a question i cant solve

Thread Starter

needshelp

Joined Mar 23, 2004
2
here is the question..if anyone can solve it id be grateful:

Consider a battery of voltage V which charges a capacitor of capcitance C through a resistor of resistance R. Derive an expression for the work done by the battery in charging the capacitor. Careful: in the process there is both energy dissipated in the resistor and stored in the capacitor. [Hint: Find the power output of the battery as a function of time, and integrate it to find the work done.]

thanks alot :eek:
 

haditya

Joined Jan 19, 2004
220
hi
Work done= (power dissapated across resistor*time)+(energy stored in capacitor)

Current through the circuit is given by
i=[E e^(-t/RC)]/R
where E is the emf of the source
Power dissapted across the resistor is (i^2)R
The energy stored in a capacitor is given by 0.5(dqV)
where dq is the charge on the capacitor at any instant of time and V is the voltage drop across the capcitor at the same instant of time.
We know that dq=i dt and V=E[1-(e^-t/RC)]

In dt time energy dissapated accross the resistor is = (i^2)R dt

Therefore the expression for work done by the battery in time dt is
dw = (i^2)R dt + 0.5dqv
dw = {[E e^(-t/RC)]/R}^2 (R dt) +{[E e^(-t/RC)]/R }{E[1-(e^-t/RC)]}dt

integrating this expression give the work done
some simplification of the expressions is necessary before integrating which i am avoiding in order to save on steps ( btw its your homework!!!!)

hope this helps you
 

Dave

Joined Nov 17, 2003
6,970
Good answer haditya, however one small point:

The energy stored in the capacitor is given by 0.5(qV)

Correct me if you think I'm wrong, but if I remember:

Energy stored W = ∫ vi dt = ∫ v dq

i.e. dq = i dt

From v = qC

∫ q/C dq from the limits 0 to Q gives
W = 0.5 Q²/C = 0.5 CV²
 

haditya

Joined Jan 19, 2004
220
Yes.. what you are saying is true... what i have done is the following:-

Energy stored in the cap is dqv, where dq and v are the charge stored on the cap and "potential diff across the cap" respectively at an arbitrary time t
Then i have written the voltage v and charge q as a function of t so as to facilitate integration with respect to time
 

Dave

Joined Nov 17, 2003
6,970
Originally posted by haditya@Mar 28 2004, 05:51 AM
Yes.. what you are saying is true... what i have done is the following:-

Energy stored in the cap is dqv, where dq and v are the charge stored on the cap and "potential diff across the cap" respectively at an arbitrary time t
Then i have written the voltage v and charge q as a function of t so as to facilitate integration with respect to time
Ok, I see what you have done and although I haven't checked through with the original question I'm sure its right ;)
 
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