Basic Electrical Lecture Notes

faris

Elecrical Engg Department

Al-Ameen Engg College, Shoranur

BASIC ELECTRICAL LECTURE NOTES

Faris.k.k Basic Electrical Lecture Notes

IMPORTANT TERMS RELATED WITH ELECTRIC FIELD

Electric Field:-

The space around a charge where its effect can be felt is called electric field.

Electric Flux:-

Electric flux through a surface is the number of electric lines of force passing normally through the surface.

Potential:-

Work done by an external force in moving a unit charge from infinity to a point, without any acceleration against the electrostatic force.

Potential Difference:-

Work done by an external force in moving a unit charge from one point to another point, without any acceleration against the electrostatic force.

Voltage (V):-

Voltage is the difference in potential provided for helping the movement of charges. Its unit is volt (V).

Electro Motive Force (EMF):-

The energy required to do the work on a unit charge in order to move the charges is called EMF. Its unit is volt (V).

Current (I):-

It is the rate of flow of charges. Its unit is ampere (A).

Resistance (R):-

It is the opposition offered to the flow of charges. Its unit is ohm (Ω).

OHM’S LAW

“At constant temperature, potential difference across the ends conductor is directly proportional to the current flowing through that conductor”. i.e.

V α I

V=IR

Where ‘R’ is the proportionality constant, called resistance.

EEE Department Al-Ameen Engineering College


KIRCHHOFF’S LAWS

Kirchhoff’s voltage law (KVL):-

“The sum of voltage drops around a closed loop is equal to the sum of the voltage sources of that loop”

Or

“The algebraic sum of the product of current & resistance in each of the conductor in any closed path in a network plus the algebraic sum of the voltages in that path is zero”

Kirchhoff’s current law (KCL):-

“In any electrical network, current entering in to a junction is equal to current leaving the junction”

SOLUTION OF SERIES-PARALLEL CIRCUITS WITH ‘DC’ EXCITATION

Problem-1:-

Two coils connected in series have a resistance of 9 Ω and when in connected parallel have a resistance of 2Ω. Find the resistance of each coil?

When coils are connected in series, effective resistance = 9 Ω …………(1)

When coils are connected in parallel, effective resistance = 2 Ω …………(2)

Put eq(1) in eq(2), we get

Put eq(3) in eq(1), we get

(

= 3Ω or 6Ω , therefore = 6Ω or 3Ω.

Problem-2:-

The effective resistance of two resistors connected in parallel is 8Ω. The resistance of one of the resistor is 12Ω. Calculate the resistance of other coil and the effective resistance of the two resistors connected in series?



When resistances are connected in parallel, effective resistance = 8 Ω …………(1)

The resistance of one of the resistor is 12Ω, = 12Ω …………………………………………………… (2)

Put eq(1) in eq(2), we get (12* ) / (12 + = 8

12 = 8 * (12 +

12 - 8 = 96

= 24Ω.

Therefore the resistance of other resistor is = 24Ω

When resistances are connected in series, effective resistance = 36Ω.

Problem-3:-

A current in a circuit is due to a potential drop of 10V applied to a resistor of resistance 100Ω. What resistance would permit the same current to flow if supply voltage were 100V?

Case-1 Current produced in a 100Ω resistor due to 10V is

I = V/R = 10/ 100 =0.1A

Case-II Value of resistance would permit the same current (i.e. is 0.1A) if voltage is 100V,

R = V/I = 100/0.01 = 1000 Ω

Problem-4:-

Calculate voltage that must be connected across terminal ‘ab’ such that the voltage across 2Ω resistor is 10V.

Current passing through 2Ω resistor is , I = V / R = 10 / 2 = 5A.



Total resistance in the circuit is

= 2 + 1 +

= 7 Ω

Therefore voltage across ‘ab’ is v = I * = 5 * 7 = 35V

Problem-5:-

Write the mesh current equations in the circuit shown in figure and determine the currents?

By KVL in mesh A-B-E-F-A

-6

9 - 3 = 42 …………………………………(1)

By KVL in mesh B-C-D-E-A

-4 + 10 + 0 + 3 = 0

-3 + 7 = 10 …………………………….(2)

Eq (2)*3 + eq (1), we get

18 = 72, = 72/18 = 4A. ……(3)

Put eq (3) in eq(1), we get

9 - 3*4 = 42 , = 54/9 = 6A ……(4)



Problem-6:-

Using Kirchhoff’s law, find the current through 10 Ω resistor?

