CBSE NCERT NOTES CLASS 12 PHYSICS CHAPTER 7

ALTERNATING CURRENT

AC voltage and AC current

When the current and voltage are changing like sine functions, they are called alternating voltage and alternating current.

The alternating voltage is represented as

And the alternating current is represented as

where vm and im are the amplitudes of the oscillating potential difference and current respectively and ω is its angular frequency.

AC voltage applied to a resistor

Let the alternating potential difference, be v = vm sin ωt, where vm is the amplitude of the oscillating potential difference and ω is its angular frequency.

Applying Kirchhoff’s loop rule , we get,

Where im = $\frac{{\mathrm{v}}_{\mathrm{m}}}{\mathrm{R}}$ = amplitude of current.

Both v and i reach zero, minimum and maximum values at the same time. That is, the voltage and current are in phase with each other.

The applied voltage and the current vary sinusoidally. The sum of the instantaneous values over one complete cycle is zero, and the average voltage and average current is zero.

AC power dissipated in a resistive circuit

The instantaneous power dissipated in the resistor is

p = i2 R = im2 R sin2 ωt

The average value of power over a cycle is

$\stackrel{‾}{\mathrm{P}}$ = < i2 R > = <im2 R sin2 ωt>

where the bar over a letter (here, p) denotes its average value and < > denotes taking average of the quantity inside the bracket. Since, im2 and R are constants,

$\stackrel{‾}{\mathrm{P}}$ = im2 R <sin2 ωt>

Since

sin2 ωt = $\frac{1}{2}$ (1– cos 2ωt),

we have

<sin2 ωt> = $\frac{1}{2}$ (1– <cos 2ωt>)

and since

<cos 2ωt> = 0

<sin2 ωt> = $\frac{1}{2}$

Therefore,

Root mean square (RMS) current Irms or effective current I, is defined as

$\mathrm{I}=\sqrt{\stackrel{‾}{{\mathrm{i}}^{2}}}=\sqrt{\frac{{{\mathrm{i}}_{\mathrm{m}}}^{2}}{2}}$

And average power as

$=\frac{{\mathrm{V}}^{2}}{\mathrm{R}}$

Root mean square (RMS) voltage or effective voltage

$\mathrm{V}=\sqrt{\stackrel{‾}{{\mathrm{v}}^{2}}}=\sqrt{\frac{{{\mathrm{v}}_{\mathrm{m}}}^{2}}{2}}$

For V = 220,

vm = $\sqrt{2}$ V = 220 × 1.414 = 311 V

Representation of AC current and voltage by rotating vectors – PHASORS

A phasor is a vector which rotates about the origin with angular speed ω.

The vertical components of phasors V and I represent the sinusoidally varying quantities v and i. The magnitudes of phasors V and I represent the amplitudes or the peak values vm and im of these oscillating quantities.

Phasor diagram for purely resistive circuit

The projection of voltage and current phasors on vertical axis, i.e., vm sin ωt and im sin ωt, respectively, represent the value of voltage and current at that instant.

As they rotate with frequency ω, we can generate the curves of v or i vs ωt are generated.

Phasors V and I for the case of a resistor are in the same direction for all times.

That is, the phase angle between the voltage and the current is zero, i.e., they are in phase.

AC voltage applied to an inductor

Although inductors have appreciable resistance in their windings, we shall assume that this inductor has negligible resistance or 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, ∑ E(t) = 0 , and since there is no resistor in the circuit,

Integrating both sides, we get,

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,

$-\mathrm{cos}\mathrm{\omega }\mathrm{t}=\mathrm{sin}\left(\mathrm{\omega }\mathrm{t}-\frac{\mathrm{\pi }}{2}\right)$

We get,

$\mathrm{i}=\frac{{\mathrm{v}}_{\mathrm{m}}}{\mathrm{L}}\mathrm{sin}\left(\mathrm{\omega }\mathrm{t}-\frac{\mathrm{\pi }}{2}\right)$

Or

Where, im = $\frac{{\mathrm{v}}_{\mathrm{m}}}{\mathrm{L}}$, is the amplitude of the current.

