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A choke coil of resistance 5 Omega and i...

A choke coil of resistance `5 Omega` and inductance 0.6 H is in series with a capacitance of `10 (mu)F`. If a voltage of 200 V is applied and the frequency is adjusted to resonance, the current and voltage across the inductance and capacitance are `(I_0), (V_0) and (V_1)` respectively. we have

A

`(I_0)=40A`

B

`(V_0)=9.8 kV`

C

`(V_1)=9.8 kV`

D

`(V_1)=19.6 kV`

Text Solution

Verified by Experts

The correct Answer is:
A, B, C
the frequency
`f=(1)/(2 pi) sqrt(1)/(sqrt(LC))=(1)/(2 pi) sqrt((1)/(0.6 xx10 xx10^(-6)))=65 Hz`
The current at resonance is
`(I_0)=(200 v)/(5 Omega) =40 A`
the quantity `(V_0)-(I_0) sqrt(R^(2)+(omega L)^(2))`
`=40 sqrt(25^(2) + (0.6 xx 2 pi xx 65)^(2))=9.8 kV`
`V_(1)=(I_0)/(omega C) =(40 xx 10^(5))/(2 pi xx 65) = 9.8 kV`
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