In the given circuit diagram, initially the switch S is opened. Find heat dissipated in the circuit (in mJ) after switch S in closed. It is given C=2μF.

When switch S is opened the potential difference across the group of capacitors is zero. Hence the initial charge in both the capacitors will be zero.When the switch S is closed, potential difference across both of them is 30 V. Charge on each of them is q=(2μF)(30V)=60μCThe heat developed in the circuit H=Energy released by 60V battery -Energy absorbed by both 30 V battery +Energy stored in both capacitors=1800μJ=1.8mJ