Inductive reactance

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

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.

Phasor diagram for an inductive circuit

The inductive reactance is directly proportional to the inductance and to the frequency of the current.

The current lags the voltage by $\frac{\mathrm{\pi }}{2}$ or one-quarter $\left(\frac{1}{4}\right)$ cycle and reaches its maximum value later than the voltage by one-fourth of a period $\left(\frac{\mathrm{T}}{4}\right).$

AC power in an inductive circuit

The instantaneous power supplied to the inductor is

The average power over a complete cycle will be,

• Since <sin 2ωt> = 0, the average power supplied to an inductor over one complete cycle is zero.

Magnetization and demagnetization of an inductor

We can plot the graphs of v, i, p etc., shown. There are 4 stages in complete cycle, as shown in the graph.

Period 0-1

Current i through the coil entering at A increases from zero to a maximum value. Flux lines are set up i.e., the core gets magnetized. The voltage and current are both positive. So their product p is positive. ENERGY IS ABSORBED FROM THE SOURCE.

Period 1-2

Current in the coil is still positive but is decreasing. The core gets demagnetized and the net flux becomes zero at the end of a half cycle. The voltage v is negative (since $\frac{\mathrm{d}\mathrm{i}}{\mathrm{d}\mathrm{t}}$is negative). The product of voltage and current is negative, and ENERGY IS BEING RETURNED TO SOURCE.

Period 2-3

Current i becomes negative i.e., it enters at B and comes out of A. Since the direction of current has changed, the polarity of the magnet changes. The current and voltage are both negative. So their product p is positive. ENERGY IS ABSORBED.

Period 3-4

Current i decreases and reaches its zero value at 4 when core is demagnetized and flux is zero. The voltage is positive but the current is negative. The power is, therefore, negative. ENERGY ABSORBED DURING THE CYCLE 2-3 IS RETURNED TO THE SOURCE.

• NET ENERGY ABSORBED DURING THE COMPLETE CYCLE IS ZERO.

AC voltage applied to a capacitor

Let the voltage across the source be v = vm sin ωt. Using the Kirchhoff’s loop rule, , and since there is no resistor in the circuit,

$\mathrm{v}=\frac{\mathrm{q}}{\mathrm{C}}$

Therefore,

Capacitive reactance

The term $\frac{1}{\mathrm{\omega }\mathrm{C}}$ plays the role of resistance. It is called capacitive reactance and is denoted by

Taking,

The equation becomes,

$\mathrm{i}={\mathrm{i}}_{\mathrm{m}}\mathrm{sin}\left(\mathrm{\omega }\mathrm{t}+\frac{\mathrm{\pi }}{2}\right)$

In the case of a capacitor, the current leads the voltage by $\frac{\mathrm{\pi }}{2}$ or one-quarter $\left(\frac{1}{4}\right)$ cycle and reaches its maximum value before the voltage by one-fourth of a period $\left(\frac{\mathrm{T}}{4}\right).$

AC power in a capacitive circuit

The instantaneous power supplied to the capacitor is

The average power over a complete cycle will be,

Since <sin 2ωt> = 0, the average power supplied to a capacitor over one complete cycle is zero.

Charging and discharging of a capacitor

We can plot the graphs of v, q, i and p, as follows,

There are 4 stages in complete cycle, as shown in the graph above,

Period 0-1

The current i flows as shown and from the maximum at 0, reaches a zero value at 1. The plate A is charged to positive polarity while negative charge q builds up in B reaching a maximum at 1 until the current becomes zero. The voltage vc = q/C is in phase with q and reaches maximum value at 1. Current and voltage are both positive. So p = vci is positive. ENERGY IS ABSORBED FROM THE SOURCE DURING THIS QUARTER CYCLE AS THE CAPACITOR IS CHARGED.

