Question

A 5.00 μF μ F capacitor is initially charged to a potential of 17.0 V V . It is then connected in series with a 4.00 mH m H inductor. What is the total energy stored in this circuit? Express your answer in joules. What is the maximum current in the inductor? Express your answer in amperes.

Answer 1

The energy stored in the circuit can be found using the equation:

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

where E is the energy stored, C is the capacitance, and V is the potential difference across the capacitor.

Substituting the given values, we get:

E = (1/2) * 5.00 μF * (17.0 V)^2 = 2.83 mJ

Therefore, the total energy stored in the circuit is 2.83 millijoules.

The maximum current in the inductor can be found using the equation:

I = (1/L) * √(2E)

where I is the maximum current, L is the inductance, and E is the energy stored in the circuit.

Substituting the given values, we get:

I = (1/4.00 mH) * √(2 * 2.83 mJ) ≈ 1.68 A

Therefore, the maximum current in the inductor is approximately 1.68 amperes.

Step-by-step explanation: