In the given circuit, a charge of $+80 \mu \mathrm{C}$ is given to the upper plate of the $4 \mu \mathrm{~F}$ capacitor. Then in the steady state, the charge on the upper plate of the $3 \mu \mathrm{~F}$ capacitor is

JEE Advanced · 2012 · 08 Apr 2012 · Paper 2 · Capacitance

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

(A) $+ 32 μ C$
(B) $+ 40 μ C$
(C) $+ 48 μ C$ ✓ Correct
(D) $+ 80 μ C$

Answer: (C)

Solution

In steady state, the voltage across the

2 μ F

and

3 μ F

capacitors must be equal. Let

q

be the charge on the

3 μ F

capacitor. Then the charge on the

2 μ F

capacitor is

80 - q

. Setting the voltages equal:

\frac{80 - q}{2} = \frac{q}{3}

. Solving:

3 (80 - q) = 2 q

, giving

240 = 5 q

, so

q = 48 μ C

JEE Advanced · 2012 · 08 Apr 2012 · Paper 2 · Single correct option

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