Period 1-2

The current i reverses its direction. The accumulated charge is depleted i.e., the capacitor is discharged during this quarter cycle. The voltage gets reduced but is still positive. The current is negative. Their product, the power is negative. THE ENERGY ABSORBED DURING THE $\frac{1}{4}$ CYCLE 0-1 IS RETURNED DURING THIS QUARTER.

Period 2-3

As i continues to flow from A to B, the capacitor is charged to reversed polarity i.e., the plate B acquires positive and A acquires negative charge. Both the current and the voltage are negative. Their product p is positive. THE CAPACITOR ABSORBS ENERGY DURING THIS $\frac{1}{4}$ CYCLE.

Period 3-4

The current i reverses its direction at 3 and flows from B to A. The accumulated charge is depleted and the magnitude of the voltage vc is reduced. vc becomes zero at 4 when the capacitor is fully discharged. The power is negative. ENERGY ABSORBED DURING 2-3 IS RETURNED TO THE SOURCE.

• NET ENERGY ABSORBED DURING THE COMPLETE CYCLE IS ZERO.

AC voltage applied to a series LCR circuit

If q is the charge on the capacitor and i the current, at time t, as per Kirchhoff’s loop rule,

To determine the instantaneous current i and its phase relationship to the applied alternating voltage v.

Phasor-diagram solution for LCR circuit

Let i = im sin (ωt + ϕ) where ϕ is the phase difference between the voltage across the source and the current in the circuit.

Let $\stackrel{\to }{\mathrm{I}}$ be the phasor representing the current in the circuit, represent the phasors corresponding to voltage across the inductor, resistor, capacitor and the source, respectively. We know that ${\stackrel{\to }{\mathrm{V}}}_{\mathrm{R}}$ is parallel to $\stackrel{\to }{\mathrm{I}}$, ${\stackrel{\to }{\mathrm{V}}}_{\mathrm{C}}$ is $\frac{\mathrm{\pi }}{2}$ behind $\stackrel{\to }{\mathrm{I}}$ and ${\stackrel{\to }{\mathrm{V}}}_{\mathrm{L}}$ is $\frac{\mathrm{\pi }}{2}$ ahead of $\stackrel{\to }{\mathrm{I}}$.

The length of these phasors or the amplitude of are

The phasor relation whose vertical component gives the above equation is

Since ${\stackrel{\to }{\mathrm{V}}}_{\mathrm{C}}$ and ${\stackrel{\to }{\mathrm{V}}}_{\mathrm{L}}$ are always along the same line and in opposite directions, they can be combined into a single phasor () which has a magnitude

Since V is represented as the hypotenuse of a right triangle whose sides are VR and (VC - VL),

Substituting the values, we have,

Impedence

The impedance Z in an ac circuit is the effective resistance of an electric circuit or component to alternating current, arising from the combined effects of ohmic resistance, inductive reactance and capacitive reactance. It is represented by Z and is given by,

Since phasor $\stackrel{\to }{\mathrm{I}}$ is always parallel to phasor ${\stackrel{\to }{\mathrm{V}}}_{\mathrm{R}}$, the phase angle ϕ is the angle between ${\stackrel{\to }{\mathrm{V}}}_{\mathrm{R}}$ and $\stackrel{\to }{\mathrm{V}}$ and is given by

If XC > XL, ϕ is positive and the circuit is predominantly capacitive. Consequently, the current in the circuit leads the source voltage.

If XC < XL, ϕ is negative and the circuit is predominantly inductive. Consequently, the current in the circuit lags the source voltage.

Limitations of phasor diagram solutions

This method of analysing ac circuits suffers from certain disadvantages.

The phasor diagrams say nothing about the initial condition. One can take any arbitrary value of t (say, t1, as done throughout this chapter) and draw different phasors which show the relative angle between different phasors.

The solution so obtained is called the steady-state solution. Steady state is the state when dy/dt = 0, i.e., y has reached its steady state. This is not a general solution. There is a transient solution which exists even for .

