Electronics - RLC Circuits and Resonance - Discussion
Discussion Forum : RLC Circuits and Resonance - General Questions (Q.No. 30)
30.
What is the current through the capacitor in the given circuit?
Discussion:
5 comments Page 1 of 1.
Simran said:
9 months ago
@All.
I am not getting this solution. Anyone, please explain to me.
I am not getting this solution. Anyone, please explain to me.
Afash said:
1 decade ago
Very simple:
Xc = 1/2*pi*f*C.
So,
Xc = 1000000/2*Pi*47*1000*0.001.
So you will get 3386.275.
Then I = 12/3386.275.
I=3.543mA.
Xc = 1/2*pi*f*C.
So,
Xc = 1000000/2*Pi*47*1000*0.001.
So you will get 3386.275.
Then I = 12/3386.275.
I=3.543mA.
Om Prakash Bharti said:
1 decade ago
Xc = 1/wc = 1/2*pi*f*C.
= 1/2*3.14*47000*0.001*10^-6.
= 3386.27.
= 1/2*3.14*47000*0.001*10^-6.
= 3386.27.
Tommy said:
1 decade ago
How do you get Xc = 3386.27. Thanks.
Gaurav said:
1 decade ago
Vc=I*Xc
In parallel circuit Voltage is same at all the points.
V=12V, Xc=3386.27. I=12/3386.27
I=3.53mA
In parallel circuit Voltage is same at all the points.
V=12V, Xc=3386.27. I=12/3386.27
I=3.53mA
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