First slide

Redistribution of charges and concept of common potential

Question

In the circuit shown switch S is kept open for long time. Switch is closed at t = 0, find total heat produced after closing the switch

Moderate

A

40 $μ J$
B

66.4 $μ J$
C

79.6 $μ J$
D

78.4 $μ J$

Solution

Before closing the switch
$Q = C_{e q} V = (\frac{C_{1} C_{2}}{C_{1} + C_{2}}) (V_{1} - V_{2})$
$Q = \frac{6}{5} \times 1 μ C$
Initial energy stored capacitors = $\frac{Q^{2}}{2 C_{1}} + \frac{Q^{2}}{2 C_{2}}$
$= \frac{{(\frac{6}{5})}^{2}}{2 \times 3} + \frac{{(\frac{6}{5})}^{2}}{2 \times 2} = \frac{3}{5} μ J$
After closing the switch
$Q_{2} = 10 μ C$
$Q_{1} = 18 μ C$
Final energy stored in capacitor
$U = \frac{1}{2} \times 3 {(6)}^{2} + \frac{1}{2} \times 2 {(5)}^{2}$
$U = 79 μ J$
Work done by batteries = $6 (Q_{1} - Q) + 5 (Q + Q_{2})$
$\begin{array}{l} = 156.8 μ J \\ H = work done by battery - change in potential energy of capacitors \\ = 156.8 - (79 - \frac{3}{5}) \\ = 78.4 μ J \end{array}$

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