rate at which energy is delivered in a rc circuit

Question

A 3 mega ohm resistor and an uncharged 10^(-6) F capacitor are connected in a single loop circuit with a constant source of 4 volt . Show that at one second after the connection is made , the rate at which energy is being delivered by the source is 16/3 e^-1/3 10^-6 J/s.

Answer 1

The solution for current as a function of time in a charging RC circuit (and actually a discharging one as well) is given by:

 

i(t) = (ε/R)e^-t/RC

 

where ε = battery EMF (voltage), R = resistance in ohms, and C = capacitance in farads.  

 

Note that this follows what we expect the circuit to do.  At the start, t=0, the capacitor is uncharged and current is flowing freely, so i should just be given by V=iR through the resistor, with V = ε.  After a long time (t --> ∞), the capacitor is fully charged, which stops current flow and results in a current of zero.

 

Since you are given R, C, and ε, you can completely evaluate the right hand side of this equation at t = 1 s.  Then the power being delivered by the source is just P = IV, where V is the battery voltage again.

 

Using this information, I obtained what was desired to be shown.  If you would like to go into more detail, I am available for online appointments.  Just let me know!