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Question: Two capacitors C<sub>1</sub> = 2μF and C<sub>2</sub> = 6μF in series, are connected in parallel to a...

Two capacitors C1 = 2μF and C2 = 6μF in series, are connected in parallel to a third capacitor C3 = 4μF. This arrangement is then connected to a battery of e.m.f. = 2 V, as shown in the fig. How much energy is lost by the battery in charging the capacitors ?

A

22×106J22 \times 10 ^ { - 6 } J

B

11×106J11 \times 10 ^ { - 6 } J

C

D

(163)×106J\left( \frac { 16 } { 3 } \right) \times 10 ^ { - 6 } J

Answer

11×106J11 \times 10 ^ { - 6 } J

Explanation

Solution

Equivalent capacitance

Ceq=C1C2C1+C2+C3=2×68+4=5.5μFC _ { e q } = \frac { C _ { 1 } C _ { 2 } } { C _ { 1 } + C _ { 2 } } + C _ { 3 } = \frac { 2 \times 6 } { 8 } + 4 = 5.5 \mu F