Representation of AC Current and Voltage by Rotating Vectors – Phasors

We have seen many electrical circuits which have resistors connected

across an AC source, inductors across an AC source, capacitors across

an AC source and also the combination of any two or all three of these

components connected across an AC source. In case of a resistor, the

current across a resistor is in phase with the voltage source. But in

case of an inductor or a capacitor, the current either lags or leads the

voltage source by any certain value. Now here is where we use the

concept of phasors to relate the current and voltage.

Phasors

A phasor is a vector that is used to represent a sinusoidal

signal. When we have more than one sinusoidal signal with the

same frequency, different phase and different amplitude then

we can use this phasor diagram to represent the phase

difference between these sinusoidal signals.

The magnitude of the phasor represents the peak value of the

voltage and the current.

The term “phasor” can also be applied to impedance and

related complex quantities that are not time dependent.

The international standard is that phasors always rotate in the

counterclockwise direction

We know that any sinusoidal signal in a general form can be

represented as:

A(t)=

A

m

sin(ωt+ϕ)

Where Am=Peak Amplitude, ω = Angular Frequency and ϕ= Phase

Shift. Now to represent the above sinusoidal signal using phasor.

Now, this phasor is nothing but a vector which rotates around its

origin at a constant speed of rad/s in an anticlockwise direction.

Source: Wikipedia

So now as you can see in the above diagram as the vector rotates in an

anticlockwise direction at a speed of rad/s and if we take the

projection of this vector in Y-axis then we get the instantaneous value

of this sinusoidal signal. If we dot all the values which are possible for

this phasor then we can reproduce the sinusoidal signal.

Rules for Drawing a Phasor Diagram

Rule 1: The length of the phasor is directly proportional to the

amplitude of the wave depicted.

Rule 2: In circuits which have combinations of L, C & R in

Series it is customary to draw the phasor representing Current

horizontally and call this the Reference phasor. This is because

the current in a series circuit is common to all the components.

Rule 3: In parallel circuits, where L, C and R are connected in

parallel the phasor representing the Supply Voltage is always

drawn in the Reference direction. This is because in a parallel

circuit it is the supply voltage that is common to all

components.

Rule 4: The direction of rotation of all phasors is considered to

be Anticlockwise.

Rule 5: In any one diagram, the same type of value(RMS, peak

etc)is used for all phasors, not a mixture of values.

Phasor Representation

There are three ways of phasor representation in mathematical form:

Polar Form: Suppose we have a phasor which has an amplitude

of Vm and makes an angle with the horizontal axis. So in the

polar form, we can represent it as Vm.

Rectangular Form :In this form we can represent any phasor as

complex number like A+iB

r=

(

A

2

+

B

2

)

−

−

−

−

−

−

−

−

√

ϕ=arctan(

B

A

)

Exponential Form : Here we represent the phasor in the form of as

V m

e jϕ

Solved Questions for You

Question: The direction in which the phasor diagram rotates.

Solution: Anticlockwise.

Question: Mention the basic representation of phasor diagram

Solution:

Polar Form

Rectangular Form

Exponential Form

Question: Which property of a sine wave does the length of a phasor

represent?

Solution: Amplitude

Question: In a phasor diagram, the frequency of each of the waves

shown by the phasors is:

a. The same for all waves, and shown by the point of origin of the

phasors.

b. Different for each wave, and proportional to the different

lengths of the phasors.

c. Different for each wave, and shown by the change in angle of

the phasors.

d. Not shown.

Solution: (d) Not shown.

AC Voltage Applied to a Resistor

Resistance is the opposition to the flow of current. AC is a current that

changes its polarity. Hence, as we shall see a Resistor does offer

resistance to AC. Here we will quantify this resistance and try to

connect its value to the value of the resistance in a DC circuit. Let us

see!

Alternating Current

When a constant voltage source or battery is applied across a resistor

current is developed in resister. This current has a unique direction

and flows from the negative terminal of a battery to positive terminal.

The magnitude of the current remains constant as well. If Direction of

current through resistor changes periodically then current is called

alternating current.

Square wave AC current

Sinusoidal AC Current

Resistor in an AC circuit

To have a sinusoidal varying alternating current we need to have an

AC voltage source because current is directly proportional to voltage.

