Question

Two capacitor $C_{1}and{C}_2$ are connected across a 60 V battery as shown in figure 1. In this situation, the final charges appearing on $C_{1}and{C}_2$ are 120mC and 180mC respectively. In another arrangement, the same capacitors $C_{1}and{C}_2$ are completely discharged and then connected with the same batteries along with a switch ‘S’ as shown in figure 2.

The total work done by both the batteries (W) and the heat dissipated in the circuit (H) after the switch is shorted are respectively,

• A. $W=720\mu J;H=360\mu J$
• B. $W=720\mu J;H=3600\mu J$
• C. $W=360\mu J;H=180\mu J$
• D. $W=3600\mu J;H=1800\mu J$

Answer 1

The correct option is A $W=720\mu J;H=360\mu J$

Work done by cells $=q_{1}E+(-q_2)E=E(q_1-q_2)720\mu J$
Change in potential energy of the circuit $=\frac{1}{2}{C}_1(V_1^2-V_{0_1}^2)+\frac{1}{2}{C}_2(V_2^2-V_{0_2}^2)=\frac{1}{2}\times 2\times ({60}^2-{72}^2)+\frac{1}{2}\times 3\times ({60}^2-{48}^2)=360\mu J$
So, heat dissipated
$=720\mu J-360\mu J=360\mu J$