Physics Alternating Current

8 - ALTERNATING CURRENT Page 1

8.1 Alternating Voltage and Alternating Current ( A. C. )

The following figure shows N turns of a coil of conducting wire PQRS rotating with a uniform angular speed with respect to the X-axis in a uniform magnetic field along the Y-axis.

Let the angle between the direction of the area vector of the coil and the magnetic field direction be zero at time, t = 0 and = t at time t = t. The magnetic flux, 0, associated with the coil at time t = 0 and t at t me t = t are given by

0 = N

B A = N A B cos 0 = N A B and

t = N A B cos t

The emf induced in t e coil according to Faradays law is

V = - dt

t d = N A B sin t = Vm sin t ... ... ... ... ... ( 1 )

where Vm = N A B is the maximum value of the induced emf.

Equation ( 1 ) shows that the induced emf versus time is a sine curve. This emf is obtained between the brushes B1 and B2 which are in contact with the slip rings A1 and A2 as shown in the above figure.

The voltage is zero at time t = 0 and varies as per the function sin t reaching maximum value Vm at time t = pipi / 2 and again zero at time t = pi / . B2 being at greater potential than B1 acts like a positive end of the voltage source during this time interval.

After time t = pi / , the potential of B1 starts to rise with respect to B2 till time t = 3 pipi //// 2 ,,,, reaches maximum in the reverse direction and again becomes zero at time t = 2pi // . This cycle keeps repeating in every time interval of T = 2pi // .

The voltage so developed is known as alternating voltage and its graph versus time is shown by a continuous line in the following figure. The arrangement to produce such a voltage is

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known as A.C. ( alternating current ) dynamo or generator.

In the adjoining figure, an alternating voltage source is connected in a circuit with resistor, R. The voltage between a and b is zero at time t = 0. Zero current will flow at this time.

The voltage in the circuit varies according as V = Vm sin t. The current as per Ohms law

will be = R

t sin Vm . The current changes

sinusoidally in the same way as the v ltage as shown by the broken line in the above graph

Here, voltage was considered changing with time as per sin t. Both the voltage and the current can be considered to be ch nging as per cos t. It is not necessary that the voltage and current should change as above. There are also other ways in which both can change periodically with time.

8.2 A. C. Circuit with Series Combination of Resistor, Inductor and Capacitor

The following figure s ows a series combination of resistor having resistance R, inductor of inductance L and a capacitor of capacitance C with an alternating voltage source of voltage changing with time as V = Vm cos t. It is assumed that the resistor has zero inductance and the inductor h s zero resistance.

Let = current in the circuit, Q = ch rge on the capacitor and

dtd

= ate of change of current at

t me t.

Hence, by Kirchhoffs law,

VR + V L + VC = V

R + L dtd

+ C

Q = Vm cos t

Putting =dtdQ

and dtd

= 2

2

dtQd

in the above equation,

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R dtdQ

+ L 22

dtQd

+ C

Q = Vm cos t

2

2

dtQd

+ L

R

dtdQ

+ LC

Q =

L

Vm cos t

This equation is known as the differential equation of Q for the series R L - C A. C. Circuit. It is similar to the differential equation of forced harmonic oscillations given as under.

2

2

dtyd

+ m

b

dtdy

+ ym

k

=

m

F0 sin t

The mechanical and electrical quantities in the above equations are com ared in the following table.

No.

Mechanical quantities

Symbol

Elec rical quantities

Symbol

1.

Displacement

y

El ctri al charge

Q

2.

Velocity dtdy

Electric current dtdQ

=

3.

Co-efficient of resistance

b

Resistance

R

4.

Mass

m

Inductance

L

5.

Force constant

k

Inverse of capacitance C1

6.

Angular frequency m

k

Angular frequency LC 1

7.

Periodic force

F0

Periodic voltage

Vm

The function of charge Q versus time t which satisfies the above differential equation of charge Q is lled its solution. Complex function is used to find the solution of the above differential equ tion

8.3 Complex Numbers ( For Information Only )

A ompl x number z = x + jy, where j = 1 -- . x is the real part and y is the imaginary p rt of the complex number.

