Question:

A capacitor with a capacitance of 10 µF is connected to a power supply with a voltage of 12 V. The capacitor is initially uncharged. After being connected to the power supply for 5 seconds, the voltage across the capacitor reaches 9 V. Calculate the charge stored in the capacitor and the energy stored in it.

Explanation:

We are given the capacitance of the capacitor (C = 10 µF), the voltage across the capacitor at time t=5 seconds (V = 9 V), and the initial voltage across the capacitor when it is uncharged (V₀ = 0 V). We need to find the charge (Q) stored in the capacitor and the energy (U) stored in it.

The charge stored in a capacitor can be calculated using the formula:

Q = C * V

The energy stored in a capacitor can be calculated using the formula:

U = (1/2) * C * V²

where Q is the charge, C is the capacitance, V is the voltage, and U is the energy.

Let's calculate the charge stored in the capacitor:

Q = C * V

= (10 µF) * (9 V)

= 90 µC

Therefore, the charge stored in the capacitor is 90 µC.

Now, let's calculate the energy stored in the capacitor:

U = (1/2) * C * V²

= (1/2) * (10 µF) * (9 V)²

= (1/2) * (10 × 10^-6 F) * (9 V)²

= 0.405 µJ

Therefore, the energy stored in the capacitor is 0.405 µJ.

Finally, the charge stored in the capacitor is 90 µC and the energy stored in the capacitor is 0.405 µJ.