The figure below shows a simple RC circuit with a 1.90-muF capacitor, a 4.40-MOh

Question

The figure below shows a simple RC circuit with a 1.90-muF capacitor, a 4.40-MOhm resistor, a 9.00-V emf, and a switch. What are the following exactly 6.50 s after the switch is closed? the charge on the capacitor the current in the resistor Your response differs from the correct answer by more than 10%. Double check your calculations. mu A the rate at which the capacitor is storing energy Your response differs from the correct answer by more than 10%. Double check your calculations. mu W the rate at which the battery is delivering energy 12.78 X Your response differs from the correct answer by more than 10%. Double check your calculations. mu W

Explanation / Answer

Current in the circuit , I = I0 [ e^(-t/RC)]

I0 = V/R = 9 V /4.40 Mohm = 2.045 uA

time costant = RC = 1.90 x 10^-6 x 4.40 x 10^6 = 8.36

I = 2.045 uA [ e^(-t / 8.36)]

at t = 6.50 s

I = 2.045uA [ e^(-6.50 / 8.36) ]

I = 0.94 uA OR 0.94 x 10^-6 A ...........Ans(B)

PD across capacitor = V - IR = 9 - (4.40 x 10^6 x 0.94 x 10^-6) =4.86 Volt

charge Q = CVc = 1.90uF x 4.86 = 9.24 uF .........Ans(A)

c) energy stored in capacitor = C V^2 / 2

Vc = V [1 - e^(-t/T)] = 9 [ 1 - e^(-t/8.36)]

energy = (1.90 x 10^-6 ) x 9^2 [ 1 - e^(-t/8.36)]^2 /2

E = (76.95 x 10^-6)[[ 1 - e^(-t/8.36)]^2 ]

rate of storing energy = dE/dt = 2 x 76.95 x 10^-6 x [ 1 - e^(-t/8.36)] x (-1/8.36) x (- e^(-t/8.36))

= 18.41 x 10^-6 [ 1 - e^(-t/8.36)] [ e^(-t/8.36) ]

putting t= 6.5 s

= 4.57 x 10^-6 J/s OR W = 4.57 uW

d) POwer of battery = VI

at t = 6.5 s

I = 0.94 x 10^-6 A

P = 9 x 0.94 x 10^-6 =8.46 x 10^-6 W = 8.46 uW

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