CBSE Class 12 · 2004 · CapacitanceShort Answer

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20 μ F

capacitor is charged by a 30 V dc supply and then connected across an uncharged

50 μ F

capacitor. Calculate
(i) the final potential difference across the combination, and
(ii) the initial and final energies. How will you account for the difference in energy?

Solution

Initial charge on the

20 μ F

capacitor:

q = 20 \times 1 0^{- 6} \times 30 = 6 \times 1 0^{- 4} C

(i) On connecting, the common potential is:

V = \frac{q}{C _{1} + C _{2}} = \frac{6 \times 1 0 ^{- 4}}{70 \times 1 0 ^{- 6}} = \frac{60}{7} \approx 8.57 V

(ii) Initial energy:

U_{i} = \frac{1}{2} \times 20 \times 1 0^{- 6} \times (30)^{2} = 9 \times 1 0^{- 3} J

Final energy:

U_{f} = \frac{1}{2} (70 \times 1 0^{- 6}) V^{2} = 2.57 \times 1 0^{- 3} J

The loss

U_{i} - U_{f}

is dissipated as heat in the connecting wires (and a little as electromagnetic radiation) during the transient redistribution of charge.

CBSE Class 12 Board · 2004 · Subjective

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CBSE Class 12 · 2004 · CapacitanceShort Answer

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