Kerala Syllabus 10th Standard Physics Solutions Chapter 3 Electro Magnetic Induction

Kerala Syllabus 10th Standard Physics Solutions Chapter 3 Electro Magnetic Induction

Kerala State Syllabus 10th Standard Physics Solutions Chapter 3 Electro Magnetic Induction

Electro Magnetic Induction Text Book Questions and Answers

Text Book Page No. 45

Sslc Physics Chapter 3 Kerala Syllabus Question 1.
You know that electrical energy can be converted into many other forms. Write down some examples.
Answer:

Text Book Page No. 46

Electromagnetic Induction Class 10 Kerala Syllabus Question 2.
Record your observations in the table given below.

Answer:

Text Book Page No. 47

Sslc Physics Chapter 3 Notes Kerala Syllabus Question 3.
Using magnet of high strength, and increasing the number of turns in the solenoid, the experiment is repeated. On the basis of the experiments complete the table (3.2).

Answer:

Physics Chapter 3 Class 10 Kerala Syllabus Question 4.
→ Why did the galvanometer needle deflect in the experiment?
Answer:
Whenever there is a change in the magnetic flux linked with a coil, an emf is induced in the coil.

→ Which were the instances in which there was a flow of current through the solenoid?
Answer:
Whenever there is a relative motion between the magnet and the solenoid, there is flow of electricity.

→ Which were the instances in which the current increased?
Answer:
Number of turns increased.Strong Magnet used Magnet / Solenoid moves with greater speed.

Text Book Page No. 48

Sslc Physics Chapter 3 Solutions Kerala Syllabus Question 5.
Observe the figure given below.
(The figure indicates the two stages of doing the experiment)

→ On which instance will the magnetic flux linked with the solenoid be less?
Answer:
Magnet is moved away from it solenoid.

→ On which instance will the magnetic flux linked with the solenoid be greater?
Answer:
When magnet is moved into the solenoid.

→ On which instance does a change in magnetic flux linked with the solenoid occur? (while it is moving/ while it is stationary)
Answer:
while it is moving.

Electromagnetic Induction Class 10 Questions And Answers Kerala Syllabus Question 6.
What may be the factors affecting the induced emf?
Answer:

• Number of turns of the coiled conductor.
• Magnetism.
• Movement of magnet and solenoid.

Class 10 Physics Chapter 3 Solutions Kerala Syllabus Question 7.
Which are the factors on which the direction of induced current in electromagnetic induction depend?
Answer:

• Direction of magnetic field
• Direction of conductor
• Direction of induced current

Text Book Page No. 49

Physics Chapter 3 Class 10 Kerala Syllabus Question 8.
Table 3.3

Answer:

Text Book Page No. 50

10th Physics Chapter 3 Kerala Syllabus Question 9.
The current obtained from the cell is unidirectional and is of the same magnitude. But what are the peculiarities of the current obtained by electromagnetic induction?
Answer:

• Direction changes.
• Magnitude changes.

Class 10 Physics Chapter 3 Kerala Syllabus Question 10.
We use mechanical energy in generators to move the magnet or coiled conductor continuously. In that case what shall be the energy change in a generator?
Answer:
Mechanical energy → Electrical energy

Class 10 Physics Chapter 3 Notes Pdf Kerala Syllabus Question 11.
Observe Fig 3.5(a) and write down the parts given in Fig. 3.5(b)

Answer:
ABCD: Armature.
B1, B2: Brush.
R1, R2: Slip rings.

Hsslive Guru 10th Physics Chapter 3 Kerala Syllabus Question 12.
ABCD indicates one turn of the armature coil. When the coil rotates about the axis in the clockwise direction, the portion AB moves upward and the portion CD moves downward.
Then according to the Fleming’s Right-Hand Rule.

→ What is the’direction of induced current in the portion AB? ( from A to B/ from B to A).
Answer:
from A to B.

→ What is the direction of induced current in the portion CD? (from C to D/from D to C).
Answer:
from C to D.

