A 50 µF capacitor is discharged through a 100 ko resistor. If the capacitor was initially charged to 400 V, determine the energy stored after 600 ms. A. 5.13 J B. 2.40 J C. 4.00 J D. 3.14 J

The energy stored in a capacitor is given by the formula:

E = 0.5 * C * V^2

where: E is the energy (in joules), C is the capacitance (in farads), and V is the voltage (in volts).

The voltage across a discharging capacitor after a time t (in seconds) is given by the formula:

V = V0 * e^(-t/RC)

where: V0 is the initial voltage (in volts), R is the resistance (in ohms), C is the capacitance (in farads), and e is the base of the natural logarithm (approximately equal to 2.71828).

First, we need to find the voltage across the capacitor after 600 ms. Convert 600 ms to seconds by dividing by 1000, which gives 0.6 s.

V = 400 * e^(-0.6/(1000005010^-6)) V = 400 * e^(-0.6/5) V = 400 * e^(-0.12) V ≈ 400 * 0.8869 V ≈ 354.76 V

Then, we can find the energy stored in the capacitor.

E = 0.5 * 5010^-6 * (354.76)^2 E ≈ 0.5 * 5010^-6 * 125,752.7376 E ≈ 3.1443 J

So, the closest answer is D. 3.14 J.