The general solution is the sum of the transient solution and the steady-state solution. After a sufficiently long time, the effects of the transient solution die out and the behaviour of the circuit is described by the steady-state solution.

Analytical consideration for LCR circuit

This is like the equation for a forced, damped oscillation.

Let as assume that the differential equation has the solution,

and

Substituting these values, we get,

Where

Multiplying and dividing by , we get,

Let

And

Using trigonometric identity

we get,

Comparing the the two sides, we can write,

Where,

and

Therefore, the current in the circuit is,

Where,

RESONANCE

Some systems have a tendency to oscillate at a particular frequency. This frequency is called the natural frequency of the system.

For an RLC circuit if ω is varied, then at a particular frequency ωo, XC = XL, and the impedance is minimum, i.e., , This frequency is called the resonant frequency.

${\mathrm{X}}_{\mathrm{C}}={\mathrm{X}}_{\mathrm{L}}$

At resonant frequency, the current amplitude is maximum,

Tuning mechanism of a radio or a TV

The antenna of a radio accepts signals from many broadcasting stations. The signals picked up in the antenna act as a source in the tuning circuit of the radio, so the circuit can be driven at many frequencies. But to hear one particular radio station, we tune the radio i.e. vary the capacitance of a capacitor in the tuning circuit such that the resonant frequency of the circuit becomes nearly equal to the frequency of the radio signal received. When this happens, the amplitude of the current; with the frequency of the signal of the particular radio station in the circuit; is maximum.

The resonance phenomenon is exhibited by a circuit only if both L and C are present in the circuit. Only then do the voltages across L and C cancel each other (both being out of phase) and the current amplitude is $\frac{{\mathrm{v}}_{\mathrm{m}}}{\mathrm{R}}$, the total source voltage appearing across R. This means that we cannot have resonance in a RL or RC circuit.

The amplitude of the current in the series LCR circuit is given by

and is maximum when

The maximum value of ${\mathrm{i}}_{\mathrm{m}}$ is

For values of ω other than ωo, the amplitude of the current is less than the maximum value.

Consider value of ω for which the current amplitude is $\frac{1}{\sqrt{2}}$ times its maximum value, so that the power dissipated by the circuit becomes half. There are two such values of ω, say, ω1 and ω2, one greater and the other smaller than ω0 and symmetrical about ωo.

Or ω1 = ωo + Δω and ω2 = ωo - Δω

The difference ω1ω2 = ω is called the bandwidth of the circuit.

Sharpness of resonance

The quantity $\left(\frac{{\mathrm{\omega }}_{\mathrm{o}}}{2\mathrm{\Delta }\mathrm{\omega }}\right)$ is regarded as a measure of the sharpness of resonance. The smaller the Δω, the sharper or narrower is the resonance.

The current amplitude, at ω1 = ωo + Δω will be,

Also at ω1 = ωo + Δω,

Comparing the two values of ${\mathrm{i}}_{\mathrm{m}}$, we can see,

$⇒\left({\mathrm{\omega }}_{\mathrm{o}}+\mathrm{\Delta }\mathrm{\omega }\right)\mathrm{L}-\frac{1}{\left({\mathrm{\omega }}_{\mathrm{o}}+\mathrm{\Delta }\mathrm{\omega }\right)\mathrm{C}}=\mathrm{R}$

${\mathrm{\omega }}_{\mathrm{o}}\mathrm{L}\left(1+\frac{\mathrm{\Delta }\mathrm{\omega }}{{\mathrm{\omega }}_{\mathrm{o}}}\right)-\frac{{\mathrm{\omega }}_{\mathrm{o}}\mathrm{L}}{\left(1+\frac{\mathrm{\Delta }\mathrm{\omega }}{{\mathrm{\omega }}_{\mathrm{o}}}\right)}=\mathrm{R}$

