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A capacitor of $2 F$ (practically not possible to have a capacity of $2 F$ ) is charged by a battery of $6 V$ . The battery is removed and circuit is made as shown. Switch is closed at time $t = 0$ . Choose the correct options.

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At time $t = 0$ current in the circuit is $2 A$ .

At time $t = (6 \ln 2)$ second, potential difference across capacitors is $3 V$ .

At time $t = (6 \ln 2)$ second, potential difference across $1 Ω$ resistance is $1 V$ .

At time $t = (6 \ln 2)$ second, potential difference across $2 Ω$ resistance is $2 V$ .

answer is A, B, C, D.

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Detailed Solution

At $t = 0$ , emf of the circuit $= P D$ across the capacitor $= 6 V$ .

$∴ i = \frac{6}{1 + 2} = 2 A$

Half life of the circuit

$= (\ln 2) τ_{C} (\ln 2) C R = (6 \ln 2) s$

In half life time all values get halved.

For example,

$V_{C} = \frac{6}{2} = 3 V$

$i = \frac{2}{1} = 1 A$

$∴ V_{1 Ω} = i R = 1 V$

$V_{2 Ω} = i R = 2 V$

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