AC Voltage Applied to an Inductor: Concepts, Formulae, Videos, Examples

We will start with making an Inductor, let’s take an insulated wire and a pencil. Take the wire and wind it around the pencil and then slide the pencil away from the wire. We will be left with the coiled wire. This coil is an inductor which is a very important circuit element. Let’s study this circuit element in an AC circuit.

Inductor with AC biasing

So now we come to the topic which is AC Voltage applied to an Inductor. This inductor can block or oppose the alternating current flowing through it. We will take two cases. In the first case, we have an inductor which is connected to a DC supply whereas in the second case we have an inductor which is connected to an AC supply.

In the first case, we see that a constant current flows through this inductor and the bulb connected to it glows brightly. Hence the inductor simply behaves like a coil of wire. But in the second case, the bulb does not glow as brightly as the first one, this happens because the inductor opposes the flow of alternating current. Hence the bulb is dim. Now we will use Lenz’s Law to explain opposition to the flow of AC current through an Inductor:

Inductor

Recall that an alternating current varies in magnitude as well as in direction. Now when the current increases from zero to peak value during this time interval the current through the coil increases which means the magnetic field linked to the coil also increases. Hence there will be an induced EMF and the direction of the induced EMF is given by the Lenz’s Law as:

\(E=-\frac{\mathrm{d}\phi }{\mathrm{d} t}\)

In case during a given time interval, the current decreases from a maximum value to zero, the magnetic field linked with the coil decreases. Hence, the induced EMF is given by :

\(E=-L\frac{\mathrm{d} i}{\mathrm{d} t}\)

Since\({\phi =L\times i}\)

The Voltage required by the AC source is given by:

\(V= L\times \frac{\mathrm{d}i }{\mathrm{d} t}\)

To maintain the current, supplied voltage must be equal to the reverse Emf. So the Voltage applied to the coil is therefore given by:

\(V= L\times \frac{\mathrm{d}i }{\mathrm{d} t}\)

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Relation between Current and Voltage across the Coil

\(V= L\times \frac{\mathrm{d}i }{\mathrm{d} t}\)

\(\frac{\mathrm{d} i}{\mathrm{d} t}= \frac{V}{L}\)

\(\frac{\mathrm{d} i}{\mathrm{d} t}= \frac{V_{m}sin\omega t}{L}\)

Where, V_m = peak value of the voltage and ω = angular frequency. On integrating the above expression we get:

\(i=\int \frac{V_{m}sin\omega t}{L} dt\)

\(i=\frac{-V_{m}cos\omega t}{\omega L}\)

\(i=\frac{V_{m}sin\left ( \omega t – \frac{\pi }{2} \right )}{\omega L}\)

\(i=i_{m}sin\left ( \omega t – \frac{\pi }{2} \right )\)

where i_m = peak value of current. So, we may conclude that the current lags the voltage by 90 degrees, which means the current reaches the same value as the voltage after a quarter cycle.

Expression for Inductive Reactance

\(V= V_{m}sin\omega t\)

\( i=i_{m}sin\left ( \omega t – \frac{\pi }{2} \right )\\ i=-i_{m}cos\omega\)

\( i_{m}= \frac{V_{m}}{\omega L}\\ i_{m}= \frac{V_{m}}{X_{L}}\)

Where X_L = Inductive Reactance
Inductive reactance\({X_{L}}=\omega \times L\)
Inductive reactance\(X_{L}=2\pi f L\)

The SI unit of Inductive Reactance is Ohm which is same as that of the resistance. The inductive reactance is proportional to the angular frequency. So when frequency increases the value of the reactance also increases and vice versa. Therefore if you have a DC current for which the value of f = 0 then X_L= 0.

In other words, no opposition is offered to the flow of DC. Therefore the lamp glows brightly when the DC source is applied because of extremely low or zero resistance.

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