In the following circuit, the switch S is closed at t = 0. The charge on the capacitor C1 as a function of time will be given by (Ceq=(C₁C₂)/(C₁+C₂)) (a) C₁E[1-exp (-tR/C₁)] (b) C₂E[1-exp (-t/RC₂)] (c) CeqE[1-exp (-t/RCeq)] (d) CeqEexp (-t/RCeq)

Answer: (c). C₁, C₂, R and E are all in one series loop, so the two capacitors act as one capacitor Ceq = (C₁C₂)/(C₁+C₂). Series capacitors carry the same charge, so the charge on C₁ equals the charge on Ceq. Charging an uncharged capacitor: q = CeqE(1 - e-t/RCeq) So the answer is (c) CeqE[1-exp (-t/RCeq)].

In the following circuit, the switch S is closed at t = 0. The charge… | JEE Main 2018 (16 Apr, Shift 1)

Q37JEE Main 2018 · 16 APR · SHIFT 1Difficult

Capacitors in DC Circuits & RC

In the following circuit, the switch S is closed at t = 0. The charge on the capacitor C1 as a function of time will be given by $\left(C_{eq}=\frac{C_1C_2}{C_1+C_2}\right)$

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Answer: (c)

$C_1$ , $C_2$ , $R$ and $E$ are all in one series loop, so the two capacitors act as one capacitor $C_{eq} = \frac{C_1C_2}{C_1+C_2}$ .
Series capacitors carry the same charge, so the charge on $C_1$ equals the charge on $C_{eq}$ .
Charging an uncharged capacitor: $q = C_{eq}E\left(1 - e^{-t/RC_{eq}}\right)$
So the answer is (c) $C_{eq}E\left[1-\exp\left(-t/RC_{eq}\right)\right]$ .

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