Q140 / 160 · TS EAMCET 2010 · Physics

Two capacitors of capacities $1 \mu \mathrm{F}$ and $C \mu \mathrm{F}$ are connected in series and the combination is charged to a potential difference of $120 \mathrm{~V}$. If the charge on the combination is $80 \mu \mathrm{C}$, the energy stored in the capacitor of capacity $C$ in $\mu \mathrm{J}$ is

  • A
    1800
  • Correct answer
    B
    1600
  • C
    14400
  • D
    7200

Solution

Capacitances $1 \mu \mathrm{F}$ and $C \mu \mathrm{F}$ are connected in series, then $C_{\text {eq }}=\frac{C}{1+C}$ Given, $V=120 \mathrm{~V}$ and $q=80 \mu \mathrm{C}$ $\because \quad q=C_{\mathrm{eq}} V$ $80=\frac{C}{C+1} \times 120$ or $\quad C=2 \mu \mathrm{F}$ The energy stored in the capacitor of capacity $C$ $U=\frac{1}{2} \frac{q^2}{C}$ $=\frac{1}{2} \times \frac{\left(80 \times 10^{-6}\right)^2}{2 \times 10^{-6}}$ $=\frac{1}{2} \times \frac{80 \times 10^{-6} \times 80 \times 10^{-6}}{2 \times 10^{-6}}$ $U=1600 \mu \mathrm{J}$

Hence, the correct answer is option: 'B'.

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