( ) The complex number can be represented by a point in a complex plane with x-axis representing real numbers and y-axis representing imaginary numbers. Point P in the figure ( next page ) represents the complex number z = x + jy. The x-coordinate of P gives the real part of z and y-coordinate gives its imaginary part. The magnitude of complex number is equal to r, i.e., l z l = r = 22 y x ++ .

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Now, x = r cos and y = r sin

z = r cos + j r sin = r ( cos + j sin )

= l z l e j ( Q e j == cos + j sin )

( 2 ) z* = x - jy is the complex conjugate of the complex number z.

( 3 ) For complex number z,

z

1 =

z

1

*z

*z = 2

z l*z

l = 22

y xjy x -

+

= 22 y x x

++ - 22

y x y

j+

The real part of z

1 = 22

y x x

+ and the imaginary part of

z

1 = - 22

y x y

j+

.

The real part of the complex number z is denoted by Re ( z ) and the imaginary part by m ( z ).

8.4 Solution of the differential equation

The equation for series L-C-R circ it is R + L dtd

+ C

Q = Vm cos t

R + L dtd

+ C

dt Vm cos t where Q = dt

t cos L

V dt

LC1

LR

dt d m ==++++++

Replacing the current by complex current i and cos t by e j t, and solving the differential equation of complex quantities, the real part of complex current i will give the equation of rea current as a function of time, t.

Thus, e differential equation of complex quantities is

t jm e L

V dt

LC1

LR

dt d

=++ iii

... ... ( 1 ), where R, L, C and t are real quantities.

Let i = i m e j t be a trial solution.

t je j dt d

= mii

and dt =

je t jmi

Putting the values of i, dt d i

and dt in equation ( 1 ) above, we have

t je j mi + LR

i m e j t

+ LC1

je t jmi

= t jm e

LV

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im ( j + LR + C L j 1 ) = LVm im ( j L + R + C j 1 ) = Vm im ( j L + R - C j ) = Vm

== j

jj

j1

-2Q

im =

+

C L j R

V

1 -

m

Putting this value of im in the trial solution, we have

i =

+

C L j R

e V

1 -

t jm

This equation shows that the resistance off ed by an inductor and a capacitor are j L and - j // C which are known as inductive eactance and capacitive reactance respectively. Their magnitudes are L and 1 / C respec ively. The inductive and capacitive reactance are represented by symbols Z L and Z C while their magnitudes are equal to X L and X C.

Z L = j L, Z C = - j / C, X L = L, X C = 1 / C.

The summation of Z L, Z C and R is called the impedance ( Z ) of the series L-C-R circuit. The unit of impedance is ohm

Z = R + Z L + Z C = R + j ( L - 1 / C )

i = Ze V t jm

= ZV

The above equation represents Ohms law for complex current, complex voltage and impedan e. Impedance is also complex which can be expressed as Z = l Z l e j .

i =

e Z

e V j

t jm

=

)))) -t ( je Z

V =

Z V

[ cos ( t - ) + j sin ( t - ) ]

where, l Z l = 2

2C

1 L R -

+

Now, = Re ( i ) = 2

2

m

C1

L R

t ( cos V

-

) -

+

=

Z t ( cos V ) -m

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The equation shows that the current in the circuit changes according to cos ( t - )) and lags the voltage by phase as shown in the following graph.

In the second graph above, point H shows the compl x impedance, Z = R + j ( L - 1 / C ) in the complex plane where x-coordinate of the poin is R which is the real part of the complex impedance and y-coordinate of the poin is L - 1 // C which is the imaginary part of the complex impedance.

From the figure, tan = ODHD

= R

C1

L -

and l Z l = OH = (( )) (( ))))22 DH OD = 2

2C

1 L R -

++

8.5 Different Cases

The rules regarding the quivalent resistances for series and parallel connection of R are also applicable to jL and - // C. Using the above geometrical construction, the relationship between voltage and c rren for various circuits can be derived.

( 1 ) A. C. Circuit Containing only Inductor:

The impedan e Z = j L = j X L is represented by the point A in the complex plane as shown in he figure.

