Determine the energy stored in the 40-μF capacitor.

Question

Quizplus · Accepted Answer

The formula for energy stored in a capacitor is given by:
Energy = 0.5 x C x V^2
where C is the capacitance and V is the voltage across the capacitor.

In this case, the capacitance is given as 40 μF. The voltage across the capacitor is not given, so we cannot directly calculate the energy stored. However, we can use the information provided in the circuit diagram.

The 10 Ω resistor and the 20 Ω resistor are in parallel, so we can find their equivalent resistance as:
1/R = 1/10 + 1/20 = 1/6
R = 6 Ω

The total resistance in the circuit is then:
Rtot = R + 30 = 36 Ω

The voltage across the entire circuit is given as 12 V, so we can use Ohm's law to find the current through the circuit:
I = V/Rtot = 12/36 = 0.333 A

Since the 20 Ω resistor has twice the resistance of the 10 Ω resistor, it will drop twice the amount of voltage. Therefore, the voltage across the 20 Ω resistor is:
V20 = 2/3 x 12 V = 8 V

The voltage across the 10 Ω resistor is then:
V10 = 1/3 x 12 V = 4 V

Finally, the voltage across the 40 μF capacitor is equal to the voltage across the 10 Ω resistor, so:
Vcap = V10 = 4 V

Now we can calculate the energy stored in the capacitor:
Energy = 0.5 x C x Vcap^2 = 0.5 x 40 x 10^-6 x 16 = 0.00032 J = 0.32 mJ

Therefore, the best answer choice is C) 2.0 mJ.

Determine the Energy Stored in the 40-μF Capacitor