t 30sin 5t
+
-AC
0.2F 1H 2cos10t30sin 5t+
-
v0(t)
+
-AC
-j1 j530 90
+
-
O.C.
1 11
5(0.2)CZ jj C j
5LZ j L j
10V
Example
+
-AC
-j1 j530 90
+
-
O.C.
1 11
5(0.2)CZ jj C j
5LZ j L j
10V
AC
Z1
Z2VS V0
+
-
+
-
Use voltage division
1 20
1 2S
Z
Z Z
V V
2
( 1)( 5) 51.25
1 5 4
j jZ j
j j j
1 8Z
10
1.25 1.25 9030 90 30 90
8 1.25 8.097 8.881
j
j
V
10 4.631 171.1 V
10 ( ) 4.631cos 5 171.12 4.631sin 5 81.12v t t t
Example
For a parallel combination of Y's we have
YV I
1 8 2 0.1 0.125 1.90eq i j j j Y Y
20
2 01.05 86.24
1.904 86.24
V
20 ( ) 1.05cos 10 86.24v t t
Yeq
+
-
20V I
20
IV
Y
S
j2 -j/10 2 0 +
-
10(0.2) 2CY j C j j
1 110
10LY jj L j
20V
1.904 86.24eq Y
By superposition
Example
+
-AC
0.2F 1H 2cos10t30sin 5t+
-
v0(t)
10 ( ) 4.631sin 5 81.12v t t
20 ( ) 1.05cos 10 86.24v t t
0 ( ) 4.631sin 5 81.12 1.05cos 10 86.24v t t t
1 20 0 0( ) ( ) ( )v t v t v t
University End Examination Questions
Using super position theorem find the current in 2 ohms resistor. Verify the result with other method
Using super position theorem, find the current I through (4 + j3) impedance
Using super position theorem find Vab in the figure