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OA forms an angle pipi // 2 with the real axis indicating that the current lags the voltage by pi // 2 as shown in the above figures. Further OA = L = l Z l. So the equation for current can be written as

= L

2t cos V -m

pipi

=

L

m

X2

t cos V -

pi

( 2 ) A. C. Circuit containing only a capacitor:

The impedance Z = - j / C = - j X C is represented by the point F in the complex plane as shown in the figure.

OF forms an angle - pi / 2 with the eal axis indicating that the current leads the voltage by pi // 2 as shown in the above figures. Further OF = 1 / C = l Z l. So the equation for current can be written as

=

C1

2t cos V m

pi++

=

C

m

X2

t cos V

pi+

( 3 ) A. C. Circui containing R and L in series:

The impedance Z = R + j L is represented by the po nt H n the complex plane as shown in the figure.

OH l Z l = 222 L R ++ = 2L2 X R +

From the figure, = tan -- 1 [RL ] = tan -- 1 [

RXL ]

= L R

t ( cos V222

m ) -+

Here the current lags the voltage by a phase , as given by the above equation for .

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( 4 ) A. C. circuit with R and C connected in series:

The impedance Z = R - j / / C = R - j XC is represented by the point H in the complex plane as shown in the figure.

OH = l Z l = 222

C1

R

+ = 2C2 X R +

Here, is negative and is given by

= tan - 1

R C1

= tan -- 1

R XC

and

=

C1

R

t ( cos V

222

m )

+

++ =

X R

t ( cos V2

C2

m ) ++

++

Here the current leads the voltage by a phase , s giv n by the above equation for .

( 5 ) An A. C. circuit having L and C connected in series:

The impedance Z = j L - j / C = j X L - j XC is represented by the point G in the compl x plane as shown in the figure.

l Z l = L - 1 / C = X L - XC

Here L > 1 /// / C, and current lags voltage by a phase pipi / 2.

== pipi / 2 and =C

1 L

t ( cos V ) m

2222pi

-

-

If L < 1 / C, current leads voltage by a phase pipi //// 2 and

= - pi / 2 and =C

1 L

t ( cos V ) m

22pi

+

--

( 6 ) A. C. circuit containing parallel combination of L and C, connected in series with R:

If Z1 = resultant impedance of the parallel combination of the inductor and the capacitor,

1Z1

=

LC Z1

Z1

+ = L j

1 j

1

C

++

-

=

L1

C j -

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Z 1 =

L 1

Cj1

--

= -

L 1

C

j --

equivalent impedance of the circuit,

Z = Z1 + R = R -

L 1

C

j -

If C > L

1

, the impedance Z can be

represented in the complex plane as shown in the graph.

Here, is negative and given by

tan = ODHD

=

L 1

C

1

R -

l Z l = 22

L1

C

1 R

+

-

and =

22

m

L1

C

1 R

t ( cos V )

+

+

-

8.6 r.m.s. Values of Alternating Voltage and Current

Special ammeters and v ltmeters are used to measure alternating voltage and current. These meters measure the r. s. ( root mean square ) value of the voltage and current.

Root mean square of a quantity varying periodically with time means the square root of the mean of the squar s of the quantity, taken over a time equal to the periodic time.

For V = Vm cos t,

the av r ge value of V 2 < V 2 > = < Vm2 cos2 t >

= Vm2 2t2 cos 1 ++

= Vm2 2t2 cos

21

+

The average value of 1 / 2 is 1 / 2 and that of cos 2t is zero for one periodic time.

< V 2 > = 2

V 2m

V r.m.s. = >< 2V = 2

Vm Similarly, r.m.s. =

2m

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8.7 Series Resonance

The current in L C R series circuit is given by

= 2

2

m

C1

L R

t ( cos V

-

) -

+

= m cos ( t - ), where, m = 2

2

m

C1

L R

V

-

++

Now, r.m.s. = 2

m =

22

m

C1

L R

2V

-

+

= 2

2C

1 L R

.sm. V

-

+

= Z

.s.m.r V

Thus, value of r.m.s. varies with . If = 0 is taken such that 0L = 1 // 0C , then

r.m.s. = R.s.m.r V

which is the maximum va ue of r.m.s.

