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Find out the phase relationship between voltage and current in a pure inductive circuit. - Physics

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प्रश्न

Find out the phase relationship between voltage and current in a pure inductive circuit.

दीर्घउत्तर

उत्तर

Consider a circuit containing a pure inductor of inductance L connected across an alternating voltage source. The alternating voltage is given by the equation.

υ = Vm sin ωt …(1)

The alternating current flowing through the inductor induces a self-induced emf or back emf in the circuit. The back emf is given by

Back emf, ε -L `"di"/"dt"`

By applying Kirchoff’s loop rule to the purely inductive circuit, we get


AC circuit with inductor

υ + ε = 0

Vm sin ωt = L`"di"/"dt"`

di = L`"V"_"m"/"L"` sin ωt dt

i = `"V"_"m"/"L" int` sin ωt dt = `"V"_"m"/"L"_omega`  (-cos ωt) + constant

The integration constant in the above equation is independent of time. Since the voltage in the circuit has only time dependent part, we can set the time independent part in the current (integration constant) into zero.

`[(cos omega"t" = sin(pi/2 - omega"t")),(- sin (pi/2 - omega"t") = sin (omega"t" - pi/2))]`

i = `"V"_"m"/"L"_omega sin (omega"t" - pi/2)  or ` i = `"I"_"m" sin(omega"t" - pi/2)`  ....(2)

where `"V"_"m"/"L"_omega = "I"_"m"`, the peak value of the alternating current in the circuit. From equation (1) and (2), it is evident that current lags behind the applied voltage by `pi/2` in an inductive circuit.
This fact is depicted in the phasor diagram. In the wave diagram also, it is seen that current lags the voltage by 90°.

Inductive reactance XL:

The peak value of current Im is given by Im = `"V"_"m"/"L"_omega`. Let us compare this equation with Im = `"V"_"m"/"R"` from resistive circuit. The equantity ωL Plays the same role as the resistance in resistive circuit. This is the resistance offered by the inductor, called inductive reactance (XL). It is measured in ohm.

XL = ωL

The inductive reactance (XL) varies directly as the frequency.

XL = 2πfL …….. (3)

where ƒ is the frequency of the alternating current. For a steady current, ƒ= 0. Therefore, XL = 0. Thus an ideal inductor offers no resistance to steady DC current.


Phasor diagram and wave diagram for AC circuit with L

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पाठ 4: Electromagnetic Induction And Alternating Current - Evaluation [पृष्ठ २६१]

APPEARS IN

सामाचीर कलवी Physics - Volume 1 and 2 [English] Class 12 TN Board
पाठ 4 Electromagnetic Induction And Alternating Current
Evaluation | Q III. 20. | पृष्ठ २६१

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