Calculating Vo in an AC Circuit Using Superposition Theorem

Michael_0039 · Sep 24, 2019

This is the circuit:

AC circuit schematic with a 6 ∠0° A current source and 2 Ω resistor R1, a 12 ∠0° V voltage source, a capacitor of -j1 Ω, an inductor j2 Ω, a 2 Ω resistor R2, and a 1 Ω resistor R3, with mesh currents I1 and I2 and output voltage Vo across R3.

I try with KVL:

AC circuit schematic with three marked loops: a 6 0° A current source and 2 Ω resistor R1 on the left, a 12 0° V voltage source, a −j1 Ω capacitor, a j2 Ω inductor, a 2 Ω resistor R2 and a 1 Ω resistor R3 on the right, with output voltage Vo across the right terminals.

Handwritten mesh current analysis: Loop 1 gives 0 = (5 - j1)·I1 + (-j1)I1 + 3I2 (eq. 1), Loop 2 gives 12 + j6 = (j1)I1 - 3I2 (eq. 2), and Loop 3 gives j2 I1 - (3 + j2) I2 = 18 (eq. 3), followed by solving for I1 = (18 + (3 + j2)I2)/j2 and substituting into eq. 2.

Handwritten KVL mesh equations for an AC circuit: j2(12+j6) = (j18 + j1(3+j2))I2 - 3I2, simplified to j2(12+j6) = I2(-5+j22), then I2 = 1,243/13.17° A, with a question mark circled.

But the given answer is 6.66L33.67 Volts. So, I'm confused, any suggestions ?

alan123hk · Sep 24, 2019

May I ask a question ?
I assume that you wrote a zero on the left side of the first loop equation, but why ?

The Electrician · Sep 24, 2019

The 6 amp current source injects 6 amps into the node at the top of the 12 volt voltage source, and the 2 ohm resistor doesn't change that fact. Injecting 6 amps into a voltage source doesn't change the voltage of the 12 volt source. The net result of what I've said so far is that you can get rid of the current source and the resistor in series with it; those things have no effect on Vo. Now you have only two meshes to solve.

Michael_0039 · Sep 25, 2019

alan123hk said:

May I ask a question ?
I assume that you wrote a zero on the left side of the first loop equation, but why ?

Because in this loop there is no Voltage source. The same on loop 3.

Michael_0039 · Sep 25, 2019

Thanks for your answers!

I try with Superposition theorem, I think this process is correct:

Handwritten superposition working for an AC circuit: part 1 shows the voltage source shorted, part 2 shows the current source opened with loop equations for I1 and I2 and the output Vo.

Handwritten circuit analysis on paper: loop equations with complex-number currents I1 and I2, I2 = 6.66 at 33.7 degrees A, and Vo = Vo' + Vo'' = 6.66 at 33.7 degrees V, with a CamScanner watermark.

Calculating Vo in an AC Circuit Using Superposition Theorem

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