The L C R series circuit is aid to e in resonance when the r.m.s. value of the current becomes maximum for a particular frequency, 0, of the voltage source.

Now, 0 = LC1

is known as the natural angular frequency or resonant angular frequency

of the given L C R series circuit.

Resonance is achieved when the imaginary part f the impedance is zero.

The figure shows the r.m.s. versus c rves for two different values of R R1 < R2 ). The resonance curve becomes sharper as the value of R reduces.

Q - Factor:

The sharpness of the resonance curve of the series L C R circuit is measured in terms of a quality known as the Q factor.

The maximum power 2 r.m.s. . R is obtained during resonance. The power

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reduces to half of the maximum power when the r.m.s. is equal to 2

) max ( r.m.s.. This

situation is shown in the graph where it is found that there are two values of angular frequency, 1 and 2 for which power is reduced to half the maximum power.

( 2 - 1 ) = is known as the half power bandwidth. The sharpness of the r sonance increases with the decrease in the value of the half power bandwidth,

The Q factor is defined as Q =

0.

Larger the value of Q, sharper will be the resonance curve.

It can be proved that the value of = LR

. Also we know hat 0 = LC1

.

Q =

0 =

CL

R1

Thus, the Q factor depends on the compon nt o the circuit. It gives the information about the tuning of the circuit as well as the selectivity of the circuit.

To tune a known frequency source, like a T V. r a radio, one has to select the right value of the inductor or the capacitor. Resona ce is obtained only when both L and C are present because at the time of resonance hey cancel reactance of each other. Hence, resonance never occurs in case of R-L or R C cir uit.

8.8 Phasor Method

Consider a harmonic function = m cos ( t + ).

A vector of magnitude m is drawn from the origin of a coordinate system a an angle of t + with the X-axis as shown in the figure.

The phase ( t + ) changes with time, i.e., the angle between he vector shown in the figure and the X-axis keep on changing with time. The vector rotates with the angular frequency in the XY plane. Such a rotat ng vector is known as phasor.

The X-component of the vector at any instant t gives the value of m cos ( t + ) which is the instantaneous value of the current. To add several functions like 1 cos ( t + 1 ), 2 cos ( t + 2 ), .... etc., one has to take the algebraic summation of the X-components of the respective phasors. One can deal with the sine function by taking the Y-component of the vector.

Now, consider the example of addition of two harmonic functions,

11 = 1 m cos ( t + 1 ) and 2 = 2 m cos ( t + 2 ).

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The figure shows the vectors representing 1 and 2 and their resultant vector . The amplitude of the resultant vector = OP and its phase angle is using the law of triangle for addition of vectors. The magnitudes of 1, 2 and are given by 1m, 2m and m respectively.

The angle between 1 and 2 is 2 - 1 as can be seen from the figure.

Now,

m2 = 1m

2 + 2m

2

+ 2 1m 2m cos ( 2 - 1 )

= 1m2

+ 2m2 + 2 1m 2m cos

where, = 2 - 1

8.9 Use of Phasor in an A. C. Circuit

An A. C. Circuit containing only resisto :

In an A. C. circuit containing only a resistor, the phase difference between the voltag an current is zero ( = 0 ). Here, current is d wn along the X-direction and as = 0, voltage is also drawn in the same direction as shown in the figure.

An A. C. Circuit ontaining only inductor:

Here, current lags the voltage V by pi // 2 radian, or in other words, th voltage leads the current by pi / 2 radian. Hence, if is rep esented along the X-direction, V will be along the Y-direction as shown in the figure.

A A. C. circuit containing only capacitor:

Here, current leads the voltage V by pi / 2 radian, or in other words, the voltage lags the current by pi // 2 radian. Hence, if is represented along the X-direction, V will be along the negative Y-direction as shown in the figure.

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L - C -- R series A. C. circuit:

The phasor diagram of each of the component is shown in the figure. Here, since L, C and R are in series, the same current flows through all the three components. If V is the applied voltage, then

V = VL + VC + VR

where, VL, VC and VR are the voltage drop across the inductor, the capacitor and the resistor respectively. If the current, , flowing through the above circuit is represented along the X-direction, then the phasor diagram of the series circuit will be as shown in the next figure.

