A 8µF capacitor is charged by a 400V supply through 0.1MΩ resistance. The time taken by the capacitor to develop a potential difference of 300V is :
(Given log₁₀4 = 0.602)

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2.2 sec

1.1 sec

0.55 sec

0.48 sec

answer is D.

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Detailed Solution

$\begin{array}{l} q = CV; q = 8 \times 10^{- 6} \times 300; \end{array}$

$q = q_{0} (1 - e^{- t / CR})$

$\begin{array}{l} 8 \times 10^{- 6} \times 300 = 8 \times 10^{- 6} \times 400 (1 - e^{- \frac{t}{0 .8}}) \end{array}$

$e^{- \frac{t}{0 .8}} = (\frac{1}{4}); t = 0 .8 \log (4) \Rightarrow t = 0 .8 \times 0 .602$
= 0.48sec

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