JEE Mains · Physics · STD 12 - 2. Electric potential and capacitance
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 \(\left( {{C_{eq}}\, = {\kern 1pt} \,\frac{{{C_1}{C_2}}}{{{C_1} + {C_2}}}} \right).\)
- A \({C_{eq}}E\,[1 - \exp ( - t/R{C_{eq}})]\)
- B \({C_1}E\,[1 - \exp ( - tR/{C_1})]\)
- C \({C_2}E\,[1 - \exp ( - t/R{C_2})]\)
- D \({C_{eq}}E\,\exp ( - t/R{C_{eq}})\)
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, \(\mathrm{Q}=\mathrm{Q}_{0}\left(1-\mathrm{e}^{-t / \mathrm{R} C}\right)\)…
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