A 800 μF capacitor is charged by a 200 V battery. After some time the capacitor is disconnected from the battery and connected in parallel across another uncharged identical capacitor. The energy stored by the combination will be :
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 8 J
When the charged capacitor is joined to an identical uncharged one, charge is conserved and shares equally, but energy is not conserved: part of it is lost as heat in the connecting wires.
\( C = 800\,\mu\mathrm{F} = 8\times10^{-4}\,\mathrm{F} \)
\( V = 200\,\mathrm{V} \)
\( Q = CV = 8\times10^{-4}\,\mathrm{F} \times 200\,\mathrm{V} = 0.16\,\mathrm{C} \)
The combination has capacitance 2C and carries the same charge.
\( \displaystyle V' = \frac{Q}{2C} = \frac{0.16\,\mathrm{C}}{1.6\times10^{-3}\,\mathrm{F}} \)
\( V' = 100\,\mathrm{V} \)
\( \displaystyle U' = \frac{1}{2}(2C)V'^{2} = \frac{1}{2} \times 1.6\times10^{-3}\,\mathrm{F} \times (100\,\mathrm{V})^{2} \)
\( U' = 8\,\mathrm{J} \)
The energy before connection was \( \frac{1}{2}CV^{2} = 16 \) J, so half of it is lost.
Shortcut: With the charge fixed, \( U = Q^{2}/2C \); doubling the capacitance halves the energy, so \( U' = 16/2 = 8 \) J.