It can be seen from the figure that

V 2 = ( VL - VC ) 2 + VR2

If m is the maximum current, then

VR = m R, VL = m X L and VC = m X C

V 2 = m2 ( X L - X C ) 2 m2 R2

V = m 2R 2) X X ( CL - +

For maximum current, V = V m

m = 2R 2) X X (

V

CL

m

- +

= l l ZVm

If = phase angle between the applied voltage and current phasors and VL > VC , then the cu ren lags the voltage by a phase . If VL < VC , then current would have led the voltage.

From the above graph,

tan = R

CLV

V V - =

R X X

m

CmLm -

=

R X CL X -

8.10 L C Oscillations

If the two ends of a charged capacitor are connected by a conducting wire or a resistor, then it gets discharged and the energy stored in the form of electric field between its two plates gets dissipated in the form of joule heat.

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Now, consider an inductor having negligible resistance connected to a charged capacitor as shown in the figure. Such a circuit is called an L-C circuit.

Let Q0 be the charge on the capacitor on the initially charged capacitor to which an inductor is connected.

Let Q = charge on the capacitor at time t = t, and = circuit current during discharge of capacitor.

applying Kirchhoffs law to the closed circuit, we have

CQ

dtd

L- + = 0

But = dtdQ

- ( negative sign indicates that he charge on the capacitor decreases with time. )

2

2

dtQd

L + CQ

= 0

2

2

dtQd

= - LCQ

Comparing the above equation with the differential equation of the simple harmonic motion,

2

2

dtyd

= - 02 y

Q is analogous to the variab e y and 02 is analogous to LC1

. The solution to the above

equation would be,

Q = Qm sin ( 0 + ) ... ... ... ( 1 )

Here, Qm and e the constants which can be determined from the initial conditions.

Putting the value of Q = Q0 at time t = 0 in the above equation,

Q0 = Q sin ... ... ... ... ... ( 2 ) Differentiating equation ( 1 ) with respect to time t, we have,

=dtdQ

= Qm 0 cos ( 0t + )

But at time t = 0, = 0.

0 = Qm 0 cos = pi / 2 ( Q Qm and 0 are not zero )

Putting the value of = pipi //// 2 in equation ( 2 ) gives Qm = Q0

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Putting the value of = pipi //// 2 and Qm = Q0 in equation ( 1 ), we have,

Q = Q0 sin ( 0t + pipi // 2 ) = Q0cos 0t ... ... ... ( 3 )

This equation shows that the charge on the capacitor changes in a periodic manner.

Differentiating the above equation with respect to time t, we have,

= dtdQ

= - Q0 0 sin 0t ... ... ... ( 4 ) This equation shows that the current through the inductor also changes n a periodic manner.

At time t = 0, the charge on the capacitor is maximum and the c rrent through the inductor is zero. The electric field intensity and hence the energy associated with the capacitor ( Q2 / 2C ) is maximum. The energy associated with the magneti field of the inductor is zero.

With the passage of time, the charge and hence the ene gy associated with the capacitor decrease as per the equation ( 3 ). At the same time the current through the inductor and hence the magnetic field and energy associated with it ( L2 / 2 ) increase as per the equation ( 4 ). It can thus be concluded that the energy of the electric field of the capacitor gets converted into the energy of the magnetic field of the nductor.

At time t = pi / ( 20 ), Q = 0, and be mes maximum and the entire energy stored in the electric field gets converted into the ene gy s ored in the magnetic field.

At time t = pi // 0, the charge on the capacitor again becomes maximum but with reverse polarity and the current in the indu tor ecomes zero. This phenomenon of charge oscillating between the capacitor and inductor in periodic manner is known as L-C oscillations.

