Superposition Theorem

Superposition Theorem

The Superposition theorem is a basic circuit analysis technique that is only useful where there are more than one voltage or current sources in the same circuit. It states that:

The current in any element is equal to the sum of currents through it produced by each source acting independently while the other sources are replaced by their internal resistance.

It is a basic technique that allows a circuit to be analysed using just one of the voltage/current sources at a time, Then summing the results by superimposing them on top of each other (hence the name). Generally such analysis is relatively straight forward and it is enough to use Ohms law and series/parallel circuit calculations to solve them.

Example 1

This circuit has two sources. A current source of 5A, with an internal resistance of 0.5Ω and a voltage source of 9V, with an internal resistance of 0.1Ω. So the steps to analyse this circuit will be:

1. Find the current through RL based on IS1 being the only source, with VS1 represented only by it’s internal resistance.
2. Find the current through RL based on VS1 being the only source, with IS1 represented only by it’s internal resistance.
3. Sum the results to find the total current through the load resistor RL.
4. Use Ohms law to calculate the voltage across RL based on the current calculated in step 3.

Step 1 – Find I_RL contributed by IS₁

This is just a simple current divider calculation based on R₁, R₂ and R_L

\(R_{Total} = \frac{1}{ \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3}} = \frac{1}{ \frac{1}{0.5} + \frac{1}{0.1} + \frac{1}{1000}} = 0.0833\Omega \\\) \(I_{RL} = IS1 . \frac{R_{Total}}{R_L} = 5 * \frac{0.0833}{1000} = 0.0004166A \approx 417\mu A\)

Step 2 – Find I_RL contributed by VS₁

After redrawing the circuit to emphasise the connectivity of the resistor network we can see that this is only marginally more complicated than before. We can solve this by first finding the current through R₂, then the fraction of that current through R_L due to the current divider network of R₁ || R_L.

\(R_{Parallel} = \frac{R_1.R_L}{R_1 + R_L} = \frac{0.5 * 1000}{0.5 + 1000} = \frac{500}{1000.5} \approx 0.5\Omega \\\) \(I_{R2} = \frac{VS_1}{R_2 + R_{Parallel}} = \frac{9}{0.1 + 0.5} = 15A\\\) \(I_{RL} = I_{R2} . \frac{R_{Parallel}}{R_L} = 15 * \frac{0.5}{1000} = 0.0075A \approx 7.5mA\)

Step 3 – Sum results to Find IRL_Total

\(IRL_{Total} = IRL_{IS1} + IRL_{VS1} = 0.000417 + 0.0075 = 0.007917A \approx 7.9mA\)

Step 4 – Calculate V_RL

\(V_{RL} = IRL_{Total} . RL = 0.0079 * 1000 = 7.9V\)

Summarising

I_RL = 7.9mA and V_RL = 7.9V

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