→ What is the direction of induced current in the coil ABCD? ( from A to D/from D to A).
Answer:
from A to D.

→ What is the direction of induced current in the external circuit? (through the galvanometer) (from B2 to B1/ from B1 to B2.
Answer:
from B2 to B1.

Class 10 Physics Chapter 3 Notes Kerala Syllabus Question 13.
What will be the positions of AB and CD when the armature completes 180° or one-half rotation?
Answer:
AB will be near the south pole of the magnet and CD will be near the north pole.

Physics 3rd Chapter Class 10 Kerala Syllabus Question 14.
Depict this stage of rotation of the armature in the science diary. At this instance,

→ What is the direction of movement of AB?
Answer:
Direction of AB is downward.

→ What is the direction of movement of CD?
Answer:
Direction of CD is upward.

→ What is the direction of current in the armature?
Answer:
from D to A.

→ What is the direction of current through the external circuit (through the galvanometer)?
Answer:
from B1 to B2.

Text Book Page No. 52

Kerala Syllabus 10th Standard Physics Chapter 3 Question 15.

Answer:

Text Book Page No. 52

Hss Live Physics 10th Chapter 3 Kerala Syllabus Question 16.
When 50 Hz AC is used, how many times will the direction of current change in the circuit?
Answer:
100 times.

Electromagnetic Induction Hsslive Kerala Syllabus Question 17.
Current is induced when the armature of a generator rotates. Slip rings and brushes are the ways and means by which this current is brought to the outer circuit. Is this arrangement necessary if the magnet in generator is made to rotate?
Answer: No

Hss Live Guru 10th Physics Kerala Syllabus Question 18.
Record in your science diary, the other ways by which mechanical energy can be made available for the working of a generator.
Answer:
From coal, naphtha, water from dam, exciters and big batteries. Field magnets are used as electromagnets. Mechanical energy also obtained from heat released by nuclear fusion.

Sslc Physics Chapter 3 Notes Pdf Kerala Syllabus Question 19.
Now write down in your science diary how the power for operating the generator which Babu saw near the stage was generated.
Answer:
The device which converts mechanical energy into electrical energy. When the armature rotates on its axis in the magnetic field, the magnetic flux linked within the coil changes. As a result, current is induced in the coil.

Sslc Physics 3rd Chapter Notes Kerala Syllabus Question 20.
Is it possible to produce DC (Direct Current) using a generator?
Answer:
Possible. If split-ring commutator is used in a generator instead of slip rings, we will get DC.

Text Book Page No. 54

Hsslive Guru 10th Physics Kerala Syllabus Question 21.
What are the similarities between the DC motor that we saw in the previous chapter and a DC generator?
Answer:

• Permanent magnet.
• Armature.
• Brushes.

Physics 10th Class Chapter 3 Kerala Syllabus Question 22.
Connect the output of a small DC generator to a galvanometer and rotate the armature continuously.

→ How is the needle deflected?
Answer:
Same direction.

→ Is the direction of current changing?
Answer:
No.

→ Is the magnitude of current the same?
Answer:
No. emf increases and decreases.

Physics Class 10 Chapter 3 Kerala Syllabus Question 23.

Answer:

Text Book Page No. 55

Physics Class 10 Chapter 3 Notes Kerala Syllabus Question 24.
Turn on & turn off the switch continuously. What do you observe?
Answer:
Bulb glows and then goes off.

Physics Class 10 Chapter 3 Kerala Syllabus Kerala Syllabus Question 25.
If the switch is kept in the on position what do you observe?
Answer:
Bulb does not glow continuously.

Question 26.
On what occasions do the flux change?
Answer:
When the switch is made to go on or off continuously.

Question 27.
What are the occasions when current flows through the second coil?
Answer:
When the switch in the first coil is kept on or off.

Question 28.
Can you suggest a method by which change can be brought in magnetic flux without switching on and off continuously?
Answer:
If AC is given to the primary coil instead of j DC, emf will be continuously induced in the secondary coil.

