RC circuit problem?

Question

a switch is closed at t=0,connecting a battery of voltage V with a series RC circuit. (a)determine the ratio of the energy delivered to the capacitor to the total energy supplied by the source as a function of time

Answer 1

Unless I made a mistake somewhere...

Energy stored in the capacitor/Energy supplied =
Ec/Eee=
=(C/R)t/(exp(-t/RC))

Dave is on right track however does not quiet answers the question.
Let us first write equations (assuming series configuration)
Vee=Vr+Vc where
Vee - Battery voltage
Vr - Voltage across the resistor as a function of time
Vc - Voltage across the capacitor as a function of time

Vc(t)=Vee(1-exp(-t/RC))
Vr(t)= Vee exp(-t/RC)

Energy in =Energy dissipated in the resistor +
energy stored in the capacitor.

Energy = Power x time

Power supplied = Vee I(t) =
Since I(t)=Vr(t)/R=(1/R) Vee exp(-t/RC)
Power supplied =(1/R) (Vee)^2(exp(-t/RC))
Energy supplies is therefore = Ecc

Ecc=(t/R) (Vee)^2(exp(-t/RC))

Just for fun..
Power dissipated by resistor = Vr(t) I(t) =
=(1/R)(Vr(t))^2=
=(1/R) (Vee)^2(exp(-2t/RC))
Similarly
Er = (t/R) (Vee)^2(exp(-2t/RC))

Finally since amount of charge stored in the capacitor is
Q = C*V,
Ec=VeeQ=CVee^2

And finally the answer to your question:

determine the ratio of the energy delivered to the capacitor to the total energy supplied by the source as a function of time
Energy delivered to the capacitor(Ec) /Energuy in (Eee)

Energy in =Energy dissipated in the resistor +
energy stored in the capacitor.

We found that Ecc=(t/R) (Vee)^2(exp(-t/RC))
And Ec=VQ=CVee^2 so

Ec/Eee= (CVee^2)/ (t/R) (Vee)^2(exp(-t/RC))=
=(C/Rt)/(exp(-t/RC))

Answer 2

one R*C time constant is sqrt3/2 = .866 or 86.6% of Vpeak.
R*C is measured in seconds.
10K * 100uF = 1sec, etc

Oops...how about .633 or 63.3% after each time constant. sorry.