Blog #33 - Calculating Current Using the Superposition Theorem

Calculate the value of the current through resistor R3 in this dc network using the superposition theorem. The superposition theorem states: In any linear network containing more than one source of electromotive force (emf) or current, the current through any branch is the algebraic sum of the currents produced by each source acting independently.

Because voltage source EB has no internal resistance, the source EB is replaced by a short circuit. The total resistance seen at EA…we will call it RTA = 100 + (100 || 100)…written out or expanded looks like this…100 + (100)(100)/(100 + 100) = 150 Ω

The total current flowing out of EA according to Ohm’s Law is ITA and…equal to 30/150…or 200 mA. From the current-divider rule, I3A = 200 mA/2 = 100 mA & I2A = 200 mA/2 = 100 mA.

Returning to our original circuit let's look at the contribution from Voltage source EB.

Because EA has no internal resistance, the EA source is replaced by a short circuit. 

As you can see the circuit is symmetrical, so the resistance that the B power supply sees…call it, RTB…the total resistance at EB is also equal to 100 + (100 || 100)… which is 150 Ω therefore, ITB = EB/RTB = 15/150 = 100 mA. (keep in mind that the B power supply polarity is the opposite to the A power supply therefore the current will be flowing out in this direction). 

From the current-divider rule, I2B & I3B = 100 mA/2 = 50 mA each.

The algebraic sum of the component currents I3A and I3B is used to obtain the true magnitude and direction of the current through R3, which is IR3 = I3A - I3B = 100 - 50 = 50 mA (in the direction of I3A).

The superposition theorem simplifies the analysis of a linear network having more than one source of emf. This theorem may also be applied in any network containing dc or ac sources of emf.

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