APPLICATION OF ELECTROMAGNETIC FIELD AND TRANSMISSION SYSTEM

 

Subject: 

PHYSICS

[mediator_tech]

Term:

FIRST TERM

Week:

WEEK 7

Class:

SS 3

Topic:

 

 

APPLICATION OF ELECTROMAGNETIC FIELD AND TRANSMISSION SYSTEM

CONTENT

  1. Electromagnetic Field
  2. Galvanometer
  3. Conversion of Galvanometer to Ammeter
  4. Conversion of Galvanometer to Voltmeter
  5. Transformer
  6. Types of Inductances
  7. Electric Motor
  8. How to Make a Simple Electric Motor
  9. Generator

 

Electromagnetic Field

The electromagnetic field is a combination of the electric field and the magnetic field. The electric field component is produced by the voltage on a current carrying conductor while the magnetic field is produced by the flow of electric current. The two fields permeate all the space around a current-carrying conductor.

 

Galvanometer

Galvanometer

A galvanometer is an instrument used to detect and measure small electric currents. It consists of a coil of wire that acts as an electromagnet when current flows through it. When connected in a circuit, the coil interacts with another magnetic field to deflect a needle, which indicates the magnitude of the current flow.

When current flows through a current carrying conductor placed in a magnetic field, it experiences a force. This principle is applied in the design of galvanometers. The force is:

  • directly proportional to the current flowing in the conductor
  • directly proportional to the magnetic field strength of the magnetic
  • directly proportional to the length of the conductor

The magnitude of the force is given as

F=BILsinθ ——– (1)

Where;

B- is the strength of the magnetic field of the magnet

I -is the current flowing in the conductor

L -is the length of the conductor

ϴ- is the angle the conductor makes with the direction of the magnetic field

A moving coil galvanometer is an instrument used for detection of very small current. It consist of

  • magnet
  • a coil
  • a soft iron core. This is place in the centre of the coil to concentrate the magnetic line of
  • Pointer and a scale

APPLICATION OF ELECTROMAGNETIC FIELD AND TRANSMISSION SYSTEM

When current flows through the moving coil galvanometer, its coil turns to give the pointer a deflection which is directly proportional to the current flowing in. The torque experienced by the coil is given as:

Γ=BANI ——– (2)

Where Γ is the torque, B is the magnetic field strength, A is area of the coil, N is the number of turns in the coil and I is the current.

The galvanometer can be adapted to measure large current like ammeter or to measure voltage like voltmeter

 

Conversion of Galvanometer to Ammeter

To convert a galvanometer to an ammeter, resistor of low resistance (called a shunt) is connected in parallel to the galvanometer. The current to be measured is divided by the shunt, and the resultant current flows through both the galvanometer coil and the shunt. Since the resistance of the galvanometer is much smaller than that of the shunt, only a small fraction of the total current will flow through it, while most of it will flow through the shunt. As a result, the galvanometer will show a small deflection which is proportional to the total current flowing in the circuit.

Conversion of Galvanometer to Voltmeter

To convert a galvanometer to a voltmeter, resistor of high resistance (called a series resistance) is connected in series with the galvanometer coil. The additional resistance increases the total resistance of the circuit, and reduces the current flowing through the galvanometer. Since only a small fraction of the total current flows through it, it will only experience a small deflection which is proportional to the potential difference across its terminals.

Transformer

A transformer is an electrical device that transfers energy from one circuit to another through electromagnetic induction. It consists of two coils, each with a number of turns around a core made from iron or some other ferromagnetic material. The current in one coil (the primary winding) induces a magnetic field in the core and across the secondary winding, creating an electric potential difference between its terminals. This is due to Faraday’s law of induction, which states that when the magnetic field produced by current in the primary winding changes, a potential difference is induced across the secondary winding.

The ratio of the voltage-induced across the secondary coil to the current flowing in the primary coil is called the turns ratio. This depends on many factors, including whether or not there is a magnetic core and the length of wire used in winding the coils. The larger the turns ratio, the greater the voltage and current that can be transferred through it.

One common application of transformers is to step down or step up voltages for transmission between power grids. Transformers are also used in many other applications, such as power supplies for electronic devices, audio amplifiers, and so on. Overall, transformers play a critical role in modern electrical systems by helping to increase efficiency, reduce energy losses, and improve safety.

APPLICATION OF ELECTROMAGNETIC FIELD AND TRANSMISSION SYSTEM

The voltage drop across the shunt is equal to the voltage drop across the galvanometer

Vs=Vg(IIg)Rg=IsRs

Conversion of Galvanometer to Voltmeter

A galvanometer can be converted to voltmeter by connecting a resistor of high resistance in series to it.

