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Using Phasor Diagram, Derive the Expression for the Current Flowing in an Ideal Inductor Connected to an A.C. Source of Voltage,

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प्रश्न

Using the phasor diagram, derive the expression for the current flowing in an ideal inductor connected to an a.c. source of voltage, v= vo sin ωt. Hence plot graphs showing the variation of (i) applied voltage and (ii) the current as a function of ωt.

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उत्तर

Figure (a) shows an ac source connected to an inductor. Usually, inductors have appreciable resistance in their windings, but we shall assume that this inductor has negligible resistance. Thus, the circuit is a purely inductive ac circuit. Let the voltage across the source be V = Vm sin ωt. Using the Kirchhoff’s loop rule, Σε (t) = 0, and since there is no resistor in the circuit,

`v - L(di)/(dt) = 0`                                 ...(i)

where the second term is the self-induced Faraday emf in the inductor; and L is the self-inductance of the inductor. The negative sign follows from Lenz’s law.

From equation (i) we have

`(di)/(dt) = v/L = v_m/L sin ωt`                ...(ii)

Equation (ii) implies that the equation for i(t), the current as a function of time, must be such that its slope di/dt is a sinusoidally varying quantity, with the same phase as the source voltage and an amplitude given by .vm/L

To obtain the current, we integrate di/dt with respect to time:

`int_  (d"i")/(d"t") d"t" = "v"_"m"/"L" int_  sin(wt)dt`

and get

`"i" = -("v"_"m")/(w"L") cos (wt) + cons tan t`

The integration constant has the dimension of current and is time-independent. Since the source has an emf which oscillates symmetrically about zero, the current it sustains also oscillates symmetrically about zero, so that no constant or time-independent component of the current exists. Therefore, the integration constant is zero.

Using

`-cos (wt) = sin(wt - pi/(2)), "we have"`

`"i" - "i"_"m" sin(wt - pi/(2))`

Where `"i"_"m" = ("v"_"m")/(w"L")` is the amplitude of the current.

The quantity ωL is analogous to the resistance and is called inductive reactance, denoted by XL :

`"X"_"L" = w"L"`

The amplitude of the current is, then

`"i"_"m" = ("v"_"m")/("X"_"L")`

The dimension of inductive reactance is the same as that of resistance and its SI unit is ohm (Ω). The inductive reactance limits the current in a purely inductive circuit in the same way as the resistance limits the current in a purely resistive circuit. The inductive reactance is directly proportional to the inductance and to the frequency of the current.

Fig. (b) A Phasor diagram for the circuit in Fig.(a) Fog. (c) Graph of v and i versus wt.

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