Question

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

• A. $\frac{100}{17}V$
• B. $\frac{20}{17}V$
• C. $\frac{50}{17}V$
• D. $10V$

Answer 1

Answer: (C)

When the capacitors are connected in series, die resultant capacitance of combination
$\frac{1}{C}=\frac{1}{C_1}+\frac{1}{C_2}+\frac{1}{C_3}$
$=\frac{1}{1}+\frac{1}{2}+\frac{1}{5}=\frac{17}{10}\mu F$
$C=\frac{10}{17}\mu F$
The charge will be same on all the capacitors in series
$Q=CV=\frac{10}{17}\times 10=\frac{100}{17}$
The potential difference across $2.0\mu F$ capacitor
$V^{\prime }=\frac{Q}{C}$
$=\frac{100/17}{2}=\frac{50}{17}$ volt