RLC Circuits and Resonance

Q1:
What is the current through the capacitor in the given circuit?

A 3.5 mA

B 5.5 mA

C 9 mA

D 11.4 mA

ANS:A - 3.5 mA

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.

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