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In a series $L C R$ 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
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The correct answer is:
$10 / \sqrt{2} \mathrm{~V}$
$\begin{aligned} & \text { As } V_R=V_L=V_C \\ & \mathrm{R}=X_L=X_{C,} \Rightarrow Z=R \\ & V=I R=10 \text { volt }\end{aligned}$
When capacitor is short circuited,
$Z=\sqrt{R^2+X_L^2}=\sqrt{R^2+R^2}=R \sqrt{2}$
New current $I^{\prime}=V / Z=V / R \sqrt{2}=\frac{10}{R \sqrt{2}}$
Potential drop across inductance
$=I^{\prime} X_L=I^{\prime} R=\frac{10 \times R}{R \sqrt{2}}=\frac{10}{\sqrt{2}}$ volt
When capacitor is short circuited,
$Z=\sqrt{R^2+X_L^2}=\sqrt{R^2+R^2}=R \sqrt{2}$
New current $I^{\prime}=V / Z=V / R \sqrt{2}=\frac{10}{R \sqrt{2}}$
Potential drop across inductance
$=I^{\prime} X_L=I^{\prime} R=\frac{10 \times R}{R \sqrt{2}}=\frac{10}{\sqrt{2}}$ volt
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