Superposition Theorem Solved Problems

Superposition Theorem Solved Problems-59
The voltage drop across the 100Ω resistor is, (10.7m A)(100Ω)= 1.1V.The voltage drop across the 5KΩ resistor is, (1.78m A)(5KΩ)= 8.9V.Notice how the 1KΩ resistor and the 5KΩ resistor are in parallel.

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At the end, once we have analyzed the circuit from each power source separately (with that one power source in the circuit and the other removed), we add up all the currents and voltages in each part of the circuit and this will equal the total current and voltage (in that part of the circuit).

So the best way to see the superposition theorem in practice is do an example. So now we'll analyze the circuit below with 2 power sources (voltage sources).

So this is how a circuit is analyzed with the superposition.

The circuit is analyzed with one power source in it and the remaining voltage sources shorted out and the current sources replaced with an open circuit.

The voltage drop across the 100Ω resistor is, (4.5m A)(100Ω)= 0.45V.

You can know this by the fact that the 5KΩ resistor and the 100Ω resistor are in parallel.This 833Ω is now in series with the 100Ω resistor, which gives a total resistance of 933Ω.We can now figure out the total current produced by the power supply (I is, (10.7m A)(833Ω/1KΩ)= 8.9m A.With this we can now calculate the total current (I= 4.6m A(98Ω/100Ω)= 4.5m A. The voltage drop across the 1KΩ resistor is, (4.6m A)(1KΩ)= 4.6V.The voltage drop across the 5KΩ resistor is, (0.09m A)(5KΩ)= 0.45V.The voltage drop across the 1KΩ resistor is, (8.9m A)(1KΩ)= 8.9V.You could also tell the voltage across the 1KΩ resistor would be the same as across the 5KΩ resistor, since they are both in parallel with each other.Superposition allows us to do circuit analysis one circuit at a time. So, above, is how the circuit will look when the 5V power source is removed.So the above circuit is the circuit we will analyze using the superposition theorem. With one power source, it is now simpler to analyze.Adding 8.91m A and 1.78m A gives us the original 10.7m A.Now that we know the currents we can figure out the voltage with ohm's law, with the formula, voltage= current * resistance (V=IR).

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