Question

A $500\mu F$ capacitor is charged at a steady rate of $100\mu C/sec$. The potential difference across the capacitor will be $10V$ after an interval of-

• A. $5sec$
• B. $20sec$
• C. $25sec$
• D. $50sec$

Answer 1

Answer: (D)

Given,

Capacitance, $C=500\mu F$

Potential difference, $V=10V$

Rate of charging, $q=100\mu C/\sec$

$=100\times {10}^{-6}C/\sec$

To fine, $t=?$

We know that,

$Q=qt$

$\begin{array}{c}t=\frac{Q}{q}\\ =\frac{5\times {10}^{-3}C}{100\times {10}^{-6}C}\times \sec \\ =50\sec \end{array}$

$\therefore$ It will take $50sec$ to raise the potential difference .

Hence,

Option $D$ is correct.