Text Book Page No. 56

Question 29.
Examine the diagrams of step up and stepdown transformers and list out the differences in their designs.
Answer:

Text Book Page No. 57

Question 30.

Answer:

Question 31.
A transformer working on a 240 V AC supplies a voltage of 8 V to an electric bell in the circuit. The number of turns in the primary coil is 4800. Calculate the number of turns in the secondary coil.
Answer:
V_p = 240V
V_s = 8V
N_p = 4800
V_s/V_p=N_s/N_p

N_s=V_s×N_p/V_p=8×4800/240=160

Question 32.
The input voltage of a transformer is 240V AC. There are 80 turns in the secondary coil and 800 turns in the primary. What is the output voltage of the transformer?
Answer:
V_p = 240V
N_s = 80
N_p = 800
V_s/V_p=N_s/N_p

V_s=N_s×V_P/N_p=80×240/800=24V

Question 33.
If voltage and current are known, what is the formula for finding power?
Answer:
Power = Voltage x Current

Text Book Page No. 58

Question 34.
In a transformer, if the voltage of the primary is V_p and the current in the primary is I_p, voltage in the secondary is V_s and the current I_s, write down the formulae connecting them.
Answer:
Power in the primary = V_p x I_p
Power in the secondary = V_s x I_s

Question 35.
In a transformer without any loss in power, there are 5000 turns in the primary and 250 turns in the secondary. The primary voltage is 120 V and the primary current is 0.1 A. Find the voltage and current in the secondary.
Answer:
N_p = 5000
N_s = 250
V_p = 120
I_p = 0.1A

Question 36.
Categorize the following relations appropriately as step up or step down transformers.

Answer:

Question 37.
As a result of the flow of current through a solenoid, is there any chance of inducing an electric current in the same solenoid?
Answer:
Yes. By self-induction.

Text Book Page No. 59

Question 38.

→ In which circuit does the bulb give a light with low intensity?
Answer:
Second circuit

→ Why does the intensity of light de-crease in that circuit?
Answer:
Back emf more

→ In which circuit is a magnetic field developed around the solenoid?
Answer:
On both circuits

→ If so in which circuit is a continuous emf induced?
Answer:
On second circuit

→ Have you understood the reason behind the decrease in the intensity of light in the second circuit? Write it down in the science diary.
Answer:
When AC passes through a solenoid, a changing magnetic field is generated around it. Due to this, an induced emf is generated inside the solenoid. This induced emf is in a direction opposite to that applied on the coil. Hence this is a back emf. This back emf reduces the effective voltage in the circuit.

Text Book Page No. 60

Question 39.
Inductors are widely used in AC circuits. Why?
Answer:
Inductors are used in the electronic circuits, to control and decrease current without power loss.

Question 40.
If resistors are used instead of inductors, what will be the disadvantage?
Answer:
Electric energy is lost in the form of heat.

Question 41.
Inductors are not used in DC circuits. Find out the reason and write it down in the science diary.
Answer:
Back emf is not produced as the flux formed by the current has no variation. So current control by inductor in DC is not possible.

Question 42.
Which are the main parts of a moving coil microphone?
Answer:
Diaphragm, Permanent magnet and Voice coil.

Question 43.
Which is the moving part in it?
Answer:
Diaphragm and Voice coil.

Question 44.
If a sound is produced in front of a movable diaphragm, what will happen to the diaphragm?
Answer:
Vibrates corresponding to the sound signals.

Question 45.
What happens to the voice coil then?
Answer:
Vibrates.

Question 46.
What will be the result?
Answer:
Creates electric signal corresponding to the sound.

Text Book Page No. 61

Question 47.
Find out the similarities and differences between a moving coil microphone and a moving coil loudspeaker and write them down in the science diary.
Answer:
Similarities:

• Both have voice coil.
• Have permanent magnet.
• Have diaphram.

Differences:

• Loudspeaker:
  Electrical energy → sound energy.
• Microphone:
  Sound energy → Electrical energy.

Question 48.
What is the energy transformation that takes place in a moving coil microphone?
Answer:
Sound energy → Electrical energy.

