Four identical capacitors are connected with a battery of voltage V and two switches k 1 and k 2 as shown in the figure below. Initially, k 1 is closed, now if k 2 is also closed, find the heat loss.

Step-by-step solution

When k 1 is closed Q   =   C V energy U i = 1 2 CV 2 When k 2 is also closed The equivalent circuit: C eq = C + 2 C 3 = 5 C 3 Energy U f = 1 2 × 5 C 3 V 2 = 5 6 CV 2 charge supplied by battery after closing k 2 = 5 3 CV - CV = 2 3 CV the energy supplied by the battery = U f - U i + Δ H ⇒   2 3 C V 2 = 5 6 CV 2 - 1 2 CV 2 + Δ H ∴    Δ H = 1 3 C V 2