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JEE MAIN 2020
04.09.20_S2
Question
A capacitor $C$ is fully charged with voltage $V_0$. After disconnecting the voltage source, it is connected in parallel with another uncharged capacitor of capacitance $\frac{C}{2}$. The energy loss in the process after the charge is distributed between the two capacitors is
Select the correct option:
A
$\frac{1}{4} C V_0^2$
B
$\frac{1}{3} C V_0^2$
C
$\frac{1}{6} C V_0^2$
D
$\frac{1}{2} C V_0^2$
✓ Correct! Well done.
✗ Incorrect. Try again or view the solution.
Solution
$\Delta U=\frac{1}{2} \frac{C_1 C_2}{C_1+C_2} V_0^2=\frac{1}{6} C V_0^2$
Question Tags
JEE Main
Physics
Easy
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