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  • The key K (figure) is connected in turn to each of the contacts over short identical time intervals so that the change in the charge on the capacitor over each connection is small. The final charge

The key K (figure) is connected in turn to each of the contacts over short identical time intervals so that the change in the charge on the capacitor over each connection is small. The final charge qr on the capacitor is :

Option: 1

\frac{\left ( E_{1}R_{2}+E_{2}R_{1} \right )C}{R_{1}+R_{2}}


Option: 2

\frac{(E_{1}E_{2})C}{E_{1}+E_{2}}


Option: 3

\frac{\left ( E_{1}R_{1}+E_{2}R_{2} \right )C}{R_{1}+R_{2}}


Option: 4

none of these


Answers (1)

best_answer

 

Capacitance of Conductor -

Q\propto V

Q=CV

- wherein

C - Capacity or capacitance of conductor 

V - Potential.

 

 

As the key is connected to 1 and 2 frequently and at equal intervals, the two emf's E1 & E2 behave as d.c. sources in continuous contact. The potential due to the two cells is :

V = \frac{\left ( E_{1}R_{2}+E_{2}R_{1} \right )}{R_{1}+R_{2}}

Hence the charge on the capacitor is q = CV

= \frac{\left ( E_{1}R_{2}+E_{2}R_{1} \right )C}{R_{1}+R_{2}}

Posted by

HARSH KANKARIA

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