APPLICATION OF ELECTROMAGNETIC FIELD AND TRANSMISSION SYSTEM

The same amount of current flows through the shunt and the galvanometer.

VgRg=VVgRm

Transformer

Transformers are used in the transmission of electricity.

Principle: Whenever there is change in the magnetic flux linking a conductor or there is relative motion between a conductor and a magnet, electric current is induced.

The transformer consists of two coils of wire wound on opposite ends of a soft iron. When an

alternating current is flowing in one of the coil, current will be induced in the other. The alternating current in the (primary ) coil generates a changing magnetic flux. In accordance to the Faraday’s law of electromagnetic induction,

Whenever there is a change in the magnetic flux linked with a circuit, an electromotive force is induced, the strength of which is proportional to the rate of change of the magnetic flux linking the circuit.

E=−∂Ø∂t

The magnitude of the induced emf E is

  • Directly proportional to the rate of change of the flux – Directly proportional to the number of turns in the coil
  • Cross sectional area of the coil.

The direction of the emf is given by the Lenz’s law:

This states that the direction of the induced emf in such that it tends to oppose the change that is causing it.

 

Types of Inductance

1. Self inductance

This occurs when there is a change in the current flowing in a coil. This alters the flux in the coil thereby inducing an emf in the coil. This type of induction is called self induction. The emf generated is given as:

E=LδIδt

The energy stored in the inductor is given as:

energy=12LI2

APPLICATION OF ELECTROMAGNETIC FIELD AND TRANSMISSION SYSTEM

Varying the resistor R causes change in the current in the circuit. Consequently, the flux in the coil C is altered and EMF is generated. This is called self inductance

2. Mutual inductance

This is when the change in current in a coil induced emf in another coil close by:

The induced emf is given as

E2=MδI1δt

M is the mutual inductance and it is given as:

M=μ0AN1N2I

APPLICATION OF ELECTROMAGNETIC FIELD AND TRANSMISSION SYSTEM

For a transformer, the alternating current flowing in primary coil generates a changing flux link to the secondary coil. This induces a emf in the secondary coil. Below is the ideal (100% efficiency) transformer equation.

npns=IsIpnpns=VpVsnpns=VpVsIsIp

APPLICATION OF ELECTROMAGNETIC FIELD AND TRANSMISSION SYSTEM

The coil that brings current flow into the transformer is called the primary coil. The coil through which current flow out of the transformer is called the secondary coil

In practice, the efficiency of the transformer is never 100%because energy is lost in the transformer due to:

  1. Heating in the coil: this is as a result of the resistance in the primary and secondary coil of the transformer. It can be reduced by thick wire or wire with low resistance for the coils
  2. Eddy current losses in the coil: eddy current are produced by the varying flux cutting the iron They consume power from the primary coil. It can be reduced by laminating the core
  3. Hysteresis loss: This is wasted energy due to reversing of magnetization of core. It can be reduced by using a special alloy in the core of the primary coil or by using soft iron core.
  4. Flux leakage: This arises because not all the lines of induction due to current in the primary coil pass entirely through the iron core of the secondary coil. It can be reduced by using special form of coil windingis losses results in the efficiency of a transformer to be far below 100% and it is usually between 50-85%.

3. Transformer regulation

  1. This refers to how well the output voltage maintains its correct level for changes that take place in the input parameters. The most common cause of voltage fluctuations is variation in load current, but there may other reasons such as supply voltage and temperature.
  2. The primary coil of a transformer is designed with an optimum number of turns that gives the desired output voltage when there is no load on it. As the load current increases, the required number of turns in the secondary coil also changes in order to maintain the same ratio between primary and secondary voltages.

All these tend to reduce the efficiency of the transformer.

Transformer formula:

number of turns in primary coilnumber of turns in secondary coil=voltage in primary coilvoltage in secondary coil

npns=VpVs

Also,

number of turns in primary coilnumber of turns in secondary coil=current in secondary coilcurrent in primary coil

npns=IsIp

Efficiency of a transformer:

ɳ=power output at secondary coilpower input at primary coil×100ɳ=P0PI×100ɳ=IsVsIpVp×100

 

Solved Examples

1. A step down transformer operates a 2000 volt line and supplies a current of 60 A. the turns ratio of the transformer is 1/24. Calculate

(i) the secondary voltage

(ii) the primary current

(iii) the power output if the transformer is 100% efficientio:

(i) The secondary voltage is 2000/24 = 83.33 V

(ii) The primary current is 60/24 = 2.5 A

(iii) Assuming the transformer is 100% efficient, then the power output is 83.33 x 2.5 = 208.3 W.

2. A step upio:n The turns ratio is 1/24, so the primary current will be 24 times larger than secondary current. Assuming 100% efficiency, the power output will be 24 x 60 = 1470 W.