Text Book Page No. 62

Question 49.
How do we get mechanical energy for such generators?
Answer:

• Water from dam.
• Nuclear energy.
• Heat produced during the combustion of Naphta, Coal, Lignite ….etc

Question 50.
Write down the name of some power stations in Kerala.
Answer:

• Idukki – Moolamattom.
• Idukki – Pallivasal.
• Alappuzha – Kayamkulam.

Question 51.
Heat is generated in accordance with the equation H = I2Rt. In that case

→ What are the methods to reduce the heat generated?
Answer:

• Reduce current.
• Reduce resistance.
• Reduce the time taken.

→ If current (I) is reduced to half, how much will be the reduction in heat? (half, one fourth)
Answer:
one fourth.

→ If current (I) is reduced to 1/10 time show much will be the reduction in heat?
Answer:
1/100

→ How can we reduce the current without change in power? Find out on the basis of the equation P = V x I.
Answer:
by increasing voltage.

Text Book Page No. 63

Question 52.
→ What is the method to reduce the transmission loss?
Answer:
Power can be increased using a step-up transformer so that power loss can be minimized during power transmission.

→ Which type of transformer is there in a power station?
Answer:
Step-up transformer.

→ Which type of transformer is there in a substation?
Answer:
Step down transformer.

→ Which type of transformer is a distribution transformer?
Answer:
Step down transformer.

→ If a person standing on the earth touches a phase line, will she get an electric shock? Why?
Answer:
The person will get an electric shock because there is a potential difference (400V) between phase and earth.

→ Which are the lines essential for household electrification?
Answer:
Neutral, Phase and Earth lines.

Text Book Page No. 64

Question 53.
Analyze Fig.3.14 and find the answers to the questions given below.

→ To which device is the electric line reaching our home connected first?
Answer:
Watt-hour meter.

→ From where does the earth line start?
Answer:
From the main switch.

→ What is the use of a watt-hour meter?
Answer:
The electricity used is measured using a Watthour meter.

→ In which line are the fuses connected?
Answer:
Phase line.

→ What is the function of the main switch? Where is its position in the circuit?
Answer:
The electricity reaches the electrical appliances through the main switch. Therefore instead of turning each switch OFF, the main switch can be switched OFF. The main switch is placed just after the Watthour meter.

→ In the household electrical circuit, which is the third line, other than the phase and the neutral?
Answer:
Earth line.

→ What are the colors used for wires in phase, neutral and earth lines?
Answer:

• Phase – Red.
• Neutral – Black.
• Earth – Green.

→ Where is the earth wire connected in a three-pin socket?
Answer:
Pin E.

→ How are the household devices connected? Series/parallel.
Answer:
Parallel.

Question 54.
The advantages of connecting devices in parallel. Write them down.
Answer:

• Devices work according to the marked power.
• Devices can be controlled using switches as per need.
• The bulbs will get the required voltage.

Text Book Page No. 65

Question 55.
In a house, 5 CF lamps each of 20 W, works for 4hours, 4 fans each of 60 W work for 5 hours and a TV of 100 W works for 4 hours in a day. What will be the daily consumption shown by the watt-hour meter?
Answer:
4hour’s consumption of 5 CFL having 20 W
5×20×4/1000 = 0.4 kWh
5 hour’s consumption of 4 Fan having 60 W
4×60×5/1000 = 1.2 kWh
4 hour’s consumption of one T.V having 100 W
1×100×4/1000 = 0.4 kWh
Total unit = 2 kWh

Text Book Page No. 66

Question 56.
→ Which are the circumstances that lead to the flow of excess current in a house hold circuit?
Answer:
Over loading or Short circuit.

→ What happens to the circuit when there is an excess current in a circuit?
Answer:
More heat is generated. So appliances and circuit will get damaged.

→ How does a safety fuse protect a circuit?
Answer:
The current in the circuit may increase due to reasons such as short circuit, overload, excess flow of current or any problems in the insulation. As the higher temperature produced in the circuit due to these reasons makes the fuse wire to melt and the flow of current stops. Thus the circuit and the appliances are protected.

