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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 \(\ln 2=0.69]\)

  1. A \(0.30\)
  2. B \(1.44\)
  3. C \(3.33\)
  4. D \(0.69\)
Verified Solution

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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