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In a series LCR circuit, the voltage across the resistance, capacitance and inductance is $10 \mathrm{~V}$ each. If the capacitance is short circuited, the voltage across the inductance will be

(a) $10 \mathrm{~V}$

(b) $\frac{10}{\sqrt{2}} \mathrm{~V}$

(c) $10 \sqrt{2} \mathrm{~V}$

(d) $20 \mathrm{~V}$

The correct option of this question will be (b).

Solution —

As $V_{R}=V_{L}=V_{C}=10 V$

$\therefore \quad R=X_{L}=X_{C}$ and $Z=R$

and $V=I R=10 V$

When the capacitor is short circuited, the impedance of the circuit is

$Z^{\prime}=\sqrt{R^{2}+X_{L}^{2}}=\sqrt{R^{2}+R^{2}}=\sqrt{2} R$

and the current in the circuit is

$I^{\prime}=\frac{V}{Z^{\prime}}=\frac{10 V }{\sqrt{2 R}}$

$\therefore$ The voltage across the inductance is

$V_{L}^{\prime}=I^{\prime} X_{L}=\left(\frac{10 V }{\sqrt{2} R}\right) R=\frac{10}{\sqrt{2}} V$

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