These oscillations of charge esults in the emission of electromagnetic radiations which result in the decrease of ene gy associated with the L-C circuit. Such a circuit is known as the tank circuit of the os illator

8.11 Power and Energy associated with L, C and R in an A. C. Circuit

In an A. C. c rcuit voltage and current continuously change with time. In a series L-C-R circuit, the nstantaneous power

= V Vm cos t . m cos ( t - )

= Vm m cos t . cos ( t - ) = 2V mm

[ cos + cos ( 2t - ) Real power, P = Average value of instantaneous power for the entire cycle

=

+

T

0

mmT

0dt )t 2 ( cos

T1

dt cos T1

2V

-

= 2

V mm

. cos TT

= 0

T

0dt )t 2 ( cos

T1

-Q

= cos2

m

2V

m = Vr.m.s. r.m.s. cos ... ... ... ( 1 )

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Special cases:

( i ) Circuit containing only resistor:

= 0 P = Vr.m.s. r.m.s.

( ii ) Circuit containing only inductor:

= pipi // 2 cos pipi // 2 = 0 P = 0

When the current through the inductor increases, the energy from the voltage source gets stored in the magnetic field linked with the inductor and is given back to he circuit when the current through the inductor decreases. Hence power consumed by the circuit is zero. Thus, there is current in an A. C. circuit containing inductor without consuming any power.

( iii ) Circuit containing only capacitor:

Here, = - pi / 2 cos ( - pi / 2 ) = 0 P = 0.

The energy consumed in charging the capacitor is sto ed in the electric field between the plates of the capacitor and is given back to the circui when current in the circuit decreases.

( iv ) Series L-C-R circuit:

Here, cos = 2

C1

L R

R

-

++

Putting this value of cos i equation ( 1 ), we get the value of the power in the series L-C-R circuit. It is less than whe only resistance is present in he circuit. Its value is maximum when 2LC = 1.

Thus, power in A. C. circuit containing only inductor or capacitor is zero. The current flowing in such a circuit is called wattless current.

8.12 Transformer

When powe P = V has to be transmitted over a long distance from the power station to th city through the cable having resistance R, 2R amount of power gets lost in the form of hea which is very large. If this power is transmitted at a very high voltage, then the current wil be less thereby reducing the power loss. Moreover, before supplying this power to the household, its voltage has to be reduced to a proper value.

Both the increase as well as decrease of voltage can be done using a transformer. The transformers, which increase the voltage are called step-up transformers, while those which decrease the voltage are called step-down transformers. There is no appreciable loss of power in the transformers.

Principle:

Transformers work on the principle of electromagnetic induction.

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Construction:

The figure shows the construction of a step-up transformer and its circuit symbol.

Two coils of wire are wound very close to each other on the rectangular slab of iron. The rectangular slab of iron is made up of several layers of iron plates to minimize the eddy currents and power loss due to them. One of the coils called primary coil, P, is connected to the A. C. source. The other coil is called the secondary coil, S.

In the step-up transformer shown in the figure, the numb of turns of the primary coil is less than that of the secondary coil and is made up of thicker copper wire. In the step-down transformer, the arrangement is opposite to that of a s ep-up transformer.

The permeability of the material of the slab is h gh A a result, all the flux generated by the primary coil remains confined to the core and gets inked to the secondary coil. Therefore, the fluxes, 1 and 2 linked with the prima y and the secondary coils are proportional to the respective number of turns N1 and N2 o the coils.

1

21

2N turns primary of No.

N turns secondary of No.

co l primary withlinked fluxco l secondary withlinked flux

==

... ... ... ( 1 )

According to Faradays law,

the emf induced in the p imary coil, 1 = dt d 1

- and

the emf induced in the primary coil, 2 = dt d 2

-

From equation ( 1 ), we have, 2 = 112

NN

dt d 2

= dt d

NN 1

12

2 = 12

NN

1 or 12

=

12

NN

= r

is known as the transformer ratio. For a step-up transformer, r > 1.

A the power loss in the transformer is negligible, power input = power output

2 2 = 1 1 12

=

12

=

12

NN

= r

The above result is valid for an ideal transformer having no power loss. Actually, some power is lost in the primary coil in the form of heat, some in the magnetization and demagnetization of the iron core and some in the form of eddy currents formed on the surface of the iron core. As a result, the output power is less than the input power.

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Physics Alternating Current