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)

see full answer

High-Paying Jobs That Even AI Can’t Replace — Through JEE/NEET

🎯 Hear from the experts why preparing for JEE/NEET today sets you up for future-proof, high-income careers tomorrow.

An Intiative by Sri Chaitanya

2.2 sec

0.55 sec

0.48 sec

1.1 sec

answer is D.

(Unlock A.I Detailed Solution for FREE)

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

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

Watch 3-min video & get full concept clarity

No courses found

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test