A series circuit consists of two capacitors, a resistors, and an ideal voltage source. The circuit is initially at steady state. Now a conducting wire is shorted across the capacitor C as shown by dotted line. After shorting the wire, select the correct statement// statements.

Options

A. Heat developed in the circuit is `2 CV^2//3`
B. heat developed in the circuit is `CV^2`
C. work done by battery is `4CV^2//3`
D. total potential energy change in both capacitors is `2CV^2//3`

Answer

A, B, D

Explanation

a., b., d.
Before shorting the wire at steady
state the charges on each capacitors
will be equal to

`q_0 = q_1 = q_2 = (2C)/3 V`
After shorting the capacitor marked
(2) will be useless in circuit. Hence
`q'_1 = 2CV = q_f and q'_2 = 0`
The charge supplied by battery
after shorting

`Deltaq = q_(f) - q_0 = 2CV - (2CV)/3 = (4CV)/3`
Hence work done by battery `W = Deltaq V = (4CV^2)/3`
Initial potential energy of system.
`V_0 = 1/2 (2/3C) V^2 = 1/3 CV^2`
Final potential energy of system `U_(f) = 1/2 (2C)V^2 = CV^2`
Hence change in PE `DeltaU = U_(f) - U_(i)`
or `DeltaU = CV^2 - 1/3 CV^2 = 2/3 CV^2`
Work done by battery `W = DeltaU + Heat`
Heat developed `H = W-DeltaU`
or `H = 4/3CV^2 - 2/3 CV^2 = 2/3CV^2` .