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 $e.m.f.$ $=$ $2V$, as shown in the figure. How much energy is lost by the battery in charging the capacitors
Medium
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(b) ${C_{eq}} = \frac{{{C_1}{C_2}}}{{{C_1} + {C_2}}} + {C_3} = \frac{{2 \times 6}}{{2 + 6}} + 4 = 5.5\,\mu \,F$
Enrgy supplied $(E) = QV = C{V^2} = 22 \times {10^{ - 6}}\,J$
$P.E.$ stored$(U) = \frac{1}{2}{C_{eq}}{V^2} = \frac{1}{2} \times 5.5 \times {(2)^2} = 11 \times {10^{ - 6}}\,J$
$==>$ Energy lost$ = E - U = 11 \times {10^{ - 6}}\,J$
Enrgy supplied $(E) = QV = C{V^2} = 22 \times {10^{ - 6}}\,J$
$P.E.$ stored$(U) = \frac{1}{2}{C_{eq}}{V^2} = \frac{1}{2} \times 5.5 \times {(2)^2} = 11 \times {10^{ - 6}}\,J$
$==>$ Energy lost$ = E - U = 11 \times {10^{ - 6}}\,J$
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