Three capacitors of capacitance 1.0,2.0 and $5.0 \mu \mathrm{F}$ are connected in series to a $10 \mathrm{V}$ source. The potential difference across the $2.0 \mu \mathrm{F}$ capacitor is

• A

  $\frac{100}{17} \mathrm{V}$
• B

  $\frac{20}{17} \mathrm{v}$
• Correct answer
  C

  $\frac{50}{17} \mathrm{V}$
• D

  $10 \mathrm{V}$

Solution

When the capacitors are connected in series, the resultant capacitance of combination $$ \frac{1}{C}=\frac{1}{C_{1}}+\frac{1}{C_{2}}+\frac{1}{C_{3}} $$ $$ \begin{array}{l} =\frac{1}{1}+\frac{1}{2}+\frac{1}{5}=\frac{17}{10} \\ C=\frac{10}{17} \mu \mathrm{F} \end{array} $$ The charge will be same on all the capacitors in series $$ Q=C V=\frac{10}{17} \times 10=\frac{100}{17} $$ The potential difference across $2.0 \mu \mathrm{F}$ capacitor $$ V^{*}=\frac{Q}{C}=\frac{100 / 17}{2}=\frac{50}{17} \text { volt} $$

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