An AC generator or AC dynamo can be used as an AC voltage source.

Voltage V(t) is applied across resistance R. V(t) is sinusoidal voltage

with peak Vm and time period T.

T=

1

f

=

2π

ω

Where f is frequency and ω is angular frequency. This kind of circuit

is a purely resistive circuit. According to Kirchhoff’s law –

v(t)=Ri(t)

i(t)=

v(t)

R

i(t)=

V

m

sin(ωt)

R

i

m

=

V

m

R

i(t)=

i

m

sin(ωt)

Here voltage and current has same frequency and both are in same

phase.

Average Value of the Current

The average value of current can be found out by summing over the

total change in the voltage and dividing it by the number of times we

do the measurements. This can be done as:

i

avg

=

∫

T

0

i(t)dt

∫

T

0

dt

i

avg

=

1

T

∫

T

0

i(t)dt

i

avg

=

1

T

∫

T/2

0

i

m

sinωtdt−

1

T

∫

T

T/2

i

m

sinωtdt

i

avg

=

i

m

T

[

cosωt

ω

]

T/2

0

−

i

m

T

[

cosωt

ω

]

T

T/2

i

avg

=

i

m

Tω

[cosπ−cos0−cosπ+cos2π]

Hence,

i

avg

=0

Average value of a AC current over a cycle is zero because in 1st haft

of time period current is positive and in 2nd half current is negative.

Root Mean Square Value of Current

i

rms

2

=

∫

T

0

i

2

(t)dt

∫

T

0

dt

i

rms

2

=

1

T

∫

T

0

i

2

(t)dt

i

rms

2

=

1

T

∫

T

0

i

2

m

sinωt

2

dt

i

rms

2

=

i

2

m

2T

∫

T

0

(1−cos2ωt)dt

i

rms

2

=

i

2

m

2T

[t−

sin2ωt

2ω

]

T

0

i

rms

2

=

i

2

m

2T

[T−0−

sin2ωT–sin0

2ω

]

i

rms

2

=

i

2

m

2T

[T−

sin4π

2ω

]

i

rms

2

=

i

2

m

2

Hence

i

rms

=

i

m

2

–

√

Solved Example for You

Q. A circuit has an ac voltage source of 100V and 50Hz frequency is

and 1KΩ resister. Find peak and rms value of current. How much time

current will take to reach its first negative peak?

Solution: Peak voltage

V

m

=100volt

Peak current

i

m

=

V

m

R

i

m

=

100

1000

A

i

m

=0.1A

RMS value

i

rms

=

i

m

2

–

√

i

rms

=

0.1

2

–

√

=0.070A=70mA

Time taken to reach first negative peak:

t=

3T

2

=

3

2×50

=.03sec

AC Voltage Applied to a Capacitor

A pair of conductors separated by some medium can be used as a

capacitor. Here we will try and see how a Capacitor behaves when ac

is passed through it. We will study an ac circuit with a capacitor and

try and get a measure of the current and other parameters.

The Capacitor in an AC Circuit

Let us suppose I have a capacitor which is connected to a DC source

and I find that no current flows through it, so if I connect a lamp to

that circuit, then the lamp does not glow which mean no current flows

through the capacitor. This seems to make sense because we know

that there is an insulating medium present between the plates of a

capacitor so current can’t flow through it.

Now if I connect an AC source with the capacitor, I find that the

current is flowing through it, and now if I put a lamp in the circuit the

lamp glows, which means current flows in the AC circuit. So a

capacitor is an insulator in a DC circuit and a conductor in the AC

circuit!

What Happens on AC Biasing?

Now how is it that current flows in the AC source and not where the

DC source is connected? In the case of the DC source, the plates of the

capacitor acquire a positive and negative charge respectively. Now

during that very short while when the capacitor is getting charged to a

voltage which is given by :

V=

Q

C

During this very short interval the lamp will glow, but then after the

same interval of time, the current stops flowing. In the case of an AC

source, we have an alternating voltage which continuously charges

and then discharges the capacitor. While charging the capacitor the

voltage across the plates of the capacitor rises and the charge also

builds up, and when the voltage across the plates decreases the charge

will also decrease.

