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Question

A circuit of resistance 5 Ω and inductance 0.6 H is in series with a capacitance of 10 μF. If a voltage of 200 V is applied and the frequency is adjusted to resonance. The current in the circuit is I0 and voltage across the inductor and capacitor are V0, and V1 respectively. We have

A
I0=40 A
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B
V0=9.8 kV
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C
V1=9.8 kV
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D
V1=19.6 kV
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Solution

The correct options are
A I0=40 A
B V0=9.8 kV
C V1=9.8 kV

We know that
I0=VR2+(XLXC)2
Since it is given that circuit is at resosnance
XL=XCωL=1ωCω=1LCω=1036

Since XL=XC
We get
I0=VR=2005=40 A

Now for voltage across inductor(V0)

V0=i0XL
And we have XL=ωL=1036×0.6
V0=40×0.6×1036V0=9.8 kV

For voltage across capacitor (V1)
V1=i0XC
We know that XL=XC
V1=9.8 kV

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