Superposition theorem

[1], [2], [3], [4] The superposition theorem (in electrical engineering) says all branches' total voltage/current equals independent source's voltages, if other sources are off/deactivated--voltage sources are replaced by short circuits, current sources by open circuits.

It finds a load's voltage:

Step 1: Short out 1 voltage source at a time and calculate voltage per load and its direction. Open current sources if there are.
Step 2: Combine all loads' voltages: Some are series aid, some oppose.

Both superposition or mesh/loop analysis can find the voltage and current of each loads if there 1+ voltage source.

Replace voltage sources by shorts.

Replace current sources by opens.

Superposition theorem

Example 1

[4] Consider the 2 sources, 2-resistor circuit.

Find VR2.

Turn off 1 source--no matter the order: We chose to turn off the current source first which is like replacing IS by an open circuit. The result is a simple voltage divider and we can calculate the partial result due to the voltage source.

partial: incomplete

2. Go back to the initial circuit and turn off the other source--set VS to zero, the same as replacing it by a short circuit.

Shorting VS makes R1 and R2 in parallel.

V''R2 = Is (R1||R2)

= (1A)(5Ω||10Ω) = 3.33 V

3. The answer is the sum (superposition) of the 2 partial answers:

VR2 = V'R2 + V''R2 = 3.33 V + 3.33 V = 6.66 V--its the same to values from all of the other methods to analyze the same circuit.

Notes

[4] Warnings:

Superposition only works for linear circuits. (which have only sources/resistors/capacitors/inductors/linear amplifiers, etc.) Most electronic devices (diodes, transistors) are non-linea--not for superposition.
As this relies on linearity, don't add powers directly by superposition. (Power goes as V2 or I2 aren't linear.) Superposition finds total current/voltage then calculate power from that result.
When finding the partial answers, ensure to keep the same voltage polarity and current direction per case. E.g. A source may induce the current in a particular resistor to flow in 1 direction while another source causes to a current flowing in the opposite direction. You must keep the proper signs when adding to the partial results to obtain the correct total result.

Example 2

Find V60Ω.

This circuit has 2 battery, first start by 8 V then 12 V source:

V60Ω due to 8 V source

RT = 50||60 + 40 = 67.27 Ω
Branch resistance: 50||60 = 27.17 Ω
Using VDR formula: V60 = (27.17/67.27)x 8 = 3.231 V and direction

IT = 8/67.27 = 0.1189
0.1189(50||60) = 3.231 V

V60Ω due to 12 V source

RT = 40||60 + 50 = 74 Ω
Branch resistance: 40||60 = 24 Ω
V60 = (24/74)x 12 = 3.892 V and direction

V60 Ω load due to both sources:

3.231 + 3.892 = 7.123 V

Example 3

Find V220Ω and its direction.

V220Ω due to 24 V source

RT = [(240+130)||220] + 160 = 297.97 Ω
Branch resistance = (240+130)||220 = 137.97
(137.97/297.97)x 24 = 11.11

IT = 24/297.97 = 0.0805 A
V220 = 137.97x0.0805 = 11.11 V

V220Ω due to 6 V source

RT = [(130+240)||160] + 220 = 331.7
V220 (220/331.7) x 6 = 3.98 V

IT = 6/331.7 = 0.018 A
V220 = 0.018x220 = 3.96 V

V220Ω due to 8 V source

RT = 130+240+ (220||160) = 462.63
Branch resistance = 220||160 = 92.63
V220 = (92.63/462.63) x 8 = 1.6 V

V220 due to all sources: 11.11 - 3.98 + 1.6 = 8.73 V

Example 4

[4] Find the VR1.

1. Turn off VS2. The result is a voltage divider between R1 and the parallel combination of R2 and R3.

2. Now turn off VS1. Again, the result is a simple voltage divider, but now R1 is in parallel with R3. More crucial, note the polarity of--it'll be V′′ negative for this partial solution.