Then when the voltage increases in opposite direction (i.e when

voltage reverses) the capacitor gets charged in the reverse order.

Hence the plate which was positively charged will become negative

and vice versa. Therefore once again the voltage decreases and comes

back to zero and the charge on the plates of the capacitor also becomes

zero. Hence, we have :

Q=C×V

Mathematical Expressions

The current flowing in the AC source is given by:

i=

dQ

dt

Now the Alternating Voltage applied to a Capacitor is given by:

V=

V

m

sin(ω×t)

Where Vm = amplitude of voltage and ω = angular frequency. Charge

Q on capacitor is:

Q=C×V

Which can be written as:

Q=C×

V

m

×sin(ω×t)

Now we can write :

i=

dQ

dt

i=

d(C×

V

m

×sin(ω×t))

dt

i=C×

V

m

×ω×cos(ω×t)

i=

V

m

×sin(ω×t+

π

2

)

1

ω×c

We can see that the current also varies sinusoidally having a phase

difference of 90 degree.

i=

i

m

×sin(ω×t+

π

2

)

Maximum Current

The maximum current will flow when sin(θ) i.e.

sin(ω×t+

π

2

)

will be unity.

Capacitive Reactance

Take a look above, the term multiplied by the sin term denotes the

value of current im. The term in the denominator can then be treated as

some form of a resistance offered to the AC. Therefore, we have:

i

m

=

V

m

X

c

Capacitive reactance

X

c

=

1

ω×c

Therefore, we may also say that Capacitive reactance

X

c

=

1

2πfc

The SI unit of Capacitive Reactance is Ohm same as that as resistance.

The Capacitive reactance is inversely proportional to the frequency of

the alternating voltage which is applied, thus for low frequency, the

reactance is extremely high and for high frequencies, the reactance

decreases.

This also explains that when the frequency is zero then capacitive

reactance is infinite. So a capacitor does not allow a DC current to

flow through it because the capacitive reactance is infinite.

Solved Examples For You

Q. A 30 F capacitor is connected to a 240 V,60 Hz circuit. What is the

current flow in this circuit?

Solution:

X

c

=

1

2πfc

X

c

=

1

2π×(60)×(30×

10

−6

)

X

c

=88.42Ω

I=

240V

88.42Ω

I=2.71Ampere

AC Voltage Applied to an Inductor

We will start with making an Inductor, let’s take an insulated wire and

a pencil. Take the wire and wind it around the pencil and then slide the

pencil away from the wire. We will be left with the coiled wire. This

coil is an inductor which is a very important circuit element. Let’s

study this circuit element in an AC circuit.

Inductor with AC biasing

So now we come to the topic which is AC Voltage applied to an

Inductor. This inductor can block or oppose the alternating current

flowing through it. We will take two cases. In the first case, we have

an inductor which is connected to a DC supply whereas in the second

case we have an inductor which is connected to an AC supply.

In the first case, we see that a constant current flows through this

inductor and the bulb connected to it glows brightly. Hence the

inductor simply behaves like a coil of wire. But in the second case, the

bulb does not glow as brightly as the first one, this happens because

the inductor opposes the flow of alternating current. Hence the bulb is

dim. Now we will use Lenz’s Law to explain opposition to the flow of

AC current through an Inductor:

Recall that an alternating current varies in magnitude as well as in

direction. Now when the current increases from zero to peak value

during this time interval the current through the coil increases which

means the magnetic field linked to the coil also increases. Hence there

will be an induced EMF and the direction of the induced EMF is given

by the Lenz’s Law as:

E=−

dϕ

dt

In case during a given time interval, the current decreases from a

maximum value to zero, the magnetic field linked with the coil

decreases. Hence, the induced EMF is given by :

E=−L

di

dt

Since

ϕ=L×i

The Voltage required by the AC source is given by:

V=L×

di

dt

To maintain the current, supplied voltage must be equal to the reverse

Emf. So the Voltage applied to the coil is therefore given by:

V=L×

di

dt

Relation between Current and Voltage across the Coil

V=L×

di

dt

di

dt

=

V

L

di

dt

=

V

m

sinωt

L

Where, Vm = peak value of the voltage and ω = angular frequency.

