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  •   In the circuit shown, switch  is closed first and is kept closed for a long time. Now <img alt="S_1" src="https://le

  In the circuit shown, switch S_2 is closed first and is kept closed for a long time. Now S_1 is closed. Just after that instant the current through S_1 is:

 

Option: 1

\varepsilon / R_1 , towards \: \: right


Option: 2

\varepsilon / R_1 , towards \: \: left


Option: 3

zero 


Option: 4

\frac{2 \varepsilon }{R_1}


Answers (1)

best_answer

 

charging of capacitor -

q=q_{0}\left ( 1-e^{-\frac{t}{\tau }} \right )\\ \\ \tau = Rc

q_{0}=max \: charge\\ \\ \tau = Time\: constant

-

 

 

just before S_1 is closed the potential difference across capacitor 2 is 2E.

            just after S_1 is closed the potential difference across capacitors 1 and 2 are 0 and 2E respectively. Applying KVL to loop ABCD immediately after S_1 is closed.

                  E = – iR1 +  0 + 2E

            or   i = \frac{\varepsilon }{R_1} towards left 

Posted by

sudhir.kumar

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