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JEE Main 2022 · Physics · Capacitors · medium

A 4 muF capacitor is charged to 100 V and then connected through a resistor to an uncharged 6 muF capacitor (no battery in the circuit). After the…

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A 4 μF4\ \mu\mathrm{F} capacitor is charged to 100 V100\ \mathrm{V} and then connected through a resistor to an uncharged 6 μF6\ \mu\mathrm{F} capacitor (no battery in the circuit). After the charges settle, the heat dissipated in the resistor is
  1. A. 8 mJ8\ \mathrm{mJ}
  2. B. 12 mJ12\ \mathrm{mJ}
  3. C. 20 mJ20\ \mathrm{mJ}
  4. D. 30 mJ30\ \mathrm{mJ}
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Answer: (B) 12 mJ12\ \mathrm{mJ}

Charge is conserved: Q=4×10−6×100=4×10−4 CQ=4\times10^{-6}\times100=4\times10^{-4}\ \mathrm{C} shared over 10 μF10\ \mu\mathrm{F}, so the common voltage is V=QC1+C2=40 VV=\dfrac{Q}{C_1+C_2}=40\ \mathrm{V}. Initial energy =12(4×10−6)(100)2=20 mJ=\frac12(4\times10^{-6})(100)^2=20\ \mathrm{mJ}. Final energy =12(10×10−6)(40)2=8 mJ=\frac12(10\times10^{-6})(40)^2=8\ \mathrm{mJ}. Heat =20−8=12 mJ=20-8=12\ \mathrm{mJ}, which is independent of the value of the resistance. Shortcut: H=12C1C2C1+C2V2H=\dfrac12\dfrac{C_1C_2}{C_1+C_2}V^{2}.
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