On integrating the above expression we get:

i=∫

V

m

sinωt

L

dt

i=

−

V

m

cosωt

ωL

i=

V

m

sin(ωt–

π

2

)

ωL

i=

i

m

sin(ωt–

π

2

)

where im = peak value of current. So, we may conclude that the

current lags the voltage by 90 degrees, which means the current

reaches the same value as the voltage after a quarter cycle.

Expression for Inductive Reactance

V=

V

m

sinωt

i=

i

m

sin(ωt–

π

2

)

i=−

i

m

cosω

i

m

=

V

m

ωL

i

m

=

V

m

X

L

Where XL = Inductive Reactance

Inductive reactance X L

=ω×L

Inductive reactance X L

=2πfL

The SI unit of Inductive Reactance is Ohm which is same as that of

the resistance. The inductive reactance is proportional to the angular

frequency. So when frequency increases the value of the reactance

also increases and vice versa. Therefore if you have a DC current for

which the value of f = 0 then XL = 0.

In other words, no opposition is offered to the flow of DC. Therefore

the lamp glows brightly when the DC source is applied because of

extremely low or zero resistance.

Solved Example For You

Ques. If inductance of an inductor L = 15 mH and frequency f = 80

KHz. Find its Inductive Reactance?

Soln. Given L = 15mH and f = 80KHz

We know that,

X

L

=2πfL

On substituting the values of f and L we get :

X

L

=2×π×80×

10

3

×15×

10

−3

X

L

=7.5KΩ

AC Voltage Applied to a Series LCR Circuit

In this section, we will study a series combination of a Resistor, an

Inductor, and a Capacitor also known as the Series LCR circuit. We

will study the growth of the current and other quantities in this circuit.

These circuits are the fundamental components of many important

devices. Let’s study the fundamental elements of this circuit.

Series LCR circuit

When a constant voltage source or battery is connected across a

resistor, current is developed in it. This current has a unique direction

and flows from the negative terminal of the battery to its positive

terminal. The magnitude of current remains constant as well.

If the direction of current through this resistor changes periodically or

alternately, then the current is called alternating current. An

alternating current or AC generator or AC dynamo can be used as AC

voltage source.

The figure shows basic LCR series circuit where a voltage Vs is

applied across RLC series circuit. We will solve this circuit by using

vector method. Vector drawn for resistance is along the X-axis

because current and voltage are in phase in case of a purely resistive

circuit and magnitude will be R.

Inductors and capacitors will be represented by their respective

reactance. In capacitor current leads the potential by 90° hence

reactance will be along the positive Y-axis with magnitude 1/ωC and

similarly inductive reactance will be along the negative Y-axis with

magnitude ωL as current in inductor lacks by 90°.

Net Reactance ‘X’

The resultant of XL and XC in the positive Y – axis can be given as:

X=

X

C

−

X

L

=(

1

ωC

−ωL)

This is the net reactance of the circuit.

Total Impedance ‘Z’

Now we need to find total impedance of RLC circuit. And it can be

given as:

Z=

R

2

+

(

1

ωC

−ωL)

2

−

−

−

−

−

−

−

−

−

−

−

−

−

−

√

And this resultant impedance makes angle θ with X-axis which can be

give as –

tanθ=

(

1

ωC

−ωL)

R

Current in The Circuit

Current in circuit can be given as:

V

S

=

V

m

sinωt

I

S

=

V

S

Z

I

S

=

V

m

sinωt

R

2

+

(

1

ωC

−ωL)

2

√

I

S

=

I

m

sinωt

Where Im is peak current.

I

m

=

V

m

R

2

+

(

1

ωC

−ωL)

2

√

=

V

m

Z

Case 1. XC > XL

When XC > XL, resultant vector for net reactance will be along

positive Y-axis and value of θ will be positive. Hence current will lead

voltage and circuit will behave as resistive-capacitive circuit.

Case 2. XC < XL

When XC > XL, the resultant vector for net reactance will be along

negative Y-axis and value of θ will be negative. Hence voltage will

lead current and the circuit will behave as a resistive-inductive circuit.

