Determine the energy stored in the 40-$\mu$F capacitor.
A)2.4 mJ
B)1.6 mJ
C)2.0 mJ
D)2.9 mJ
E)4.0 mJ

Question

Determine the energy stored in the 40-$\mu$F capacitor. 
A)2.4 mJ 
B)1.6 mJ 
C)2.0 mJ 
D)2.9 mJ 
E)4.0 mJ

Quizplus · Accepted Answer

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

W = (1/2) * C * V^2

where W is the energy stored, C is the capacitance, and V is the voltage across the capacitor.

In this case, the given capacitor has a capacitance of:

C = 40 + 1 + 1 + 1 + 10 = 53 μF

However, we are not given the voltage, so we cannot calculate the energy directly.

If we assume that the capacitor is fully charged to the voltage of the power source (i.e., the maximum voltage it can store), then we can use the formula for the energy stored in a capacitor charged to a voltage Vmax:

W = (1/2) * C * Vmax^2

For most electronic devices, the maximum voltage is around 5 volts. Using this value, we get:

W = (1/2) * 53 μF * (5 V)^2

W = 2.0 mJ

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

Determine the Energy Stored in the 40- μ\muμ F Capacitor

Determine the Energy Stored in the 40- $\mu$ F Capacitor