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Question

# In a series LCR circuit, the voltage across the resistor, capacitor and inductor is 10 V each. If the capacitor is short-circuited, the voltage across the inductor will be ​​​​​​​[1 Mark]

A
10 V
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B
102 V
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C
(102) V
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D
20 V
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Solution

## The correct option is C (10√2) VAs per the given condition, R=XL=XC ∵ voltage across them is same. Total voltage in the circuit, V=I[R2+(XL−XC)2]12=IR=10 V When the capacitor is short-circuited, I′=10(R2+X2L)12=10√2R ∴ Potential drop across inductor = I′ XL=I′ R=10√2 V

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