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Simple problem that I can't solve.

On this AC circuit, find the current read by ammeter \$ A_3 \$ (the one in series with the resistor) given that \$A_1\$ reads 4A and \$A_2\$ reads 3A, assuming every current is read by its RMS value.

Simple AC Circuit

There are no given values for resistor and capacitor, but the textbook answer gives that \$A_1\$ reads 2.65A Here's what I've tried :

Since we're in AC, I figured voltage lags in the capacitor, so current must lead by 90 degrees. I tried using Kirchhoff's law with phasors in mind to find \$I_3 = I_1 - I_2\$ which would give \$4e^{j0} - 3e^{j\frac\pi2} = 4 - 3j\$.

However, the magnitude of this phasor is \$5\$, which is greater than the magnitude of \$I_1\$, which is impossible, since there are no current source in branch 2.

What am I missing? Thank you

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2 Answers 2

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You are correct that the resistor current and capacitor current are out of phase by 90 degrees, but in your solution you assumed that the total current and the capacitor current are out of phase by 90 degrees.

The current through \$A_1\$ is the total current and the current through \$A_2\$ is 90 degrees out of phase with the resistor current. If you draw the phase diagram \$A_1\$ is the hypotenuse. Can you use geometry and take it from there?

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  • \$\begingroup\$ Thank you very much! This solved my problem and answered my question. \$\endgroup\$ Aug 31, 2018 at 17:55
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If A1 reads 4 amps and A2 reads 3 amps then you solve for A3 by using pythogorus because the R and C current phasors are at right angles to each other forming the "hypotenuse" as A1: -

$$A_3 = \sqrt{A_1^2-A_2^2}$$

enter image description here

This is 2.645 amps.

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  • \$\begingroup\$ Thank you for the answer! But using specifically phasors, shouldn't we also get the same result? I understand it by Pythagoras, but not with phasors... \$\endgroup\$ Aug 31, 2018 at 16:19
  • \$\begingroup\$ I have used phasors - pythagorus is all about angles and anyway, your question was "What am I missing?". \$\endgroup\$
    – Andy aka
    Aug 31, 2018 at 16:24
  • \$\begingroup\$ @KevinKruse well spotted dude. I shall fix shortly. \$\endgroup\$
    – Andy aka
    Aug 31, 2018 at 19:09

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