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Question

A 100 volt AC source of frequency 500 Hz is connected to a L–C-R circuit with L=8.1 mH, C=12.5 µF and R=10 Ω , all connected in series. The potential difference across the resistance is -

A
400 V
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B
300 V
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C
200 V
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D
100 V
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Solution

The correct option is D 100 V
Given: V=100 volt, f=500 Hz, L=8.1 mH, C=12.5 µF, R=10 Ω

The impedance of L-C circuit is,

Z=R2+(XLXc)2

Here XL=2πfl=2×3.14×500×(8.1×103)=25.4 Ω

and Xc=12πfC=12×3.14×500×12.5×106=25.4 Ω

R=10 Ω

Z=(10)2+(25.425.4)2=10 Ω

Now, irms=ErmsZ=10010=10 A

VR=irms×R

VR=10×10=100 V

Hence, option A is correct.

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