A 24-V battery is attached to capacitor and a 100-ohm resistor.If 1.73 J of energy are dissipated by the resistor while the capacitor is charging,what is the capacitance of the capacitor?
A)6.0 mF
B)3.0 mF
C)8.3 mF
D)4.5 mF
E)7.2 mF

Question

Quizplus · Accepted Answer

The energy dissipated by the resistor is given by:$$E = \frac{1}{2}CV^2$$where:* E is the energy dissipated (in joules)* C is the capacitance (in farads)* V is the voltage (in volts)We can rearrange this equation to solve for C:$$C = \frac{2E}{V^2}$$Plugging in the given values, we get:$$C = \frac{2(1.73 J)}{(24 V)^2} = 6.0 mF$$Therefore, the capacitance of the capacitor is 6.0 mF.

A 24-V Battery Is Attached to Capacitor and a 100-Ohm