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  • In the circuit shown (Figure), switch  is closed first and is kept closed for a long

In the circuit shown (Figure), switch \mathrm{ S_2 } is closed first and is kept closed for a long time. Now \mathrm{S_1} is closed. Just after that instant, the current through \mathrm{S_1} is

Option: 1

\mathrm{\frac{\varepsilon}{R_1} ~towards~ right}


Option: 2

\mathrm{\frac{\varepsilon}{R_1} towards ~left}


Option: 3

 Zero


Option: 4

\mathrm{\frac{2 \varepsilon}{R_1}}


Answers (1)

best_answer

Just before \mathrm{S_1} is closed, the potential difference across capacitor 2 is \mathrm{2 \varepsilon}

Just after \mathrm{S_1} is closed, the potential differences across capacitors 1 and 2 are 0 and \mathrm{2 \varepsilon}, Respectively. Applying KVL to loop A B C D immediately after \mathrm{S_1} is closed, \mathrm{\varepsilon=-i R_1+02 \varepsilon, i=\frac{E}{R_1}} towards left

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Riya

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