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