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JEE Mains · Physics · STD 12 - Electric potential and capacitance
  • JEE Mains 2018
  • Hard
  • MCQ

In the following circuit, the switch \(S\)  is closed at \(t = 0.\)  The charge on the capacitor  \(C_1\)  as a function of time will be given by \(\left( {{C_{eq}}\, = {\kern 1pt} \,\frac{{{C_1}{C_2}}}{{{C_1} + {C_2}}}} \right).\)

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Answer & Solution

Correct Answer

(A) \({C_{eq}}E\,[1 - \exp ( - t/R{C_{eq}})]\)

Step-by-step Solution

Detailed explanation

During charging charge on the capacitor increases with time. Charge on the capacitor

\(\mathrm{C}_{1}\) as a function of time, \(\mathrm{Q}=\mathrm{Q}_{0}\left(1-\mathrm{e}^{-t / \mathrm{R} C}\right)\)

\(\mathrm{Q}=\mathrm{C}_{\mathrm{eq}} \mathrm{E}\left[1-\mathrm{e}^{-t / \mathrm{RC}_{\mathrm{eq}}}\right]\)

\(\left(\because Q_{0}=C_{e q} E\right)\)

Both capacitor will have charge as they are connected in series

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