Question

A $ 500\,\mu F $ capacitor is charged at the steady rate of $ 100\,\mu C/s $ . How long will it take to raise the potential difference between the plates of the capacitor to $10\, V$?

• A. 5s
• B. 10s
• C. 50s
• D. 100s

Answer 1

Answer: (C)

Charge, $Q=C V=500 \,\mu F \times 10\,V $
$=5 \times 10^{-3} \,C$
Now, $Q=q t$
or $t=\frac{Q}{q}$
Or $t=\frac{5 \times 10^{-3}}{100 \times 10^{-6}} s$
$=\frac{1}{20} \times 1000\,s $
$=50\, s$