Electrical Engineering - Circuit Theorems in AC Analysis - Discussion
Discussion Forum : Circuit Theorems in AC Analysis - General Questions (Q.No. 20)
20.
Referring to the given circuit, what is VTH if VS = 12
0° V?



Discussion:
13 comments Page 1 of 2.
BASAVARAJ K T said:
1 decade ago
Any one can explain this problem?
NExIaN said:
1 decade ago
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.
Sowjanya said:
1 decade ago
Answer said by @Nexian is not clear. I understand the procedure. But I can't get the answer. So, give me complete solution.
Chintan said:
1 decade ago
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.
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.
Adam said:
1 decade ago
@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
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
Jkl said:
1 decade ago
Here they didn't given to find Vth across which resistance?
Steve said:
1 decade ago
Its always found out across RL.
Kishan said:
10 years ago
Vout = [15k/(15k-j12k)]*Vsrc.
= [0.781 mag, 38.65 degree]*6.
= 4.686 mag, 38.65 degrees.
= [0.781 mag, 38.65 degree]*6.
= 4.686 mag, 38.65 degrees.
Stuti kushwaha said:
9 years ago
@Kishan you are right.
Kevin said:
9 years ago
@Adam is pretty good.
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