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AC Voltage Applied To An Inductor | AC Circuit Containing Inductor Only

AC Voltage Applied To An Inductor | AC Circuit Containing Inductor Only :- Suppose a pure inductor L (the conducting wire from which an inductor is made does have some resistance, but we assume this resistance to be negligible) is connected to an alternating voltage source, as shown in figure (a) below :

Suppose the alternating voltage is represented by the following equation :

$\displaystyle E={{E}_{0}}sin\text{ }\omega t$ …..(1)

If dI/dt is the rate of change of current in the circuit at any instant, then at that moment the induced electromotive force across the inductor L is:

$\displaystyle e=-L\frac{dI}{dt}$ …..(2)

Here, the negative sign indicates that the induced electromotive force (e) opposes the change in current. To maintain a continuous flow of electric current in the circuit, the applied alternating voltage (E) must be equal in magnitude and opposite in direction to the induced emf (e), that is,

$\displaystyle E=-\left( e \right)=-\left( -L\frac{dI}{dt} \right)$

$\displaystyle \Rightarrow {{E}_{0}}sin\text{ }\omega t=L\frac{dI}{dt}$

$\displaystyle \Rightarrow dI=\left( \frac{{{E}_{0}}}{L}sin\text{ }\omega t \right)dt$

Integrating both sides,

$\displaystyle \int{dI}=\int{\left( \frac{{{E}_{0}}}{L}sin\text{ }\omega t \right)dt}$

$\displaystyle \Rightarrow \int{dI}=\frac{{{E}_{0}}}{L}\int{\left( sin\text{ }\omega t \right)dt}$

$\displaystyle I=\frac{{{E}_{0}}}{L}\left( \frac{-cos\text{ }\omega t}{\omega } \right)+c$ …..(3)

Here, c is a constant of integration whose dimensions are those of electric current. Now, since the applied electric source is an alternating source, the current flowing in the circuit is also alternating. This means that the value of the electric current cannot be constant at any time. Therefore, the value of the constant of integration c is zero. Hence, from equation (3)…

$\displaystyle I=-\frac{{{E}_{0}}}{\omega L}cos\text{ }\omega t$

$\displaystyle \Rightarrow I=-\frac{{{E}_{0}}}{\omega L}sin\left( \frac{\pi }{2}-\omega t \right)$

$\displaystyle I=\frac{{{E}_{0}}}{\omega L}sin\left( \omega t-\frac{\pi }{2} \right)$ …..(4)

The maximum value of the current,

$\displaystyle {{I}_{0}}=\frac{{{E}_{0}}}{\omega L}$ …..(5)

Hence, the electric current flowing in the circuit,

$\displaystyle I={{I}_{0}}sin\left( \omega t-\frac{\pi }{2} \right)$ …..(6)

From equations (1) and (6),

The phase of the applied alternating electromotive force (emf) = ωt

The phase of alternating current = ωt – (π/2)

Thus, in a purely inductive (L-only) circuit, the alternating current lags behind the alternating emf by a phase angle of (π/2). This fact is illustrated in the phasor diagram in figure (b) and in the waveform diagram in figure (c).

Inductive Reactance – X_L

Comparing equation (5) with Ohm’s law, we find that the effective resistance offered in the path of the electric current by the inductor is ωL. Therefore, ωL is called the inductive resistance or inductive reactance. Hence, the inductive reactance is :

$\displaystyle {{X}_{L}}=\omega L=2\pi \nu \text{L}$ …..(7)

Where ν is the frequency of the alternating voltage source. For a given inductor,

$\displaystyle {{X}_{L}}\propto \nu$ …..(8)

Thus, the inductive reactance is directly proportional to the frequency, and therefore, the graph between X_L and ν is a straight line :-

For direct current, the frequency ν = 0. Hence, $\displaystyle {{X}_{L}}=0$ , which means that a pure inductor offers no resistance in the path of direct current.

Unit of Inductive Reactance

$\displaystyle {{X}_{L}}=\omega L\Rightarrow \frac{1}{\sec .}\left( henry \right)=\frac{1}{\sec .}\left( \frac{Volt}{Amp.\sec {{.}^{-1}}} \right)=\frac{Volt}{Amp.}=ohm$

Hence S.I. unit of inductive reactance is Ohm (Ω).

Inductive Susceptance (S_L)

The reciprocal of inductive reactance is called inductive susceptance or inductive conductance.

$\displaystyle {{S}_{L}}=\frac{1}{{{X}_{L}}}=\frac{1}{\omega L}$ …..(9)

S.I. unit of S_L is Ω^-1 or mho.

Average Power

(AC Voltage Applied To An Inductor | AC Circuit Containing Inductor Only)

The instantaneous power of a purely inductive circuit,

$\displaystyle P=E\times I=\left( {{E}_{0}}sin\text{ }\omega t \right)\times \left[ {{I}_{0}}sin\left( \omega t-\frac{\pi }{2} \right) \right]$

$\displaystyle \Rightarrow P={{E}_{0}}{{I}_{0}}\left( sin\text{ }\omega t \right)\times \left[ sin\left( \omega t-\frac{\pi }{2} \right) \right]={{E}_{0}}{{I}_{0}}\left( sin\text{ }\omega t \right)\left( -cos\text{ }\omega t \right)$

$\displaystyle \Rightarrow P=-{{E}_{0}}{{I}_{0}}\left( sin\text{ }\omega t \right)\left( cos\text{ }\omega t \right)$

$\displaystyle \Rightarrow P=\frac{-{{E}_{0}}{{I}_{0}}}{2}\left( 2\text{ }sin\omega t\text{ }cos\omega t \right)$

$\displaystyle \Rightarrow P=\frac{-{{E}_{0}}{{I}_{0}}}{2}sin2\omega t$

Now, over one complete cycle, the value of (sin 2ωt) becomes zero; therefore, over one complete cycle, the average power of the circuit is :-

$\displaystyle {{P}_{avg.}}=\langle \frac{-{{E}_{0}}{{I}_{0}}}{2}sin2\omega t\rangle =\frac{-{{E}_{0}}{{I}_{0}}}{2}\langle sin2\omega t\rangle =\frac{-{{E}_{0}}{{I}_{0}}}{2}\times 0=0$

Thus, the average power supplied to a purely inductive circuit over a complete cycle is zero.

Example 1.

(NCERT Example 7.2)

A pure inductor of 25.0 mH is connected to a source of 220 V. Find the inductive reactance and rms current in the circuit if the frequency of the source is 50 Hz.

Solution :

Inductive reactance,

$\displaystyle {{X}_{L}}=2\pi \nu \text{L}=2\times 3.14\times 50\times 25\times {{10}^{-3}}\Omega$

$\displaystyle \therefore {{X}_{L}}=7.85\Omega$

rms value of the current in the circuit,

$\displaystyle {{I}_{rms}}=\frac{{{E}_{rms}}}{{{X}_{L}}}=\frac{220V}{7.85\Omega }=28A$

Example 2.

(NCERT Example 7.5)

A light bulb and an open coil inductor are connected to an ac source through a key as shown in figure below :

The switch is closed and after sometime, an iron rod is inserted into the interior of the inductor. The glow of the light bulb (a) increases; (b) decreases; (c) is unchanged, as the iron rod is inserted. Give your answer with reasons.

Solution :

As the iron rod enters the coil, the magnetic field inside the coil magnetizes the rod, which increases the magnetic field inside the coil. Therefore, the inductance of the coil increases. As a result, the inductive reactance of the coil also increases. Thus, most of the applied AC voltage appears across the inductor, and the voltage across the bulb decreases. Hence, the brightness of the bulb reduces.

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