## What is called an Inductor?

An inductor is component used in electrical or electronic circuits to store energy. The energy stored in inductor is in the form of magnetic energy.

- An inductor impedes or opposes any change in the current flowing through it. This property of an inductor is called its inductance.
- When an AC is applied to an inductor, then the circuit is called an inductive circuit.
- If a pure inductor or ideal inductor is used, it is called a
.**Pure Inductive Circuit**

### Inductance

When any change occurred in the electric current flowing through an inductor, it opposes the change in the current by inducing an EMF which drives a current in opposite direction. This property of an inductor is called its inductance.

*Inductance of an inductor is defined as the ratio of the induced EMF to the rate of change in the current.*

If, for rate of change in current \left ( \frac {dI}{dt} \right ) , the induced EMF in an inductor is ( \epsilon ) , then its inductance will be –

L = \left [ \frac {\epsilon}{\frac {dI}{dt}} \right ]

- Inductance is a result of the induced magnetic field on the coil.

## AC current applied to an Inductor

Consider about a pure inductive circuit as shown in figure. AC applied to a pure inductor of *inductance* ( L ) .

- The AC voltage applied to inductor is given by –

V = V_0 \sin ( \omega t ) ……… (1)

### Phases of AC current in Inductor

- The
*induced EMF*across the inductor will be –

\epsilon = - L \left ( \frac {dI}{dt} \right )

Minus sign indicates that induced emf opposes the growth of *current* in the circuit.

- In a pure inductive circuit, potential drop is zero. Therefore –

V + \left ( - \frac {dI}{dt} \right ) = 0

Or, \quad \left ( \frac {dI}{dt} \right ) = \left ( \frac {V}{L} \right )

From equation (1), we get –

\left ( \frac {dI}{dt} \right ) = \left ( \frac {V_0}{L} \right ) \sin ( \omega t )

Or, \quad dI = \left ( \frac {V_0}{L} \right ) \sin ( \omega t ) dt …….. (2)

Integrating both sides of the expression, we get –

\int dI = \left ( \frac {V_0}{L} \right ) \int \sin ( \omega t ) dt

So, \quad I = \left ( \frac {V_0}{L} \right ) \left [ - \frac {\cos ( \omega t )}{\omega} \right ]

= \left ( \frac {V_0}{L \omega} \right ) \left [ - \cos ( \omega t ) \right ]

Also from trigonometric relations, we get \left ( - \cos \omega t \right ) = \sin \left ( \omega t - \frac {\pi}{2} \right ) .

Therefore, \quad I = \left ( \frac {V_0}{L \omega} \right ) \sin \left ( \omega t - \frac {\pi}{2} \right )

### Peak value of AC current in Inductor

Comparing above expression with the standard expression of alternating current [ I = I_0 \sin ( \omega t ) ] , we get the *peak value* of current as –

I_0 = \left ( \frac {V_0}{L \omega} \right )

Therefore, \quad I = I_0 \sin \left ( \omega t - \frac {\pi}{2} \right ) ………. (3)

Comparing equations (1) and (3), for AC applied to an inductor, we conclude that –

- Voltage and current have phase difference of \left ( \frac {\pi}{2} \right ) .
- Current lags behind the voltage by an angle \left ( \frac {\pi}{2} \right ) .

Figure (A) shows Time diagram and figure (B) shows *Phasor diagram* of a pure inductive circuit.

### Inductive Reactance

*Inductive reactance of a circuit is defined as the effective opposition offered by the inductor to the flow of current in the circuit.*

Comparing the relation \left [ I_0 = \left ( \frac {V_0}{L\omega} \right ) \right ] and \left [ I_0 = \left ( \frac {V_0}{R} \right ) \right ] we conclude that ( L \omega ) has the dimension of a *resistance*.

- The term \left ( L \omega \right ) is called
. It is represented by ( X_L )*Inductive Reactance*

Therefore, \quad X_L = L \omega

- In SI system the unit of inductive reactance is ohm.

**TO BE NOTED –**

**(1) ****Behavior of a pure Inductor in DC supply** –

- Inductive reactance of pure inductor is given by \left [ X_L = L \omega = 2 \pi L \nu \right ] .

So, it is a function of *frequency* ( \nu ) .

- For a DC source ( \nu = 0 ) .

Therefore, \left ( X_L = 0 \right ) . Thus a pure inductor offers no *resistance* to the flow of a DC.

- Hence DC can flow easily through an inductor coil.

**(2) Behavior of pure Inductor at high frequency –**

- Inductive reactance of pure inductor is given by \left [ X_L = L \omega = 2 \pi L \nu \right ] .

So it is function of *frequency* ( \nu ) .

- At very high frequency \left [ X_L \rightarrow \infty \right ]

Thus an inductor behaves as an open circuit at very high frequency.

## AC applied to a Capacitor

*When an alternating source is connected to an ideal capacitor, the circuit is called a pure capacitive circuit.*

Consider about a pure capacitive circuit as shown in figure.

- The capacitor is periodically charged and discharged when an alternating voltage is applied to it.

The alternating voltage applied across the capacitor is given by –

V = V_0 \sin ( \omega t ) ……… (1)

### Phase of AC current in Capacitor

Let, ( q ) is the charge on the *capacitor* at an instant.

- Then potential difference across the capacitor will be –

V_c = \left ( \frac {q}{C} \right )

But ( V_c ) is the applied voltage across the capacitor.

