When the frequency of a purely inductive circuit is decreased, what happens to the current in the circuit?

In a purely inductive circuit, the relationship between frequency and current is governed by the inductive reactance, which is determined by the formula ( X_L = 2\pi f L ), where ( X_L ) is the inductive reactance, ( f ) is the frequency, and ( L ) is the inductance.

As frequency decreases, the inductive reactance also decreases. Inductive reactance opposes the flow of alternating current (AC); therefore, a lower reactance allows more current to flow through the circuit. However, when considering the effects of a decrease in frequency, it's essential to note that the overall impedance of the circuit also plays a role.

In a purely inductive circuit with a constant voltage source, a decrease in frequency results in an increase in the current since ( I = \frac{V}{Z} ), where ( Z ) is the total impedance. Since impedance is influenced by the inductive reactance, reducing the frequency, which lowers the reactance, increases the current in the circuit up to a point.

When assessing the choices, the correct conclusion is that the current in the circuit increases when the frequency is decreased, as a lower frequency means less opposition