By KVL in mesh A-B-E-F-A,

-5

15 - 10 = 20 …………………………………(1)

By KVL in mesh B-C-D-E-A

-2 - 8 + 0 + 10 = 0

10 - 12 = 8 …………………………….(2)

Eq (1)*2 - eq (2)*3, we get

Therefore current through 10 Ω resistor is, - = 1A .

MAGNETIC CIRCUITS

Magnetic circuit may be defined as the path which followed by magnetic flux. The law of magnetic circuit is similar to those of electrical circuit.



Consider a solenoid or toroid have iron ring having a magnetic path of ‘l’ meter, area of cross section ‘A’ and a coil of

‘N’ turns carrying ‘I’ amperes wound anywhere as shown in figure

Then field strength inside the solenoid is

H =

Now B = * * H =

Therefore total flux produced

Φ = B * A

=

=

The numerator ‘NI’ produces magnetization in the magnetic circuit is known as magneto motive force (mmf) and the

denominator is called reluctance of the circuit.

Therefore,

Flux = .



Flux = = (Wb)

M.M.F

It is the work done in joules in carrying a unit magnetic pole once through the entire magnetic circuit. It measures in ampere turns (AT).

M.M.F, F = NI (AT)

FLUX

It is the amount of magnetic field produced by a magnetic source. It measures in Weber (Wb).

Flux, Φ = B * A (Wb)

RELUCTANCE

It is the opposition offered to the passage of magnetic flux through a material. It measures in .

Reluctance, S= ( )

COMPARISON OF ELECTRIC AND MAGNETIC CIRCUITS

Magnetic Circuit Electric Circuit



Flux = Current =

m.m.f (AT) e.m.f (Volt)

Flux Density B (Wb/ ) Current Density J (A/ )

Reluctance S= ( ) Resistance, R= (Ω)

Permeance (1/S) Conductance (1/R)

Once magnetic flux is set up due to energy , does not require any further energy to maintain it.

Continuous energy is required to maintain flow of electricity

Problem-7:-

An iron ring of mean length 1.2m & cross-sectional area 0.005 is wound with a coil of 900 turns. If a current of 2A in

the coil produces a flux density in the ring of 1.1T, calculate (a) mmf, (b) total flux in the ring and (c) relative permeability of iron at this flux density?

(a) mmf = NI = 900*2 = 1800AT(b) Flux , Φ = B*A = 1.1* 0.005 =5.5mWb

(c) Reluctance, S = mmf/ flux = 1800/(5.5* ) = 327272.723

Also S= , we get

=



=1.2 / (4π*10^-7*0.005*327272.723)

= 583.568

Relative permeability, = 583

Problem-8:-

A coil of insulated wire of 500 turns and a resistance of 4Ω is closely wound on an iron ring. The ring has a mean

diameter of 0.25m & a uniform cross-sectional area 700 . Calculate the total flux in the ring when a DC supply at 6V

is applied to the ends of winding. Assume = 550.

From Ohm’s law, I = V/R = 6/4 = 1.5A

= π*d = 3.14* 0.25 = 0.785m

MMF = NI = 500* 1.5 = 750 AT

Reluctance, S = = = 1.6225*

Flux, Φ = = 750/ (1.6225* = 0.462 mWb.

FARADAY’S LAWS OF ELECTROMAGNETIC INDUCTION

Faraday’s first law:-

“Whenever the magnetic field linked with a circuit changes, an emf is always induced in it”

Faraday’s second law:-

“The magnitude of induced emf is equal to the rate of change of flux linkages”

Explanation:-

Suppose a coil has N turns and flux through it changes from an initial value of ‘ ’ webers to final value of ‘ webers

in time ‘t’ seconds. Then remembering that by flux linkages mean the product of number of turns and flux linked with the coil, we have

Initial flux linkages = N ,

Final flux linkages = N ,



Therefore change in flux linkages = N ( )

Let ‘t’ b e the time taken for transition, the rate of change of flux linkages =

According to faraday’s 2nd law, this must be equal to emf induced in it.

Therefore induced emf, e = .

e = (in differential form)

LENZ’S LAW

“Electromagnetically induced emf set up a current in such a direction that magnetic effect produced by

it opposes the very cause producing it”

e = - (

Problem-9:-

A coil of 500 turns is linked by a flux of 0.4mWb. If the flux is reversed in 0.01second, find the emf induced in the coil?

N = 500 , = 0.4* mWb , = -0.4* mWb , t = 0.01 s

e = = - = 40V

Problem-10:-

The current flowing through the coil of 400 turns is suddenly changed so that the flux linking the turns increased by 6mWb in 0.2 second, calculate the induced voltage?

