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  • Find the current flowing through the resistance in the steady state as also the charge on the capacitor C. If the externally applied potential are now withdrawn, The charge on the capacitor vary as

Find the current flowing through the resistance in the steady state as also the charge on the capacitor C. If the externally applied potential are now withdrawn, The charge on the capacitor vary as a function of time as

(R = 1k\Omega , C = 10 \mu F)

Option: 1

50 \mu \mathrm{C} \times e^{-t / 6.67} \\


Option: 2

5 \mu C \times e^{-t / 6.67}


Option: 3

25 \mu \mathrm{C} \times \mathrm{e}^{-\mathrm{t} / 6.67}


Option: 4

None of the above


Answers (1)

best_answer

The current through B A=\frac{10 V-5 V}{1}=5 \mathrm{~mA}

similarly current through AC = 5 mA, & BC = 10 mA

steady state charge on \mathrm{C}=5 \mathrm{~V} \times 10 \mu \mathrm{F}=50 \mu \mathrm{C}

If the potential differences are withdrawn at time t = 0, the charge on the capacitor varies as a function of time as it discharges through the external resistance. The equivalent resistance of the circuit across AC is

=\frac{2}{3} \times \mathrm{R}=667 \Omega \text { (approx) }

The time constant \tau=(2 / 3 R) \times C

= 6.67 m sec.

The charge across the capacitor is

\mathrm{q}(\mathrm{t})=50 \mu \mathrm{C} \times \mathrm{e}^{-\mathrm{t} / 6.67 \mathrm{~ms}}

The charge across the capacitor is
\mathrm{q}(\mathrm{t})=50 \mu \mathrm{C} \times \mathrm{e}^{-\mathrm{t} / 6.67}

Posted by

Sanket Gandhi

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