Simple phasor problem with AC

Question

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.

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

Answer 1

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?

Answer 2

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: -

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

This is 2.645 amps.