# Video: Explaining Why an AC-Inductive Circuit Approximates an Open Circuit for High Frequency and Negligible Resistance

Explain why a circuit containing an inductive coil with negligible resistance, and an AC source is considered to be an open circuit at very high frequencies.

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### Video Transcript

Explain why a circuit containing an inductive coil with negligible resistance and an AC source is considered to be an open circuit at very high frequencies.

We can start on our solution by sketching out a circuit with just these elements: an AC source and an inductive coil. Considering that this circuit has negligible resistance, we might think that the current generated by the source will have no impediment to its flow even at very high frequencies.

It will be helpful though for us to recall that the impedance of a circuit; that is, itβs resistance to current flow, depends not just on resistance π, but also on a term called inductive reactance. In our case, since the resistance π is negligible, we can say that impedance is equal to inductive reactance π sub πΏ. But what is this term?

Inductive reactance is equal to two π times the frequency of oscillation of the circuit multiplied by the self-induction coefficient πΏ. Inductive reactance has units of ohms. And this is accurate because it does indeed resist the flow of current.

In our particular circuit, whatever the value of πΏ, we know that as π gets higher and higher and higher, so will π sub πΏ and so will the impedance to current flow. At very high frequencies, the impedance will be so great that current wonβt be able to flow through this circuit.

In other words, it will perform just like an open circuit: a circuit with a break in it, where no current flows. We can say then that inductive reactance increases proportionately to frequency, reducing the current.

This is why at very high frequencies such a circuit behaves like an open circuit.