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2. Electric Potential and Capacitance
medium
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
(d) Charge required to reach the capacitor upto $10\, V$ is
$Q = 500 \times {10^{ – 6}} \times 10 = 5 \times {10^{ – 3}}\,C$
Now required time $ = \frac{{5 \times {{10}^{ – 3}}}}{{100 \times {{10}^{ – 6}}}}$ $= 50\, sec$
Standard 12
Physics