In a series LCR circuit the voltage across the resistance, capacitance and inductance is 10V each. If the
capacitance is short circuited, the voltage across the inductance will be

10 V

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\frac{10}{\sqrt{2}} V

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\frac{10}{3} V

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20 V

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Solution

The correct option is B $\frac{10}{\sqrt{2}} V$
Total voltage in the circuit $V = V = I {[R + (X_{L} - X_{C})^{2}]}^{1 / 2}$
= IR = 10V
When capacitor is short circuited $I = \frac{10}{(R^{2} + X_{L}^{2})^{1} / 2} = \frac{10}{\sqrt{2} R} = \frac{10}{\sqrt{2} \times X_{L}}$
$V_{L} = L X_{L} V_{L} = \frac{10}{\sqrt{2}} V$

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