Effective capacitance of parallel combination of two capacitors C1 and C2 is 10 μF. When these capacitors are individually connected to a voltage source of 1V, the energy stored in the capacitor C2 is 4 times that of C1. If these capacitors are connected in series, their effective capacitance will be :
Solution
C<sub>1</sub>
+ C<sub>2</sub>
= 10 ......(1)
<br><br>${1 \over 2}{C_2}{V^2} = 4 \times {1 \over 2}{C_1}{V^2}$
<br><br>$\Rightarrow$ C<sub>2</sub>
= 4C<sub>1</sub> .....(2)
<br><br>From (1) ans (2),
<br><br>C<sub>1</sub>
= 2 and C<sub>2</sub>
= 8
<br><br>For series combination
<br><br>C<sub>eq</sub> = ${{{C_1}{C_2}} \over {{C_1} + {C_2}}}$ = ${{8 \times 2} \over {8 + 2}}$ = 1.6 $\mu$F
About this question
Subject: Physics · Chapter: Electrostatics · Topic: Electric Potential and Capacitance
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