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  • The switch in the circuit diagram (Figure) is closed at =0. The charge on capacitors at any time is <img alt="" src="https://cdn.entrance360.com/media/uploads/2023/08/18/f52_q18.PNG" styl

The switch in the circuit diagram (Figure) is closed at =0. The charge on capacitors at any time is

Option: 1

\mathrm{ q_{(t)}=E C\left(1-e^{-2 t / 3 R C}\right)}


Option: 2

\mathrm{ q_{(f)}=E C\left(1-e^{-t / 2 R C}\right)}


Option: 3

\mathrm{ q_{(t)}=E C\left(1-e^{-t / 3 R C}\right)}


Option: 4

\mathrm{ q_{(f)}=E C\left(1-e^{-3 t / 2 R C}\right)}


Answers (1)

best_answer

Applying KVL, we get

\begin{aligned} &\mathrm{\frac{I}{2}=\frac{d q}{d t} \Rightarrow I=2 \frac{d q}{d t}}\\ &\begin{aligned} &\mathrm{ E-I R-\frac{1}{2} R-\frac{q}{C}=0 }\\ &\mathrm{ E-\frac{3}{2} I R-\frac{q}{C}=0 }\\ &\mathrm{ E-\frac{3}{2} \times 2 \frac{d q}{d t} R-\frac{q}{C}=0 }\\ &\mathrm{ q_{(f)}=E C\left(1-e^{-1 / 3 R C}\right)} \end{aligned} \end{aligned}\begin{aligned} &\mathrm{\frac{I}{2}=\frac{d q}{d t} \Rightarrow I=2 \frac{d q}{d t}}\\ &\begin{aligned} &\mathrm{ E-I R-\frac{1}{2} R-\frac{q}{C}=0 }\\ &\mathrm{ E-\frac{3}{2} I R-\frac{q}{C}=0 }\\ &\mathrm{ E-\frac{3}{2} \times 2 \frac{d q}{d t} R-\frac{q}{C}=0 }\\ &\mathrm{ q_{(f)}=E C\left(1-e^{-1 / 3 R C}\right)} \end{aligned} \end{aligned}

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Rishi

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