First slide

Redistribution of charges and concept of common potential

Question

A capacitor of capacitance $C = 3 μF$ is first charged by connecting it across a 10V battery by closing key $K_{1}$ . It is then allowed to get discharged through $2 Ω$ and $4 Ω$ resistor by closing the key $K_{2}$ and opening key $K_{1}$ .The total energy dissipated in the $2 Ω$ resistor is equal to

Moderate

A

$0.5 mJ$
B

$0.05 mJ$
C

$0.15 mJ$
D

$0.10 mJ$

Solution

Before $K_{1}$ opened the energy stored over capacity $C_{1} = \frac{1}{2} C V^{2}$
$= \frac{1}{2} \times 3 \times 10^{- 6} \times 10^{2} = 150 μJ$
Now this energy will be dissipated $(E = i^{2} R t)$ in the two resistors in the ratio of $R_{1} : R_{2}$ as they are in series (current flowing through the resistors is same) when switch $K_{2}$ is closed.
$∴ 150 μJ$ is lost at the ratio $2 : 4 = 1 : 2$
In $2 Ω, \frac{1}{3} r d$ of $150 μJ = 50 μJ$ and
In $4 Ω, \frac{2}{3} r d$ of $150 μJ=100 μJ$ will be lost.
Therefore, the correct answer is ( B )

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