KCL for AC Waveforms: Can I Get the Right Answer?”

jaus tail · Jan 23, 2018

Homework Statement

Circuit schematic with a 5 ∠ 0° A current source on the left and a 10 ∠ 60° A current source on the right, a 3 Ω resistor and a branch labelled -j₂ Ω in the middle, with current i₁(t) marked. Options (A) to (D) are listed below.

[/B]

Homework Equations

Can I use KCL at the central node?

The Attempt at a Solution

I tried KCL and got: 5cos0 + j 5sin0 + 10 cos 60 + j 10 sin 60 = (-I₁)
5 + 5 + j 10 *1.732 / 2 = (-I₁)
10 + j 5 * 1.732 = -I₁
This gives 13.33 angle (220.89)

Book answer is A
Is the 5 and 10 Ampere in figure, RMS or Maximum?

cnh1995 · Jan 23, 2018

Instantaneous values have no phase. So in that sense, the options seem meaningless.

jaus tail said:

Is the 5 and 10 Ampere in figure, RMS or Maximum?

In India, I guess they are taken as peak quantities unless specified otherwise.
I agree with your calculations.

Maybe they have given instantaneous values of the magnitudes and phases of the current sources (in an incorrect way) and they want you to find the magnitude and phase of I1 at that instant.

KCL for AC Waveforms: Can I Get the Right Answer?”

Homework Statement

Homework Equations

The Attempt at a Solution

Discussion

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