Case 2. XC = XL

When XC = XL, the resultant vector for net reactance will be zero and

value of θ will be zero. Hence current and voltage will be in the same

phase and circuit will behave as purely resistive circuit and peak

current will be maximum.

Resonance

Peak current is maximum when:

1

ωC

−ωL=0

1

ωC

=ωL

ω=

1

LC

√

=2πf

f=

1

2π

LC

√

This frequency is known as resonance frequency.

Solved Example For You

Q1: An LCR series circuit with inductance 1mH, capacitance 100mF

and resistance 1K Ohm are connected to AC voltage source. Find the

frequency of source for which current through the resistor is

maximum.

Ans: Maximum current will be drawn in circuit in case of resonance

frequency which is given by:

f=

1

2π

LC

√

f=

1

2π

1×

10

−3

×100×

10

−3

√

=

100

2π

≃16Hz

Power in AC Circuit: The Power Factor

So we have seen that DC and AC are two different entities. The power

delivered to the circuit by a DC is a result of the flow of charges or in

other words the electric potential. Here we will see what the power

factor in an AC circuit is. We use AC in our household circuits and the

concept of the power consumed in an AC circuit is very interesting.

Let us see!

Mathematical Analysis

Suppose a Voltage V is applied to a LCR circuit, where V is give by:

V=

V

m

×sinωt

The current in this case is written by:

I=

I

m

×sin(ωt+ϕ)

Where, Vm = Voltage Amplitude, Im = Current Amplitude, ω =

Angular Frequency, ø = Phase Constant

(Source: Wikipedia)

Now, Current Amplitude is related to Voltage Amplitude as:

I

m

=

V

m

Z

Where, Z = Impedance of circuit and is given by:

Z=

(

X

L

−

X

C

)

2

+

R

2

−

−

−

−

−

−

−

−

−

−

−

−

−

−

√

tanϕ=

X

L

−

X

C

R

ϕ=arctan

X

L

−

X

C

R

Power Consumption

Now we will use all the above formulae to derive Power in AC circuit.

We know that P=VI. So we can write power consumed in an AC

circuit is:

P=(

V

m

×sinωt)×(

I

m

×sin(ωt+ϕ))

Note: Since P= V x I. So, If V=0 or I=0 then P=0. Also If V & I both

are positive then P will be positive and If anyone of either V & I is

negative then power will be negative and If V&I both are negative

then P will be positive.

Now Finding the Average Value of Power

P

avg

=(

V

m

×sinωt)×(

I

m

×sin(ωt+ϕ))

Using trigonometric identity:

2sinAsinB=cos(A−B)–cos(A+B)

We get:

P

avg

=Average of

V

m

I

m

2

⌊cosϕ–cos(2ωt+ϕ)⌋

$$P_avg=Average \ of \ \fracV_mI_m2\cos\phi-Average \

of \ \fracV_mI_m2\cos \left ( 2\omega t+\phi \right )$$

P

avg

=

V

m

I

m

2

cosϕ

Average of

V

m

I

m

2

cos(2ωt+ϕ)=0

P

avg

=

V

m

I

m

2

cosϕ

P

avg

=

V

m

2

–

√

×

I

m

2

–

√

×cosϕ

P

avg

=

V

rms

×

R

rms

×cosϕ

Resistive Circuit

Now for a Resistive Circuit we know that, Φ = 0 which implies cos Φ

= 1

P

avg

=

V

rms

×

R

rms

Inductive Circuit

But for a Inductive Circuit we know that, Φ = 90º which implies cos Φ

= 0. This is as we know that voltage across the inductor leads the

current by 90 degrees.

P

avg

=0

Capacitive Circuit

But for a Capacitive Circuit we know that, Φ = -90º which implies cos

Φ = 0. This is as we know that voltage across the inductor lags the

current by 90 degrees.

P

avg

=0

Power Factor

The power factor of an AC electrical power system is defined as the

ratio of the real power flowing to the load to the apparent power in the

circuit. It is dimensionless quantity and in the closed interval of -1 to

1.

Power Factor=

True Power

Apparent Power

Suppose a graph in which Voltage leads the current in a LCR circuit

by .We can resolve this voltage into two components that will be:

VIcosϕ−resistivepower

VIsinϕ–reactivepower

Reactive power is the power loss in the circuit which takes place due

tpo reactive components. So,

VIsinϕ

is referred to as wattles component of the voltage because it does not

produce any power loss.

