In the circuit shown in fig. the cell is ideal with emf 15V. Each resistance is of `3Omega.` The potential difference across the capacitoro in steady state is

Options

A. 0
B. 9 V
C. 12 V
D. 15 V

Answer

C

Explanation

As the capacitors are identical, each of them finally have
chargel Q//2.
Initial energy of the system in `E_(1) = (Q^(2))/(2C)`
Final energy of the system is `E_(f)=2[((Q//2)^(2))/(2C)]=(Q^(2))/(4C)`
Heat produced = loss in energy `=E_(i) - E_(f) = (Q^(2))/(4C)`