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Inductive reactance is a key concept in AC circuit analysis, especially when dealing with inductors in the circuit. It measures how much an inductor opposes the change in current within an AC circuit.
This opposition is especially more noticeable in circuits with higher frequencies. The formula to calculate this reactance is:- \(X_L = 2\pi f L\)
where:

• \( X_L \) is the inductive reactance in ohms (\(\Omega\)),
• \( f \) is the frequency of the alternating current in hertz (Hz), and
• \( L \) is the inductance in henrys (H).

This formula shows that the reactance increases with both frequency and inductance. For instance, in the given exercise, with a frequency of 620 Hz and an inductance of 8.2 mH, the inductive reactance is 31.894 ohms. This reactance is what makes the circuit less eager to let current through, acting much like resistance in a DC circuit does but in a time-varying manner.

Ohm's law for AC circuits extends the familiar linear relationship of voltage, current, and resistance into alternating current. In AC circuits, the role of resistance is played by impedance, which can be more complex due to reactance components. Specifically, for an inductive circuit, we focus on inductive reactance.
The formula for Ohm's law in the context of RMS values (which stand for root mean square) is given by:- \(I_{rms} = \frac{V_{rms}}{X_L}\)
where:

• \( I_{rms} \) is the RMS current,
• \( V_{rms} \) is the RMS voltage, and
• \( X_L \) is the inductive reactance.

In our exercise, with an RMS voltage of 10 V and an inductive reactance of 31.894 ohms, the RMS current comes out to be approximately 0.3138 A. This current value is crucial in determining how the circuit will respond to the input AC voltage.

When analyzing AC circuits, the peak current is an essential quantity, as it represents the maximum current flowing through the circuit. Unlike DC circuits, AC circuits have varying currents; thus, understanding peak values gives insight into the highest electrical demand on the circuit.
The peak current \(I_0\) can be found from the RMS current using the relationship:- \(I_0 = \sqrt{2} \times I_{rms}\)
where \( I_{rms} \) is the current in RMS terms. The factor \(\sqrt{2}\) arises due to how sinusoidal waveforms, like standard AC, have their peak and average levels related mathematically.
In the given scenario, using the RMS current of 0.3138 A, the peak current calculates to approximately 0.4435 A. Knowing this peak current is important for assessing the performance and capacity requirements of the AC circuit under consideration.