Solved Example for You

Question: Two loads of 10KW each, are operating at a power factor

0.8 lagging (each).What is their combined power factor?

Solution: In the above question, it is not specified whether the loads

are connected in series or parallel, our answer does not depend

whether it is connected in series or parallel because in both the case

the answer will be same.

$$Power \ Factor=\fracTrue \ PowerApparent \ Power$$

Apparent Power=

True Power

Power Factor

For Load1:

Apparent Power=

10KW

0.8

Apparent Power=12.5KVA

For Load2:

Apparent Power=12.5KVA

Combined Power Factor:

Combined Power Factor=

Total True Power

Total Apparent Power

Combined Power Factor=

10+10

12.5+12.5

Combined Power Factor=0.8lag

LC Oscillations

Have you ever noticed that the small LED lights of some devices like

chargers keep on glowing even after these devices are switched off?

What causes the bulbs to glow? These circuits have capacitors and

inductors. The energy keeps oscillating in the circuit even after the

battery is disconnected. The combination of an inductor and a

capacitor creates an LC oscillator circuit. Let us learn more.

LC Oscillator

Whenever we connect a charged capacitor to an inductor the electric

current and charge on the capacitor in the circuit undergoes LC

Oscillations. The process continues at a definite frequency and if no

resistance is present in the LC circuit, then the LC Oscillations will

continue indefinitely.

This circuit is known as an LC oscillator. Let’s take a capacitor with

capacitance C and an inductor with inductance L. The capacitor is

fully charged with charge Qo. What we do here is to connect the

capacitor and the inductor end to end.

Assuming the inductor and capacitor to be ideal (meaning resistance

will be zero in the overall circuit). Initially, the capacitor C of the LC

circuit carries a charge Qo and current I in the Inductor is zero.

Therefore at time T = 0, the charge on the capacitor will be:

q(T=0)=

Q

o

Current Flowing:

I=0

At time T = t, the capacitor now begins to discharge through the

inductor. The current begins to flow in an anti-clockwise direction.

Therefore the charge of the capacitor decreases, but the energy of the

inductor increases. The energy gets transferred from the capacitor to

the inductor.

At this stage, there is the maximum value of the current in the

inductor. Then the relationship between the current and the charge will

be:

I=−

dq

dt

…………..(1)

The negative sign is added because as the time passes from 0 to t the,

charge on the plates of capacitor decreases i.e. charge decreases with

respect to time and thus the dq/dt obtained will be negative and this is

why we add a negative sign to make a current positive.

Applying Energy Conservation

Since, Heat Loss = 0 and Kinetic Energy = 0. Total energy which is

constant is given by:

1

2

×

q

2

C

+

1

2

×L

I

2

=

E

o

Now initially we had charge Qo and current flowing was zero I=0. So,

the total energy at T = 0, will only be the energy stored in the

capacitor and energy stored in the inductor will be zero.

E

o

=

1

2

Q

2

o

C

Now on equating the value of total energy we get:

1

2

×

q

2

C

+

1

2

×L

I

2

=

1

2

Q

2

o

C

On differentiating with respect to t we get:

1

2

2q

C

dq

dt

+

1

2

L×2I

dI

dt

=0

On further calculation we get the equation as:

L

dI

dt

=

1

C

q

Substituting the value of I from equation 1 we get:

d

2

q

d

t

2

=−

1

LC

q…………..(2)

And now double differentiating equation 1 with respect to t we get:

d

2

I

d

t

2

=−

1

LC

I…………..(3)

We can see that the above both the equation 2 and 3 of Charge and

Current both represent the Simple Harmonic Motion. That is both

charge and current are oscillating as a simple harmonic waves with

respect to time.

