A 24-V battery is attached to a 3.0-mF capacitor and a 100-ohm resistor.What is the energy dissipated as heat inside the resistor while the capacitor is charging.
A)0.036 J
B)0.30 J
C)0.86 J
D)1.7 J
E)5.76 J

Question

Quizplus · Accepted Answer

The time constant for the circuit is given by τ = RC = (100 Ω)(3.0 mF) = 0.3 s
After one time constant, the capacitor will have charged to approximately 63% of its final voltage. Therefore, the final charge on the capacitor is given by:
Q = CV = (3.0 × 10^-3 F)(24 V) = 0.072 C
The energy stored in the capacitor is given by:
E = (1/2)CV^2 = (1/2)(3.0 × 10^-3 F)(24 V)^2 = 2.59 J
The energy dissipated by the resistor is equal to the energy stored in the capacitor, since all the energy transferred from the battery to the circuit ultimately ends up as heat in the resistor. Therefore:
E = 2.59 J
The closest answer choice is C = 0.86 J.

A 24-V Battery Is Attached to a 3