Two capacitors {C_1} = 2,mu ,F and {C_2}, = ,6,mu ,F in series, are connected in parallel to a third capacitor {C

Q: 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

b (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$

SECTION - A [PHYSICS - MCQ]

GSEB - English Medium

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

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

Medium

STD 11 Science
NEET
STD 12 - 2. Electric potential and capacitance
Physics

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