A capacitor of capacitance C = 6.5 F is initially uncharged. It is connected in

Question

A capacitor of capacitance C = 6.5 F is initially uncharged. It is connected in series with a switch of negligible resistance, a resistor of resistance R = 11.5 k, and a battery which provides a potential difference of VB = 25 V.

17% Part (a) Calculate the time constant for the circuit in seconds.
17% Part (b) After a very long time after the switch has been closed, what is the voltage drop VC across the capacitor in terms of VB?
17% Part (c) Calculate the charge Q on the capacitor a very long time after the switch has been closed in C.
17% Part (d) Calculate the current I a very long time after the switch has been closed in A.
17% Part (e) Calculate the time t after which the current through the resistor is one-third of its maximum value in s. 17% Part (f) Calculate the charge Q on the capacitor when the current in the resistor equals one third its maximum value in C.

Explanation / Answer

a)we know that, time constant in RC circuit is:

tau = RC = 11.5 x 10^3 x 6.5 x 10^-6 = 0.075 s

Hence, tau = 0.075 s

b)After a very long time, the drop across the capacitor is equal to that of the battery voltage.

Vc = Vb

c)Charge after velong long time of closing the switch will be:

Q = C Vb

Q = 6.5 x 10^-6 x 25 = 162.5 x 10^-6 C = 162.5 muC

Hence, Q = 162.5 muC

d)The current after a velong long time becomes zero in the circuit.

I = 0

e)Imax = V/R = 25/11.5 x 10^3 = 2.17 x 10^-3 A

1/3 Imax = 0.72 x 10^-3

We know that, I at any time during charing is:

I = I(max)e^-t/tau

0.72 x 10^-3 = 2.17 x 10^-3 e^-t/tau

0.33 = e^-t/tau

taking log both sides:
-1.11 = -t/0.075

t = 0.083 s

f)Q = C Vb (1 - e^-t/tau)

Q = 6.5 x 10^-6 x 25 ( 1 - e^-0.083/0.075) = 16.44 x 10^-6 C

Hence, Q = 16.44 muC

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