Superposition

To solve a circuit by superposition first "turn off" all independent voltage and current sources except for one and calculate the voltages and currents you want to find in your circuit. Then repeat this for each voltage or current source, having only one source at a time. To get the final voltage or current at a point in your circuit simply add all of the voltages/currents you found at that point from each source.

To "turn off" a voltage source replace it with a straight piece of wire (short circuit). To "turn off" a current source replace it with a break in the wire (open circuit).

Find the current through R2 in the circuit below.

First we'll find the current due to the voltage source V1 by replacing V2 with a short circuit. With V2 shorted R2 and R3 are in parallel and their combination is in series with R1 so the total current flowing out of V1 is: \(I_{V1} = \frac{28}{400 + \frac{20000}{300}} = \frac{28}{400 + 66.67} = 0.060A\) Then by the current divider rule the current flowing through R3 because of V1 is: \(I_{R3V1} = I_{V1} \times \frac{R2}{R2 + R3} = 0.06 \times \frac{100}{300} = 0.02A\) Now we do the same but for V2 so we replace V1 with a short circuit and the total current leaving battery V2 is: \(I_{V2} = \frac{7}{100 + \frac{80000}{600}} = \frac{7}{100 + 133.33} = 0.03A\) Then by the current divider rule the amount of current flowing through R3 because of V2 is: \(I_{R3V2} = I_{V2} \times \frac{R1}{R1 + R3} = 0.03 \times \frac{400}{600} = 0.02A\) Therefore the total current flowing through R3 is given by the sum of these two currents, or: \(I_{R3} = I_{R3V1} + I_{R3V2} = 0.02A + 0.02A = 40mA\)

Superposition only works with voltage and current, it does not work with power. To find the power through an element due to several power sources find the individual voltages and/or currents like normal and then use the final, total voltage and/or current to calculate the power.

Calculate the power dissipated by resistor R3 in the circuit below.

This is the same circuit as we analysed above. To find the current used up by R3 we use its total current \(I_{R3} = 0.04A\) and its resistance \(R_3 = 200\Omega\) to find: \(P_{R3} = I_{R3}^2 \times R_3 = 0.0016 \times 200 = 320mW\) Note that we used superposition to find the current and then our usual power formula to THEN get the power.