$⇒{\mathrm{\omega }}_{\mathrm{o}}\mathrm{L}\left(1+\frac{\mathrm{\Delta }\mathrm{\omega }}{{\mathrm{\omega }}_{\mathrm{o}}}\right)-{\mathrm{\omega }}_{\mathrm{o}}\mathrm{L}\left(1-\frac{\mathrm{\Delta }\mathrm{\omega }}{{\mathrm{\omega }}_{\mathrm{o}}}\right)=\mathrm{R}$

The sharpness of resonance is given by = $\frac{{\mathrm{\omega }}_{\mathrm{o}}\mathrm{L}}{\mathrm{R}}$

Quality factor

The ratio Q = $\frac{{\mathrm{\omega }}_{\mathrm{o}}\mathrm{L}}{\mathrm{R}}$ is called quality factor of the LCR circuit.

Q may also be written as Q = $\frac{1}{{\mathrm{\omega }}_{\mathrm{o}}\mathrm{C}\mathrm{R}}$

• For Q to be large, R has to be small and L has to be large.

• If the resonance is less sharp, the maximum current is less and the circuit is close to resonance for a larger range Δω of frequencies and the tuning of the circuit will not be good. So, less sharp the resonance, less is the selectivity of the circuit or vice versa.

Power in AC circuit - the power factor

For usual notations of v and i, the instantaneous power p supplied by the source is given by,

The average power over a cycle is given by the average of the two terms in R.H.S. The second term is time-dependent. Its average is zero (the positive half of the cosine cancels the negative half).

Therefore,

The average power dissipated depends not only on the voltage and current but also on the cosine of the phase angle ϕ between them. The quantity cos ϕ is called the power factor of the circuit.

Case (i) Purely resistive circuit

If the circuit contains only R, it is called resistive. In that case ϕ = 0, cos ϕ = 1. There is maximum power dissipation.

Case (ii) Purely inductive or capacitive circuit

If the circuit contains only an inductor or capacitor, we know that the phase difference between voltage and current is $±\frac{\mathrm{\pi }}{2}$. Therefore, cos ϕ = 0, and no power is dissipated even though a current is flowing in the circuit. This current is called wattless current.

Case (iii) LCR series circuit

In an LCR series circuit, power dissipated is given by above equation, where

So, ϕ may be non-zero in a RL or RC or RCL circuit.

Even in such cases, power is dissipated only in the resistor.

Case (iv) Power dissipated at resonance in LCR circuit

At resonance XC – XL = 0, and ϕ = 0. Therefore, cos ϕ = 1 and P = I2Z = I2R. That is, maximum power is dissipated in a circuit (through R) at resonance.

LC oscillations

A capacitor and an inductor can store electrical and magnetic energy, respectively. When a capacitor (initially charged) is connected to an inductor, the charge on the capacitor and the current in the circuit exhibit the phenomenon of electrical oscillations similar to oscillations in mechanical systems

Let a capacitor be charged qm (at t = 0) and connected to an inductor.

The moment the circuit is completed, the charge on the capacitor starts decreasing, giving rise to current in the circuit. Let q and i be the charge and current in the circuit at time t.

Since $\frac{\mathrm{d}\mathrm{i}}{\mathrm{d}\mathrm{t}}$ is positive, the induced emf in L will have polarity as shown, i.e., vb < va.

Therefore,

$\frac{\mathrm{d}\mathrm{i}}{\mathrm{d}\mathrm{t}}=-\frac{{\mathrm{d}}^{2}\mathrm{q}}{\mathrm{d}{\mathrm{t}}^{2}}$

This is like the equation of the form,

$\frac{{\mathrm{d}}^{2}\mathrm{x}}{\mathrm{d}{\mathrm{t}}^{2}}+{{\mathrm{\omega }}_{\mathrm{o}}}^{2}\mathrm{x}=0$

Which is the equation for a simple harmonic oscillator.

The charge, therefore, oscillates with a natural frequency,

${\mathrm{\omega }}_{\mathrm{o}}=\frac{1}{\sqrt{\mathrm{L}\mathrm{C}}}$

and varies sinusoidally with time as,

where qm is the maximum value of q and ϕ is a phase constant.

