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

Answer 1

Answer: (C)

$\frac{50}{17}V$

$C_{eq}=\frac{10}{17}\mu F,Q=\frac{100}{17}\mu C$
$\therefore$ Potential difference across $2\mu F$ capacitor
$=\frac{\frac{100}{17}\mu C}{2\mu F}=\frac{50}{17}V$