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# 9) For the circuit shown in the figure, the switch S is initially open and the...

9) For the circuit shown in the figure, the switch S is initially open and the capacitor is uncharged. The switch is then closed at time t 0. What is the time constant of the circuit? How many seconds after closing the switch will the energy stored in the capacitor be equal to 49.1 x 10-3 J? The capacitance is 89 x 10-6 F, the resistor is 0.56 x 106 ohms, and the voltage is 40. V

Given,

C = 89 x 10^-6 F ; R = 0.56 x 10^6 Ohms ; V = 40 V ; U = 49.1 x 10^-3 J

We know that, time constant is given by:

tau = RC

tau = 0.56 x 10^6 x 89 x 10^-6 = 49.84 s

Hence, tau = 49.84 s

the energy is:

U = 1/2 Q^2/C

Q = sqrt (2 U C)

Q = sqrt (2 x 49.1 x 10^-3 x 89 x 10^-6) = 0.0029 C

We know that,

Q = C V (1 - e^-t/RC)

0.0029 = 89 x 10^-6 x 40 (1 - e^-t/49.84)

0.81 = 1 - e^-t/49.84

e^-t/49.84 = 1 - 0.81 = 0.19

-t/49.84 = -1.66

t = 1.66 x 49.84 = 82.73 s

Hence, t = 82.73 s

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