Question

2. A 12. 0 �٠resistor is connected in series with a 1.75 �� capacitor and a battery with ��� 13.50 �. Before the switch is closed at time � = 0 �, the capacitor is uncharged. (a) What is the time constant? (b) What fraction of the final charge �K is on the capacitor at � = 48.5 �? (c) What fraction of the initial current �R is still flowing at � = 48.5 �?

Answer 1

What we have?

`R=12M\Omega=12*10^6\Omega\\c=1.75\mu F = 1.75*10^(-6)F\\V=13.50V\\`

a) We obtain the time constant

`\tau = Rc\\\tau= (12.10^6)(1.75*10^(-6))\\\tau=21s`

b) We have the time and this time is t=48.5s

With the charge on capacitor Q(t) whe can calculate the final charge, so

`Q(t)=Q_f(1-e^(-t/Rc))\\Q(t)=Q_f(1-^(48.5/21))\\Q(t)= Q_F(0.900691)`

Through the relation

`Q(t)=Q_f(0.9)=0.9Q_f\\(Q(t))/(Q_f)=0.9`

We conclude that the fraction of final charge Qf is on the capacitor at t=48.5s is 0.9

c) To calculate the current we have,

`I(t) = I_0*e^(-t/Rc)\\I(48.5)=I_0*e^(-48.5/2)\\I(t)=I_0(0.099308)\\I(t)=0.099I_0\\(I(t))/(I_0)=0.099`

The fraction of the initial current I:0 is still flowing at t=58.5 is 0.099