An a.c. voltage E = E₀ sin ωt is applied across an inductor L. Obtain an expression for current I.

Solution

Let, a pure inductance L connected across a source of alternating emf given by E = E₀ sin ωt

The rate of change of current in the circuit is

\frac{dI}{dt} .

(If I is the current through the circuit).

∴ Instantaneous induced emf across inductance = -L

\frac{dI}{dt}

(By Kirchhoff's law)

∴ The net emf is

E - L \frac{dI}{dt} = 0

E = L \frac{dI}{dt}

\Rightarrow

dI = \frac{E}{L} dt = \frac{E_{0}}{L} \sin ωt dt

\int dI = \frac{E_{0}}{L} \int \sin ωt dt

I = - \frac{E_{0}}{ωL} \cos ωt = - I_{0} \cos ωt

where,

I_{0} = \frac{E_{0}}{ωL} is the peak value of alternating current .

Here, ωL has the units of resistance and it is called inductive reactance.

Current, I in the circuit is expressed as,

I = – I₀ cos ωt
= I₀ sin (ωt –

π

/2).

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