Question

Two capacitors $C_1=2\mu F$ and $C_2=6\mu F$ in series, are connected in parallel to a third capacitor $C_3=4\mu F$ . This arrangement is then connected to a battery of e.m.f.=2 V, as shown in figure. The energy lost by the battery in charging the capacitors is :

• A. $22\times {10}^{-6}J$
• B. $11\times {10}^{-6}J$
• C. $\left(\frac{32}{3}\right)\times {10}^{-6}J$
• D. $\left(\frac{16}{3}\right)\times {10}^{-6}J$

Answer 1

The correct option is B $11\times {10}^{-6}J$

total charge flown is $Q=C\cdot V$
$=\frac{11}{2}\times {120}^{-6}\times 2$
$Q=11\times {10}^{-6}$
Now energy lost by battery in charging $=\frac{1}{2}QV$
$=\frac{1}{2}\times 11\times {10}^{-6}\times 2J$
$=11\times {10}^{-6}J$