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  • Suppose that the circuit shown in the adjacent figure was connected to a source of emf and switched on. After a long time, the circuit is switched off. The charge on the capacitors as a function of

Suppose that the circuit shown in the adjacent figure was connected to a source of emf and switched on. After a long time, the circuit is switched off. The charge on the capacitors as a function of time after the external circuit is switched off is

Option: 1

\mathrm{\left\{\left(\frac{C_{1}}{C_{2}}+1\right) \tau-C_{2} V_{0}\right\} e^{-t / \tau}+\left(\frac{C_{1}+C_{2}}{C_{2}}\right) \tau}


Option: 2

\mathrm{\left\{\left(\frac{C_{1}}{C_{2}}+1\right) \tau-C_{1} V_{0}\right\} e^{-t / \tau}+\left(\frac{C_{1}+C_{2}}{C_{2}}\right) \tau}


Option: 3

\mathrm{\left\{\left(\frac{C_{1}}{C_{2}}+1\right) \tau-C_{1} V_{0}\right\} e^{-t / \tau}+\left(\frac{C_{1}+C_{2}}{C_{1}}\right) \tau}


Option: 4

None of the above


Answers (1)

best_answer

Suppose that the potential difference applied between \mathrm{A} and \mathrm{B} is \mathrm{V_{0}}
The charges on \mathrm{C_{1}} and \mathrm{C_{2}} are respectively,

\mathrm{q_{1}^{0}=C_{1} V_{0}} and \mathrm{q_{2}^{0}=C_{2} V_{0} \ldots (i)}

After the external potential is switched off, let I be the current in the circuit while \mathrm{q_{1}(T)} and \mathrm{q_{2}(t)} be the charges on \mathrm{C_{1}} and \mathrm{C_{2}}
Kirchoff's law gives
\mathrm{\quad \frac{q_{1}}{C_{1}}+2\left(\frac{d q_{1}}{d t}\right) R+\frac{q_{2}}{C_{2}}=0\quad \cdots(ii)}
\mathrm{i=\frac{d q_{1}}{d t}=\frac{d q_{2}}{d t}}

\mathrm{\text { or, } q_{1}=q_{2}+\text { constant }}
\mathrm{ =q_{2}+a(\text { say) }\quad \cdots (ii)}

\mathrm{ \text { equation }(I) \text { gives }}
\mathrm{q_{1}\left(\frac{1}{C_{1}}+\frac{1}{C_{2}}\right)+\frac{q}{C_{2}}+2\left(\frac{d q_{1}}{d t}\right) R=0}
\mathrm{\text { or } \frac{d q_{1}}{d t}+\frac{1}{2 R}\left(\frac{1}{C_{1}}+\frac{1}{C_{2}}\right) q_{1}=-\frac{-a}{C_{2}}}
\mathrm{\Rightarrow q_{1}=A_{1} e^{-t / \tau}-\frac{a}{C_{2}} \times \tau}

\mathrm{\text { where } \frac{1}{\tau}=\frac{1}{2 R}\left(\frac{1}{C_{1}}+\frac{1}{C_{2}}\right)}
\mathrm{ q_{2}=q_{1}-a}
\mathrm{ =A_{1} e^{-t / \tau}-a \frac{\tau}{C_{2}}+1}
\mathrm{ \text { Initially, } q_{(0)}(0)=-C_{1} V_{0} \text { and } q_{2}(0)=C_{2} V_{0} \\ }
\mathrm{ a=q_{1}(0)-q_{2}(0)=-\left(C_{1} V_{0}+C_{2} V_{0}\right) }
\mathrm{=-\left(C_{1}+C_{2}\right) V_{0} }
\mathrm{q_{1}(t)=A_{1} e^{-t / \tau}+\frac{C_{1}+C_{2}}{C_{2}} \times \tau }
\mathrm{\therefore \quad A_{1}=\left(\frac{C_{1}}{C_{2}}+1\right) \tau-C_{1} V_{0} }
\mathrm{\left.\therefore \quad\left(\frac{C_{1}}{C_{2}}+1\right) \tau-C_{1} V_{0}\right\} e^{-t / \tau}+\left(\frac{C_{1}+C_{2}}{C_{2}}\right) \tau }

 

Posted by

Ritika Harsh

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