In the given circuit, the two-way switch S is initially kept at position 1 for a long time, fully charging the capacitor C_1 = 4\ using a 50\ V battery. The switch is then shifted to position 2, disconnecting the

Question

Quantrex Academy · Accepted Answer

Correct answer: B. 3 × 10^ -3 \ J

When the switch is at position 1, capacitor

C_1

is charged to potential

V_1 = 50\ \text{V}

.

When the switch is shifted to position 2,

C_1

is connected across the initially uncharged capacitor

C_2

(

V_2 = 0

).

The electrostatic energy dissipated as heat during the redistribution of charge is given by the formula:

\Delta U = \dfrac{1}{2} \dfrac{C_1 C_2}{C_1 + C_2} (V_1 - V_2)^2

Substituting the given values

C_1 = 4\ \mu\text{F}

,

C_2 = 6\ \mu\text{F}

,

V_1 = 50\ \text{V}

, and

V_2 = 0\ \text{V}

:

\Delta U = \dfrac{1}{2} \times \dfrac{(4 \times 10^{-6}) \times (6 \times 10^{-6})}{(4 + 6) \times 10^{-6}} \times (50 - 0)^2

\Delta U = \dfrac{1}{2} \times \dfrac{24 \times 10^{-12}}{10 \times 10^{-6}} \times 2500

\Delta U = \dfrac{1}{2} \times 2.4 \times 10^{-6} \times 2500

\Delta U = 1.2 \times 10^{-6} \times 2500

\Delta U = 3000 \times 10^{-6}\ \text{J}

\Delta U = 3 \times 10^{-3}\ \text{J}

Answer:

3 \times 10^{-3}\ \text{J}

In the given circuit, the two-way switch S is initially kept at...

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