JEE Advanced · 2013 · 02 Jun 2013 · Paper 1 · Capacitance

In the circuit shown in the figure, there are two parallel plate capacitors each of capacitance

C

. The switch

S_{1}

is pressed first to fully charge the capacitor

C_{1}

and then released. The switch

S_{2}

is then pressed to charge the capacitor

C_{2}

. After some time,

S_{2}

is released and then

S_{3}

is pressed. After some time,

(A) the charge on the upper plate of

C_{1}

2 C V_{0}

.
(B) the charge on the upper plate of

C_{1}

C V_{0}

.
(C) the charge on the upper plate of

C_{2}

is 0.
(D) the charge on the upper plate of

C_{2}

- C V_{0}

Answer: (B), (D)

Solution

Step 1: When

S_{1}

is closed,

C_{1}

charges to

2 V_{0}

, so charge on

C_{1} = 2 C V_{0}

.

Step 2: When

S_{2}

is pressed (

S_{1}

open),

C_{1}

and

C_{2}

share charge:

V_{co mm o n} = \frac{2 C V _{0}}{C + C} = V_{0}

Charge on

C_{1} = C V_{0}

(upper plate positive)

Charge on

C_{2} = C V_{0}

(upper plate positive)

Step 3: When

S_{3}

is pressed (

S_{2}

open),

C_{2}

is connected in reverse polarity. The charge on

C_{2}

remains

C V_{0}

but with upper plate negative.

Therefore:

Hence the correct answers are (B) and (D).

JEE Advanced · 2013 · 02 Jun 2013 · Paper 1 · One or more correct options