In the given circuit, a charge of +80 μ C is given to the upper plate of the 4 μ F capacitor. Then in the steady state, the charge on the upper plate of the 3 μ F capacitor is

Correct answer: C. +48 μ C The total charge on plate A will be 80 μ C . 2 μ F and 3 μ F capacitors are in parallel. Therefore, C_ e q =2+3=5 HF Charge on capacitor of 3 μ F capacitance q=(3)/(5) × 80=48 μ C

JEE Advanced 2012Paper 2

Exam: JEE Advanced
Year: 2012
Paper: Paper 2
Subject: Physics
Chapter: Capacitance

JEE Advanced 2012 · Paper 2 · Physics · Capacitance

In the given circuit, a charge of +80 μ C is given to the upper...

In the given circuit, a charge of

+80 \mu C

is given to the upper plate of the

4 \mu F

capacitor. Then in the steady state, the charge on the upper plate of the

3 \mu F

capacitor is

Options

A $+32 \mu \mathrm{C}$
B $+40 \mu \mathrm{C}$
C $+48 \mu \mathrm{C}$
D $+80 \mu \mathrm{C}$

Correct answer

C. $+48 \mu \mathrm{C}$

Step-by-step solution

The total charge on plate

A

will be

80 \mu \mathrm{C}

2 \mu F

and

3 \mu F

capacitors are in parallel. Therefore,

C_{e q}=2+3=5 \mathrm{HF}

Charge on capacitor of

3 \mu \mathrm{F}

capacitance

q=\frac{3}{5} \times 80=48 \mu C

In the given circuit, a charge of +80 μ C is given to the upper plate of the 4 μ F capacitor. Then in the steady state,

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