N = 400 , - = 6* mWb , t = 0.2 s

e = = - = -12V.

DYNAMICALLY INDUCED EMF



Here magnetic field is stationary & the conductor or coil cut across it. By changing the position of conductor, flux linked with circuit changes. Therefore emf is induced dynamically.

Generators works based on principle of dynamically induced emf. In generators, emf is induced in a conductor housed in revolving armature lying within a strong magnetic field.

Dynamically induced emf,

e = B v sinθ

Where B = flux density in Weber/ ,

= length of conductor in meter

V = velocity in m/s

Problem-11:-

A conductor of length 1m moves at right angles to a uniform magnetic field of flux density 1.5 Wb/ with a velocity of

50 m/s. Calculate emf induced in it?. Find also the value of induced emf when the conductor moves at an angle of 30° to the direction of field?

Case –I B = 1.5 Wb/ , = 1m, v = 50 m/s, θ = 90°

e = B v sinθ = 1.5*1*50*sin90 = 75V.

Case-II B = 1.5 Wb/ , = 1m, v = 50 m/s , θ = 30°

e = B v sinθ = 1.5*1*50*sin30 = 37.5V

STATICALLY INDUCED EMF



Here the conductor or coil is stationary & flux linked with it changes by simply increasing or decreasing the current producing this flux. This type of emf is obtained by changing the strength of current there by changing the flux linkages with the coil. There are two types of statically induced emf

(a) Self induced emf ,(b) Mutually induced emf

Self induced emf: - The emf induced in one coil due to change in current in the same coil is called self induced emf.

Self induced emf is directly proportional to rate of change of current producing it. That is

e α - , e = -L

e = -L

Where ‘L’ is the proportionality constant called self inductance

Mutually induced emf: - The emf induced in one coil due to change in current the other coil is called mutually induced emf.

Mutually induced emf in one coil ( coil-2) due to change in current in other coil (coil-1) is directly proportional to

rate of change of current . That is

,

Where ‘M’ is the proportionality constant called mutual inductance.

SELF INDUCTANCE (L)

Consider a coil of ‘N’ turns carrying a current ‘ i ’, which produces a flux links with that coil.

From Faraday’s law, induced emf in the coil is

e = - ( …………………….(1)

If ‘L’ is the self inductance, induced emf in the coil is

e = -L ……………………….(2)



Compare eq(1) & eq(2), we get

- ( -L

L = N

For a coil with non-magnetic core, flux is directly proportional to current . That is

=

In such a case inductance is,

L = N

If Φ= , we get

L =

Its unit is ‘Henry’ (H).

MUTUAL INDUCTANCE

Consider two coils 1&2 with N1&N2 turns respectively. The coils are located physically close to one another so that a flux

produced by current in coil-1 links with coil-2 as shown in figure.

From Faraday’s law, induced emf in the coil-2 due to change in flux is

…………………………….(1)

If ‘M’ is the mutual inductance, induced emf in the coil-2 due to current flow in coil-1 is

…………………………(2)



Compare eq(1) & eq(2), we get

=

M =

For a coil with non-magnetic core, flux is directly proportional to current . That is

=

In such a case mutual inductance is,

M = …………………(3)

If = , we get

M= ……………………..(4)

Therefore

M = or M=

Its unit is ‘Henry’ (H).

COEFFICIENT OF COUPLING

Consider two magnetically coupled coils A & B having turns respectively. Their individual self

inductances are . Therefore,

and ………………….(1)

Suppose a fraction of flux ( i.e ) is linked with coil-2.



Therefore mutual inductance ‘M’ is M = …………….…….(2)

Similarly a fraction of flux ( i.e ) is linked with coil-1.

Therefore mutual inductance ‘M’ is M = ………………….(3)

From eq(1), we get ……………..(4)

From eq(2) and eq(3), we get = …………….(5)

Compare eq(5) and eq (6), we get

=

=

=

Put k= √( ), we get

k =

k =

The constant ‘k’ is called coefficient of coupling & is defined as the ratio of mutual inductance actually present between the coils to the maximum possible value.

Problem-12:-EEE Department Al-Ameen Engineering College


An air-cored toroid has 450 turns and mean diameter of 300mm and a cross sectional area of 300m .

Determine self inductance of coil & the average voltage induced in it when a current of 2A is reversed in 40ms?

N = 450 , d = 300mm, l= πd = π * 300*10^-3 = 0.9424 m

, A = 300*10^-6

S = = =

Self inductance L =


Basic Electrical Lecture Notes

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