Initially `K_(1)` is closed, now if `K_(2)` is also closed, find heat dissipated in the resistances of connecting wires

Options

A. `1/2 CV^(2)`
B. `2/3 CV^(2)`
C. `1/3 CV^(2)`
D. `1/4 CV^(2)`

Answer

C

Explanation

When `K_(1)` is closed upper circuit is connected to source

Q = CV
Energy `U_(1) = 1/2 CV^(2)`
When `K_2` is also closed complete circuit is connected to source.

Therefore, the Equivalent circuit

`C_("equi") = C + (2C)/3 = (5C)/3`
`U_(f) = 1/2 (5C)/3 V^(2) = (5C)/6 V^(2)`
Charge supplied by battery after closing `K_2` is
`Q = 5/3 CV - CV = 2/3 CV`
Energy supplied by battery `= U_(f) - U_(i) + H`
`2/3 CV^(2) = 5/6 CV^(2) -1/2 CV^(2) + H`
`H = 1/3 CV^(2)`