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

Correct answer: C. 10 / 2 ~V aligned & As V_R=V_L=V_C & R=X_L=X_ C, Z=R & V=I R=10 volt aligned When capacitor is short circuited, Z= R^2+X_L^2 = R^2+R^2 =R 2 New current I^ =V / Z=V / R 2 = 10 R 2 Potential drop across inductance =I^ X_L=I^ R= 10 × R R 2 = 10 2 volt

AIIMS 2009

Exam: AIIMS
Year: 2009
Subject: Physics
Chapter: Alternating Current

AIIMS 2009 · Physics · Alternating Current

In a series L C R circuit, the voltage across the resistance,...

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

Options

A $10 \mathrm{~V}$
B $10 \sqrt{2} \mathrm{~V}$
C $10 / \sqrt{2} \mathrm{~V}$
D $20 \mathrm{~V}$

Correct answer

C. $10 / \sqrt{2} \mathrm{~V}$

Step-by-step solution

\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

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