Q142 / 200 · WBJEE 2014 · Physics

Three capacitors $3 \mu \mathrm{F}, 6 \mu \mathrm{F}$ and $6 \mu \mathrm{F}$ are connected in aeries to a source of $120 \mathrm{V}$. The potential difference, in volt, across the $3 \mu \mathrm{F}$ capacitor will be

  • A
    24
  • B
    30
  • C
    40
  • Correct answer
    D
    60

Solution

The combination of three charges in series $$ \begin{array}{l} \frac{1}{C}=\frac{1}{C_{1}}+\frac{1}{C_{2}}+\frac{1}{C_{3}} \\ \frac{1}{C}=\frac{1}{3}+\frac{1}{6}+\frac{1}{6} \Rightarrow C=\frac{6}{4}=1.5 \mu \mathrm{F} \end{array} $$ The charge of this circuit $$ \begin{array}{l} q=C V=1.5 \times 120 \\ q=180 \mu \mathrm{C} \end{array} $$ The potential difference across the $3 \mu \mathrm{F}$ $$ q=C V $$ $$ V=\frac{q}{C}=\frac{180}{3}=60 \mathrm{V} $$

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

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