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2. Electric Potential and Capacitance
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A $500\,\mu F$ capacitor is charged at a steady rate of $100\, \mu C/sec$. The potential difference across the capacitor will be $10\, V$ after an interval of.....$sec$
A
$5$
B
$20$
C
$25$
D
$50$
Solution
$Q=\mathrm{C V}$
$\frac{\Delta Q}{\Delta t}=C \frac{\Delta V}{\Delta t}$
$100 \times 10^{-6}=500 \times 10^{-6} \times \frac{10}{\Delta t}$
$\Delta t=50 \mathrm{\,sec}$
Standard 12
Physics
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