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  • Two capacitors having capacitance <img alt="\mathrm{C}_{1}$ and $\mathrm{C}_{2}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7BC%7D_%7B1%7D%24%20and%20%24%5Cmathrm%7BC%7D_%7B2%7D

Two capacitors having capacitance \mathrm{C}_{1}$ and $\mathrm{C}_{2} respectively are connected as shown in figure. Initially, capacitor C_{1} is charged to a potential difference \mathrm{V} volt by a battery. The battery is then removed and the charged capacitor C_{1} is now connected to uncharged capacitor C_{2} by closing the switch S. The amount of charge on the capacitor C_{2} , after equilibrium, is :

 

Option: 1

\begin{aligned} &\frac{C_{1} C_{2}}{\left(C_{1}+C_{2}\right)} V \\ \end{aligned}


Option: 2

\frac{\left(C_{1}+C_{2}\right)}{C_{1} C_{2}} V \\


Option: 3

\left(C_{1}+C_{2}\right) V \\


Option: 4

\left(C_{1}-C_{2}\right) V


Answers (1)

best_answer


By charge conservation,

\mathrm{q_{1}+0= Q_{1}+Q_{2}}
\mathrm{q_{1}= Q_{1}+Q_{2}\, \rightarrow (1)}

At steady state (equilibrium)
\mathrm{Common \, Potential = V_{0}= \frac{C_{1}V_{1}+C_{2}V_{2}}{C_{1}+C_{2}}}
                                               \mathrm{ V_{0}= \frac{C_{1}V_{}}{C_{1}+C_{2}}}
\mathrm{\left.\begin{matrix} Q_{1}= C_{1}V_{0}\\ Q_{1}= \frac{C_{1}^{2}V}{C_{1}+C_{2}} \end{matrix}\right|\begin{matrix} Q_{2}= C_{2}V_{0}\\ Q_{2}= \frac{C_{1}C_{2}V}{C_{1}+C_{2}} \end{matrix}}
The correct option is (1)

Posted by

Ajit Kumar Dubey

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