Electrical Engineering - Circuit Theorems in AC Analysis - Discussion

@Adam is exactly right.

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

Can someone tell me why Vsrc=6?

@Adam is pretty good.

@Kishan you are right.

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

= [0.781 mag, 38.65 degree]*6.

= 4.686 mag, 38.65 degrees.

Its always found out across RL.

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

@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.