Since q = qm at t = 0, we have cos ϕ = 1 or ϕ = 0.

Therefore,

Therefore, the current in the circuit is,

ENERGY OSCILLATIONS IN LC CIRCUIT

At any instant the power in LCR circuit is given by,

Initially the capacitor has the maximum qm. Figure (a) shows a capacitor with initial charge qm connected to an ideal inductor.

The electrical energy stored in the charged capacitor is,

Since, there is no current in the circuit, energy in the inductor is zero. Thus, the total energy of LC circuit is,

At t = 0, the switch is closed and the capacitor starts to discharge as shown in figure (b). As the current increases, it sets up a magnetic field in the inductor and thereby, some energy gets stored in the inductor in the form of magnetic energy

As the current reaches its maximum value im, (at t = $\frac{\mathrm{T}}{4}$) as in figure (c), all the energy is stored in the magnetic field

The maximum electrical energy equals the maximum magnetic energy.

The capacitor now has no charge and hence no energy. The current now starts charging the capacitor, as in figure (d).

This process continues till the capacitor is fully charged (at t = $\frac{\mathrm{T}}{2}$) as in figure (e). But it is charged with a polarity opposite to its initial state. The whole process keeps repeating till the system reverts to its original state.

Thus, the energy in the system oscillates between the capacitor and the inductor.

The LC oscillation is similar to the mechanical oscillation of a block attached to a spring.

The above discussion of LC oscillations is not realistic for two reasons,

1. Every inductor has some resistance. The effect of this resistance is to introduce a damping effect on the charge and current in the circuit and the oscillations finally die away.

2. Even if the resistance were zero, the total energy of the system would not remain constant. It is radiated away from the system in the form of electromagnetic waves.

Analogies between mechanical and electrical quantities

 Mechanical system Electrical system Mass m Inductance L Force constant k Reciprocal capacitance $\frac{1}{\mathrm{C}}$ ${\mathrm{\omega }}_{\mathrm{o}}$ = $\sqrt{\frac{\mathrm{k}}{\mathrm{m}}}$ ${\mathrm{\omega }}_{\mathrm{o}}$ = $\frac{1}{\sqrt{\mathrm{L}\mathrm{C}}}$ Displacement x Charge q Velocity v = $\frac{\mathrm{d}\mathrm{x}}{\mathrm{d}\mathrm{t}}$ Current i = $\frac{\mathrm{d}\mathrm{q}}{\mathrm{d}\mathrm{t}}$ Force F = ma = m $\left(\frac{\mathrm{d}\mathrm{v}}{\mathrm{d}\mathrm{t}}\right)$= m $\left(\frac{{\mathrm{d}}^{2}\mathrm{x}}{\mathrm{d}{\mathrm{t}}^{2}}\right)$ Emf ɛ = - L -L$\left(\frac{{\mathrm{d}}^{2}\mathrm{q}}{\mathrm{d}{\mathrm{t}}^{2}}\right)$ Mechanical energy Electromagnetic energy

Forced oscillations vs. driven LCR circuit

 Forced oscillations Driven LCR circuit Equation Equation Displacement, x Charge on capacitor, q Time, t Time, t Mass, m Self inductance, L Damping constant, b Resistance, R Spring constant, k Inverse capacitance, $\frac{1}{\mathrm{C}}$ Driving frequency, ωd Driving frequency, ω Natural frequency of oscillations, ω Natural frequency of LCR circuit, ωo Amplitude of forced oscillations, A Maximum charge stored, qm Amplitude of driving force, Fo Amplitude of applied voltage, vm

Transient current

An electric current which varies for a small finite time, while growing from zero to maximum or decaying from maximum to zero, is called a transient current.

Growth of current in an inductor

When connected to a voltage source, growth of current in an LR circuit at any instant of time t is given by

Decay of current in an inductor

When the voltage source is removed, decay of current in an inductor at any time t is given by

Here $\frac{\mathrm{L}}{\mathrm{R}}$ = τ, is called the time constant of an LR circuit.

