JEE Mains · Physics · STD 12 - 2. Electric potential and capacitance
A capacitor of capacitance \(C=1\, \mu \,{F}\) is suddenly connected to a battery of \(100\, volt\) through a resistance \({R}=100\, \Omega\). The time taken for the capacitor to be charged to get \(50 \,{V}\) is \(....\,\times \,10^{-4}\,s.\) [Take \(\ln 2=0.69]\)

- A \(0.30\)
- B \(1.44\)
- C \(3.33\)
- D \(0.69\)
Answer & Solution
Correct Answer
(D) \(0.69\)
Step-by-step Solution
Detailed explanation
\(V=V_{0}\left(1-e^{-\frac{t}{R C}}\right)\) \(50=100\left(1-e^{-\frac{t}{R C}}\right)\) \(t=0.69 \times 10^{-4} \;sec\)
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