Solution of Simple Harmonic Motion Equation

In general charge as a function of time in SHM will be given as:

q(t)=

Q

o

sin(ωt+ϕ)

At t = 0, q(0) = Qo. Therefore putting the value of t=0 and q= Qo in

above equation we get:

Q

o

=

Q

o

sin(0+ϕ)

sinϕ=1

ϕ=

π

2

So charge as a function of time will be:

q(t)=

Q

o

sin(ωt+

π

2

)

q(t)=

Q

o

cos(ωt)

Now differentiating charge with respect to time so we will get current

as a function of time:

i(t)=−

dq(t)

dt

Solved Examples for You on LC Oscillator

Q1: In a LC circuit we have a inductor of L = 20mH and a capacitor

of capacitance 50F. Initially charge on the plate of capacitor is 10mC.

What is the total electric field energy stored in the capacitor and also

mention whether that energy in the capacitor is going to decrease with

time?

Solution: Given, charge Q=10 mC=10 x 10-3 C and capacitance C =

50 x 10-6 F. The electric field energy stored in capacitor is :

U

E

=

1

2

×

Q

2

C

U

E

=1J

Q2: In an oscillating LC circuit the maximum charge on the capacitor

is Q. The charge on the capacitor when the energy is stored equally

between the electric and magnetic field is:

Solution: Let Q denote maximum charge on capacitor. Let q denote

charge when energy is equally shared

1

2

×

1

2

Q

2

C

=

1

2

q

2

C

Q

2

=2

q

2

q=

Q

2

–

√

Transformers

Have you heard about Transformers before? Yes. You also might be

knowing about the purpose of using transformers. In simple language,

we can say that a transformer is a voltage controller device (a device

which controls voltage by increasing or decreasing it). Transformers

are basically used for the transmission of electrical energy. The main

principle operation of a transformer is mutual inductance between two

circuits which is linked by a common magnetic flux.

Transformers

A basic transformer consists of two coils that are electrically separate

and inductive but are magnetically linked through a path of reluctance.

The working principle can be better understood by the above-given

figure.

Source: Wikipedia

As shown in the above diagram an electrical transformer consists of

two coils named as the primary coil and secondary coil which is

wounded on a soft iron core. The soft iron core is laminated to

minimize eddy currents. Both the coil in the transformer has high

mutual inductance.

A mutual electro-motive force is induced in the transformer from the

alternating flux that is set up in the laminated core, due to the coil that

is connected to a source of alternating voltage. Most of the alternating

flux developed by this coil is linked with the other coil and thus

produces the mutual induced electro-motive force.

Induced EMF

The so produced electro-motive force can be explained with the help

of Faraday’s laws of Electromagnetic Induction given:

e=−

dϕ

dt

e

p

=−

d

ϕ

p

dt

e

s

=−

d

ϕ

s

dt

By using the above equations we get :

e

s

=

N

s

N

p

e

p

The ratio Ns Np is called transformation ratio. The transformer

formula is given by:

V

p

V

s

=

N

p

N

s

Where, Vp = primary voltage, Vs = secondary voltage, Np = number

of turns in the primary, Ns = number of turns in the secondary.

Types of Transformers

There are two types of transformers:

Step-up Transformer

Step-down Transformer

Step-up Transformer converts a low voltage into a high voltage. The

number of turns in the primary coil is less than the number of turns in

the secondary coil, i.e. Np < Ns.

Step-down Transformer converts a high voltage when current

decreases into a low voltage when current increases. The number of

turns in the primary coil is greater than the number of turns in the

secondary coil, i.e. Np > Ns. In short, a transformer carries the

operations shown below:

1. Transfer of electric power from one circuit to another.

2. Transfer of electric power without any change in frequency.

3. Transfer with the principle of electromagnetic induction.

4. The two electrical circuits are linked by mutual induction.

Solved Examples For You

Q1: The number of primary and secondary windings is 100 and 300

respectively. The primary voltage is given by 200V, determine the

secondary voltage.

Solution: Given:

N

p

=100

N

s

=300

V

p

=200V

The transformer formula is given by,

V

p

V

s

=

N

p

N

s

V

s

=

V

p

×

N

s

N

p

V

s

=200×

300

100

V

s

=600V

Q2: Why do we use a soft iron core for the construction of a

transformer?

Solution: This is because the hysteresis curve of a soft iron core is

extremely thin and thus having less energy loss. That is whatever

energy is transferred from primary to secondary coil there will be the

minimum loss if we use a soft iron core.