Electrical Engineering - Circuit Theorems in AC Analysis - Discussion

@Chintan, you are wrong. @Nexian is pretty much correct.

How to do it correctly:

1) Vth should be across RL, so you remove it and want to measure the voltage.

2) Realize that the inductor and voltage source are in parallel. So we can ignore the inductor for this problem.

3) The problem is simplified to the Capacitive Reactance and the 15k Resistor.

4) Use the voltage divider formula (thanks @Nexian!).

Vout = [15k / (15k - j12k)] * Vsrc

= [0.781 mag, 38.65 degree] *Vsrc

= 9.37 mag, 38.65 degrees

Nexian is not correct.

Here Vth is the voltage across R = 15 ohm.

Now consider Voltage source as short circuit. So Inductive reactance will be zero as its in parallel with the Short circuit line.

Now capacitor = 12k and R = 49k are in parallel. After solving them, Z = 9.639k. And angle will be easily determined.

Take inductive reactance as +12 k and capacitive reactance as -12 k irrespective of its polarity, and by using simple voltage divider formula, you can easily find voltage drop across R, which is 9.38<38.7 degree V.

Answer said by @Nexian is not clear. I understand the procedure. But I can't get the answer. So, give me complete solution.

Vout = [15k/(15k-j12k)]*Vsrc.

= [0.781 mag, 38.65 degree]*6.

= 4.686 mag, 38.65 degrees.

The problem states that Vsrc = 12. So you are correct @Adam.

Here they didn't given to find Vth across which resistance?

Any one can explain this problem?

Its always found out across RL.

Can someone tell me why Vsrc=6?