Question

Two capacitors A and B with capacities 3$\mu$F , 2$\mu$F respectively are charged to a potential difference of 100 V and 180 V respectively. The plates of the capacitors are connected as shown in the figure. With one wire from each capacitor free. The upper plate of A is positive and that of B is negative. An uncharged capacitor C with lead wires falls on the free ends to complete the circuit. Energy loss in redistribution of charges on capacitors will be :

• A. 47.5 mJ
• B. 45.6 mJ
• C. 1.8 mJ
• D. 29.4 mJ

Answer 1

Answer: (C)

1.8 mJ

Before joining,

$q_A=C_A{V}_A=100\times 3=300\mu c$

$q_B=C_B{V}_B=180\times 2=360\mu c$

Applying Kirchoff's law ,

$\frac{q_1}{3}-\frac{q_2}{2}+\frac{q_3}{2}=0$

$2q_1-3q_2+3q_3=0$ (1)

From conservation of charge

Net charge before joining = Net charge after joining

$q_1+q_2=300$ (2)

$q_2+q_3=360$ (3)

Solving (1),(2) and (3)

$q_1=90\mu C$

$q_2=210\mu C$

$q_3=150\mu C$

Energy lost = $\frac{1}{2}\frac{(90\times {10}^{-6}{)}^2}{3\times {10}^{-6}}+\frac{1}{2}\frac{(210\times {10}^{-6}{)}^2}{2\times {10}^{-6}}+\frac{1}{2}\frac{(150\times {10}^{-6}{)}^2}{2\times {10}^{-6}}=1.8mJ$