Electronic – Need help solving a problem with ac resonance circuits from the uni book

accircuit analysisresonance

schematic

simulate this circuit – Schematic created using CircuitLab

Okay I have struggling with this problem for some time and I cant figure out how to solve it.

The circuit is in resonance!

I tried solving it like this:
P = (I1^2)*R1 + (I3^2)R2 = R(I1^2 + I3^2)

R = (P)/(I1^2 + I3^2)

Then I drew phasor diagram and found I3 like this:

I – I3 = sqrt(I1^2 – I2^2)

I3 = I – sqrt(I1^2 – I2^2)

Then I plugged it into the first equation and found R and then I calculated the voltage like this:

U = I3 * R

and then the rest was easy but I got the answers different than in the book.
The solutions in the book are C = 50uF and L = 2.5mL

Anyway I dont know which part of my reasoning is wrong. I suspect that its this part : P = (I1^2)*R1 + (I3^2)R2 = R(I1^2 + I3^2)
or the way I drew the phasor diagram:
![diagram](https://imgur.com/HZrbWWN)

So any help would be greatly appreciated.

Best Answer

The trick is to find the current taken by the combined impedances of L1 || (C1 + R1). You say it's in resonance but the 3 amps through C1 compared to the 2 amps in L1 clearly means that "resonance" must mean a zero phase angle of total current and therefore a zero phase angle of current is also taken by L1||(C1 + R1).

It can't mean that \$F = \dfrac{1}{2\pi\sqrt{LC}}\$ for instance.

We can calculate the phase angle of the current through C1 and R1 to be \$\arcsin(2/3)\$ = 41.8 degrees. The phasor diagram in your question appears to be alluding to that.

enter image description here

It then follows that the joint current taken by the inductor and the capacitive/resistive path is \$3\cdot cos(41.8)\$ = 2.24 amps. This is in phase with the supply and will also be in phase with the current through resistor R2.

This means that the current through R2 (I3) must be: -

I3 = 7.24 - 2.24 amps = 5 amps (conveniently).

This is I3 in your diagram and you know that I2 is 3 amps. You also know the total power is 100 watts hence: -

$$R = \dfrac{100}{5^2 + 3^2} = 2.941 \space\Omega$$

You can now calculate the line voltage given that you know R3 = 2.941 ohms and that 5 amps flows through it. Line voltage = 14.71 volts.

Given this voltage, you know the reactance of the inductor is 14.71/2. An inductor of reactance 7.353 ohms at 5 kHz is an inductor of 234 uH.

I'm going to stop here because your question states that the inductance is 2.5 "mL" (I assume you mean mH) and my value is more than ten times lower. What value did you calculate?