- Therefore, \quad V_c = V

Or, \quad V_c = V_0 \sin ( \omega t )

Also, \quad V_c = \left ( \frac {q}{C} \right )

- Therefore, \quad \left ( \frac {q}{C} \right ) = V_0 \sin ( \omega t )

Or, \quad q = V_0 C \sin ( \omega t )

Now, \quad I = \left ( \frac {dq}{dt} \right )

= \left ( \frac {d}{dt} \right ) \left [ V_0 C \sin ( \omega t ) \right ]

= V_0 C \left ( \frac {d}{dt} \right ) \left [ \sin ( \omega t ) \right ]

= V_0 C \omega \left [ \cos ( \omega t ) \right ]

= \left ( \frac {V_0}{1 / C \omega} \right ) \cos ( \omega t )

Also from trigonometric relations, we get –

\cos ( \omega t ) = \left [ \sin ( \omega t ) + \left ( \frac {\pi}{2} \right ) \right ]

Therefore, \quad I = \left ( \frac {V_0}{1 / C \omega} \right ) \sin \left ( \omega t + \frac {\pi}{2} \right ) ……… (2)

### Peak value of AC current in Capacitor

Comparing this equation with standard equation of current \left [ I = I_0 \sin ( \omega t ) \right ] , we get the *peak value* of current as –

I_0 = \left [ \frac {V_0}{\left ( \frac {1}{C \omega} \right )} \right ]

Comparing equations (1) and (2), we conclude that –

- Voltage and current have phase difference of \left ( \frac {\pi}{2} \right ) .
- Current leads the voltage by an angle \left ( \frac {\pi}{2} \right ) .

Figure (A) shows Time diagram and figure (B) shows Phasor diagram of a pure Capacitive circuit.

### Capacitive Reactance

*The capacitive reactance of a circuit is defined as the effective opposition offered by a capacitor to the flow of electric current in the circuit.*

Comparing the relations \left [ I_0 = \left ( \frac {V_0}{1 / C \omega } \right ) \right ] and \left [ I_0 = \left ( \frac {V_0}{R} \right ) \right ] , we conclude that \left ( \frac {1}{C \omega} \right ) has the dimension of a resistance.

- The term \left ( \frac {1}{C \omega} \right ) is called
It is represented by ( X_C )**Capacitive Reactance.**

Therefore, \quad ( X_C ) = \left ( \frac {1}{C \omega } \right )

- In SI system unit of capacitive reactance is ohm.

**TO BE NOTED –**

**Behavior of a capacitor in DC supply**

- Capacitive reactance is given by the relation \left [ X_C = \left ( \frac {1}{C \omega} \right ) \right ] Or, \left [ X_C = \left ( \frac {1}{C \times 2 \pi \nu} \right ) \right ]

So, it is a function of *frequency* ( \nu ) .

- For a DC circuit, ( \nu = 0 )

Therefore, \left [ X_C = \left ( \frac {1}{0} \right ) = \infty \right ] . Thus a capacitor offers infinite opposition to the flow of DC current.

- So in DC circuit a capacitor behaves as an open circuit.

## AC applied to a Resistor

*When an AC source is connected to a resistor of pure resistance then this type of circuit is called a pure resistive circuit.*

Consider about a pure resistive circuit as shown in figure.

Let, the applied alternating voltage in the circuit is –

V = V_0 \sin ( \omega t ) …….. (1)

### Phase of AC current in resistor

Let, ( I ) is the *current* in the circuit at any instant ( t ) and ( I_0 ) is the *peak value* of the current.

- As per
*Ohm ‘s Law*, across a resistor –

I = \left ( \frac {V}{R} \right )

Therefore, \quad I = \left ( \frac {V_0 \sin ( \omega t )}{R} \right )

= \left ( \frac {V_0}{R} \right ) \sin ( \omega t )

= I_0 \sin ( \omega t ) …….. (2)

Comparing equations (1) and (2) we see that *phase difference* between voltage and current is zero i.e. they are in the same phase.

If a graph for voltage and current in the resistor is plotted with respect to time ( t ) , it is observed that –

- Voltage and current are in same phase i.e. phase difference is zero.
- Both voltage and current have zero, maximum and minimum values at the same instants as shown in figure.

Time diagram for pure resistive circuit is shown in figure.

## Power of a Resistor

Average values of alternating voltage and alternating current for full cycle are zero. But, power consumed or dissipated in a resistor is not zero.

- Instantaneous power dissipated in the resistor –

P = I^2 R = \left [ I_0 \sin ( \omega t ) \right ]^2 R

Or, \quad P = I_0^2 R \sin^2 ( \omega t )

- Therefore, Average power will be –

P = I_0^2 R \times Average of \sin^2 ( \omega t )

But, from trigonometric relation, we get –

\sin^2 \theta = \left [ \frac {1 - \cos ( 2 \theta )}{2} \right ]

Therefore, average of \left [ \sin^2 ( \omega t ) \right ] will be equal to the average of \left [ \frac {1 - \cos ( 2 \omega t ) }{2} \right ]

Now, average value of ( \cos 2 \omega t ) for full cycle can be obtained by integrating it within the limits \left ( t = 0 \right ) and \left ( t = T \right ) .

By integration, we get –

\left ( \frac {1}{T} \right ) \int\limits_{0}^{T} \cos ( 2 \omega t ) dt = 0

Therefore, average of \left [ \sin^2 ( \omega t ) \right ] is equal to \left ( \frac {1 - 0}{2} \right ) = \left ( \frac {1}{2} \right )

Hence, Average power is given by –

P = I_0^2 R \times Average of \sin^2 ( \omega t )

= \left ( \frac {1}{2} \right ) I_0^2 R

= \left ( \frac {I_0}{\sqrt {2}} \right )^2 R

= ( I_{rms} )^2 R