An 80 μC charge is given to the 4μF capacitor in the circuit shown in the figure so that the upper plate A is positively charged. An unknown resistance R is connected in the left limb. As soon as the switch S in the central limb is closed, a current of 2A flows through the 2 Ω resistor in the central limb. The capacitive time constant for the circuit is

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56 μs

8 μs

200 μs

40 μs

answer is D.

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Detailed Solution

Potential difference across the central limb = 16 V = Potential difference across 16 Ω = Potential difference across the left limb.⇒ current through 16 Ω = 1 A⇒current through the left limb = 1 A & R = 11 Ω∴ τC=162+2ohms×4×10−6F=40μs

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