Question

a 0.75 μf capacitor is charged to 60 v. it is then connected in series with a 45 ω resistor and a 110 ω resistor and allowed to discharge completely. How much energy is dissipated by the 45Ω resistor? Express your answer with the appropriate units.

Answer 1

The energy dissipated by the 45Ω resistor is 0.027 Joules.

Step-by-step explanation:

To find the energy dissipated by the 45Ω resistor, we can use the formula:

E = 0.5 * C * V^2

Where E is the energy, C is the capacitance, and V is the voltage.

First, we need to convert the capacitance from μF to F:

0.75 μF = 0.75 * 10^-6 F

Next, we can calculate the energy:

E = 0.5 * (0.75 * 10^-6 F) * (60 V)^2

Plugging in the values and simplifying, we get:

E = 0.027 J

So the 45Ω resistor dissipates 0.027 Joules of energy.