Time constant of an LR circuit is the time in which the circuit grows to 63.2% of the maximum value of current or decays to 36.8% of the maximum value of current.

Current during charging and discharging of a capacitor

The instantaneous charge on a capacitor on charging at any instant of time t is given by

Instantaneous voltage

Instantanteous current

When a capacitor is discharging, then,

Here RC = τ, is called time constant of a RC circuit.

Time constant of a RC circuit is the time in which charge in the capacitor grows to 63.8% or decay to 36.8% of the maximum charge on capacitor.

Transformer

The device which changes (or transforms) an alternating voltage from one to another of greater or smaller value is called transformer. The transformer uses the principle of mutual induction.

A transformer consists of two sets of coils, insulated from each other and wound on separate limbs of the core. One of the coils called the primary coil, also called input coil, has NP turns. The other coil is called the secondary coil or output coil and has NS turns.

When an alternating voltage is applied to the primary, the resulting current produces an alternating magnetic flux which links the secondary and induces an emf in it. The value of this emf depends on the number of turns in the secondary.

Let us consider an ideal transformer in which the primary has negligible resistance and all the flux in the core links both primary and secondary windings. Let Ф be the flux in each turn in the core at time t due to current in the primary when a voltage vp is applied to it.

Then the induced emf or voltage ɛs, in the secondary with Ns turns is

The alternating flux Ф also induces a back emf in the primary. This is

But ep = vp. If this were not so, the primary current would be infinite since the primary has zero resistance (as assumed).

If the secondary is an open circuit or the current taken from it is small, then es = vs where vs is the voltage across the secondary. Therefore,

Assumptions

1. the primary resistance and current are small

2. the same flux links both the primary and the secondary as very little flux escapes from the core,

3. the secondary current is small

If the transformer is assumed to be 100% efficient (no energy losses), the power input is equal to the power output, and since, , we have,

Or

$\frac{{\mathrm{i}}_{\mathrm{p}}}{{\mathrm{i}}_{\mathrm{s}}}=\frac{{\mathrm{v}}_{\mathrm{s}}}{{\mathrm{v}}_{\mathrm{p}}}=\frac{{\mathrm{N}}_{\mathrm{s}}}{{\mathrm{N}}_{\mathrm{p}}}$

1. Step up Transformer converts a low voltage of high current into a high voltage of low current.

Ns > Np,

Vs > Vp

and Ip > Is

2. Step-down transformer converts a high voltage of low current into a low voltage of high current.

Np > Ns,

Vp > Vs

and Is > Ip

Transformation ratio

The ratio of the voltage in the secondary coil of a transformer to the voltage in its primary coil, is referred to as the transformation ratio.

For step-up transformer, K > 1

For step-down transformer, K < 1

Energy losses in a transformer

The main energy losses in a transformer are given below

1. Flux leakage: There is always some flux leakage; that is, not all of the flux due to primary passes through the secondary due to poor design of the core or the air gaps in the core. It can be reduced by winding the primary and secondary coils one over the other.

2. Resistance of the windings: The wire used for the windings has some resistance and so, energy is lost due to heat produced in the wire (I 2R). In high current, low voltage windings, these are minimized by using thick wire.

3. Eddy currents: The alternating magnetic flux induces eddy currents in the iron core and causes heating. The effect is reduced by having a laminated core.

4. Hysteresis: The magnetization of the core is repeatedly reversed by the alternating magnetic field. The resulting expenditure of energy in the core appears as heat and is kept to a minimum by using a magnetic material which has a low hysteresis loss.

Power dissipation during transmission of electricity

The connecting wires from the power station to the device have a finite resistance Rc. The power dissipated in the connecting wires, which is wasted is Pc = I2Rc

Since,

That is the reason why electrical energy is transmitted through transmission lines at high voltage so that current is reduced and consequently the I 2R loss is cut down.