Consider two uncharged capacitors of equal capacitance 200 pF. One of them is charged by a 100 V supply and disconnected. Now this capacitor is connected to the uncharged capacitor. The amount of electrostatic energy lost in the process is:
A.1.0 × 10⁻⁶$1.0 \times 10^{-6}$ J
B.0.5 × 10⁻⁶$0.5 \times 10^{-6}$ J
C.0.5 J
D.1.0 J
Solution & Explanation
Related Formula
Δ U = (1)/(2) (C₁ C₂)/(C₁ + C₂) V²$$\Delta U = \frac{1}{2} \frac{C_1 C_2}{C_1 + C_2} V^2$$
Keywords:#electrostatic energy loss connecting capacitors#NEET 2026 Code 12 Q31#Electrostatic Potential and Capacitance NEET 2026#Energy Loss in Capacitors NEET 2026
More Electrostatic Potential and Capacitance Previous-Year Questions
Q15neet_2024_05_may_morningCombination of Capacitors
Five capacitors of capacitances C₁=C₂=C₃=C₄=10μ F$C_{1}=C_{2}=C_{3}=C_{4}=10\mu F$ and C₅=2.5μ F$C_{5}=2.5\mu F$ are connected as shown, along with a battery of 50 V.
Five capacitors arranged in a bridge loop across a 50V source.
The equivalent capacitance and the charges on each capacitor respectively are:
A. 5 μ$\mu$ F, 125 μ$\mu$ C on all capacitors
B. 5 μ$\mu$ F, 250 μ$\mu$ C on all capacitors
C. 4 μ$\mu$ F, 250 μ$\mu$ C on C₁$C_{1}$ to C₄$C_{4}$ and 125 μ$\mu$ C on C₅$C_{5}$
D. 5 μ F, 125 μ C$\mu \mathrm{F}, 125 \mu \mathrm{C}$ on C₁$\mathrm{C}_{1}$ to C₄$\mathrm{C}_{4}$ and 25 μ C$25 \mu \mathrm{C}$ on C₅$\mathrm{C}_{5}$
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