series circuit?

Question

A series circuit of 12V source of emf, a 2.00 mF capacitor and a 1000 ohms resistor and switch. When the switch is closed, how long does it take for the current to reach one half its maximum value?

a) 0.693 s
b) 1.39 s
c) 2.00 s
d) 1.69 s
e) 0.0250 s

please help, thanks

Answer 1

Are you serious?

It is b)
here is the proof

V=Vmax exp(-t/(RC))
I=V/R= (Vmax/R)exp(-t/(RC))

I(1/2 max)= Vmax/2R=(Vmax/R)exp(-t/(RC))
1/2= exp(-t/(RC))
-t/(RC)= ln(1/2)
t/(RC)= ln(2)
t= RCln(2)
t=1 E+3 ohm x 2 E-3 F ln(2)=1.39 sec

Answer 2

The relationship between instantaneous current at time t and steady-state current (Imax)for a charging capacitor is:
I=I(max)*(1-1/e^n) where n is the number of time constants.
The time constant of an RC circuit is T=RC so you can write:
I=I(max)*(1-1/e^(t/RC).
The rest is just maths.