3. A step down transformer has a voltage of 120 V at its secondary terminal and supplies a primary current of 10 A. Calculateio:n

(i) The turns ratio is 1/12, so the secondary voltage will be 120/12 = 10 V.

(ii) The primary current will be 10 x 12 = 120 A.

(iii) Assuming 100% efficiency, the power output will be 10 x 120 = 1200 W.​io:n It is important to measure the voltage and current in a transformer carefully, as they can vary significantly depending on factors such as load conditions and ambient temperature. To ensure proper performance of a transformer, it is necessary to test its input and output parameters periodically to make sure they are within acceptable limits.

Solution:

Primary voltage, Vp = 2000 V,

Secondary current Is = 60 A

Turn ratio for step down transformer nsnp=124

(Since it is a step down transformer, the np is bigger)

(i) nsnp=VsVp124=Vs2000Vs=242000=0.012V

(ii) nsnp=IpIs124=Ip60Ip=6024=2.5A

(iii) Power output P0=IsVsP0=60×0.012=0.72W

 

2. A transformer is designed to convert a 25 V supply to an output of 240 V. If the transformer is 90% efficient, calculate the current in the primary winding when the output terminal are connected to a 240 V 80 W lamp.

Solution:

Primary voltage Vp = 25 V

Secondary voltage Vs = 240 V

Power output P0 = 80 W

Efficiency η =0.9

P0Pi=η80PI=0.9PI=800.9=88.9W

But PI=IPVP88.9=IP×25IP=88.925=3.6A

EVALUATION

  1. Differentiate between self inductance and mutual inductance
  2. List the factors that reduces the efficiency of a transformer
  3. State the Faraday’s law of electromagnetic induction
  4. Describe the process by which a galvanometer is converted to an ammeter.

Magnetic Induction

magnetic induction is the process by which a changing magnetic field creates an electric current in a nearby conductor. Self-inductance and mutual inductance are two related concepts that refer to the tendency of an electric circuit to resist changes in current flow, either by locking the current in place or by storing energy as a magnetic field. Electric Motor

This is a device which can convert electric energy into rotational mechanical energy. It consists of coils of wire pivoted in a magnetic field. When current flows through the coil, it experiences a torque which makes is to spin about its axis in the magnetic field.

APPLICATION OF ELECTROMAGNETIC FIELD AND TRANSMISSION SYSTEM

Electric Motor

How to Make a Simple Electric Motor

Required Materials

  • Magnet
  • Insulated wire guage-18
  • A battery
  • A transparent cup
  • Rubber band
  • Two large paper clips
  • Connecting wire and alligator clip​nt step is to place the magnet inside the cup. Next, wrap two loops of insulated wire around the paper clips and secure them with the rubber band. Connect one end of each wire loop to a battery terminal using alligator clips or other conductors, then hold one end of each wire close to the top of the magnet. First step is to place the magnet inside the cup. Next, wrap two loops of insulated wire around the paper clips and secure them with the rubber band. Connect one end of each wire loop to a battery terminal using alligator clips or other conductors, then hold one end of each wire close to the top of the magnet.

APPLICATION OF ELECTROMAGNETIC FIELD AND TRANSMISSION SYSTEM

The insulated wire gauge is wound round the battery ten times. The end if striped. On one of the end half is coated with a permanent marker. The other half is left uncoated. On the other side, the end is stripped bare. The wire gauge is given a gentle tilt and it begins to rotate.

 

Generator

This device converts mechanical energy into electrical energy. It is a reverse of the electric motor. It contains a coil pivoted in a magnetic field. When the coil is rotated in the magnetic field electrical energy is generated.

APPLICATION OF ELECTROMAGNETIC FIELD AND TRANSMISSION SYSTEM

The D.C generator has a split ring commutator and the e.m.f generated is unidirectional

Simple D.C Generator

APPLICATION OF ELECTROMAGNETIC FIELD AND TRANSMISSION SYSTEM

The A.C generator has a slip ring commutator and the emf generated is bi-directional

Simple A.C Generator

 

EVALUATION

  1. State the difference between the ac generator and the dc generator
  2. Explain the principle of an electric motor.
  3. with the aid of diagram explain the principle of operation of a transformer
  4. A galvanometer has a resistance of 5Ω. By using a shunt wire of resistance of 0.05Ω, the galvanometer could be converted to an ammeter capable of capable of reading 2A. What value of current flows through the galvanometer?
  5. The magnetic flux in a coil having 200 turns changes at the time rate of 0.08Wbs-1 Calculate the induced emf in the coil.
  1. The difference between an ac generator and a DC generator is that an AC generator produces alternating current, while a DC generator produces direct current. The principle of operation of an electric motor involves the interaction between magnetic fields and moving conductors to generate electrical energy