A 6.44 E+06 ohm resistor and a 1.00 microfarad capacitor are connected in a single loop circuit with a seat of emf of E = 2.00 V. At 6.00 s. after the connection is made, what is the rate at which energy (in joules/second) is being delivered by the seat of emf?

Answer 1

R = 6.44*10^6 Ohms.
C = 1.00 uF = 1*10^-6 Farads.

n=t/R*C=6 / (6.44*10^6 * 1*10^-6)=0.932

Vc = E-Vr.
Vr = E/e^n. = 2/e^0.932 = 0.788 Volts.
I=V/R = 0.788/6.44*10^6=1.22*10^-7 Amps

Energy=E*I*T=(2*1.22*10^-7)*6=1.47*10^-6
Joules.

Answer 2

Correction:

Energy=E*I=2 * 1.22*10^-7=2.45*10^-7 Watts = 2.45*10^-7 Joules/s.