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An 80 \mu \mathrm{C} charge is given to the 4 \mu \mathrm{F} capacitor in the circuit shown in 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 2 A flows through the 2 \Omega resistor in the central limb. The capacitive time constant for the circuit is

Option: 1

56 \mu \mathrm{s}


Option: 2

8 \mu \mathrm{s}


Option: 3

200 \mu \mathrm{s}


Option: 4

40 \mu \mathrm{s}


Answers (1)

best_answer

Potential difference across the central limb =16 \mathrm{~V}

= potential difference across 16 \Omega= potential difference Across the left limb

Current through 16 \Omega=1 \mathrm{~A}

Current through the left limb =1 \mathrm{~A} \text{ and R}=11 \Omega

\mathrm{\therefore \tau_C=\left(\frac{16}{2}+2\right) \Omega \times 4\left(4 \times 10^{-6}\right) F=40 \mu \mathrm{s}}

Posted by

SANGALDEEP SINGH

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