Text Book Page No. 67

Question 57.
What are the differences between ordinary fuse and MCB?
Answer:
Ordinary fuse works making use of heating effect of electricity while MCB works mak-ing use of heating and magnetic effects of electricity. The current in the circuit may increase due to reasons such as short circuit, overload, excess flow of current or any problems in the insulation.

As the higher temperature produced in the circuit due to these reasons makes the fuse wire to melt and the flow of current stops. Thus the circuit and the appliances are protected. MCB automatically breaks the circuit whenever there is an excess flow of current due to short circuit or overloading. After rectifying the circuit we can switch on the MCB and make the circuit as it was.

Question 58.
What is the advantage of MCB over a safety fuse?
Answer:
MCB is more sensitive to current than fuse, In case of MCB, the faulty zone of electrical circuit can be easily identified, With MCB it is very simple to resume to the supply, MCB is reusable and hence has less maintenance and replacement cost.

Question 59.
What is the function of ELCB/RCCB in the circuit?
Answer:
ELCB helps to break the circuit automatically whenever there is a current leak due to insulation failure or any other reason. Hence a person touching the electric circuit or a device does not get an electric shock. Nowadays RCCB, which ensures more safety than ELCB is made use of. In an ELCB one end of the relay coil is connected to the outer metallic cover and the other end is earthed.

If current leaks into the earth due to insulation failure or any other reason, a potential difference is developed between the ends of the relay coil. As a result, ELCB gets tripped off if the current exceeds a particular limit. This is with the help of a relay. In RCCB there is a provision to distinguish between the phase current and neutral current and to break the circuit if needed.

Question 60.
In the figure, which are the lines that are connected to the coil of the electric iron?

Answer:
Earth, Phase and Neutral lines.

Question 61.
If the phase line comes into contact with the body of the appliance due to defects in the insulation, what happens to the person who touches the body of the appliance?
Answer:
The person will get shock.

Question 62.
How can safety be ensured using a three-pin plug?
Answer:
The pin E of a three-pin plug comes into contact with the earth line. This pin is now connected to the body of the appliance. If the body comes into contact with an electrie connection, electricity flows to the earth through the earth wire. The flow of current to the earth through a circuit of low resistance increases the current. Hence heat generated in the fuse wire increases, the fuse wire melts and the circuit is broken. This will protect the instrument and the person handling it.

Question 63.
Which line comes into contact with the pin E.
Answer:
Earthline.

Question 64.
Howdoestheearth pin differs from the other pins? Why is it made different in this way?
Answer:
The earth pin is thick and long. As it is long it first comes in contact with the earth. Similarly, it loses the contact with the earth only at the end. As it is thick it offers less resistance. It carries more current than the amount of current coming from the phase.

Question 65.
Which part of the instrument is connected to the earth line?
Answer:
The earth line is connected to the metal casing of the electrical appliances.

Question 66.
Is TV working on AC or DC?
Answer:
on AC

Question 67.
We get DC from a mobile phone battery. But we use AC for charging it. What may be the reason?
Answer:
Most of the devices that work only on DC work by converting AC into DC. Mobile charger is a device that converts AC into DC.

Question 68.
Classify the devices known to you as those working in AC and DC.

Answer:

Text Book Page No. 71

One way switch:
Use to start/stop current flow and to change the direction of current flow.

Two-way switch:
Having two or more switches in different locations to control one electrical device.

Three-pin socket:
Electricity can be supplied to electrical appliances safely.

Ceiling rose:
Ceiling rose helps in that one live wire that is needed for ceiling light or fan to go in continuation.

ELCB:
Helps to break the circuit automatically whenever there is a current leak due to insulation failure or any other reason.

Regulator:
Automatically maintain a constant voltage level.

Indicator:
A device which provides visual or remote indication of a fault on the electric power system.

RCCB:
Distinguish between the phase current and neutral current and to break the circuit if needed.

MCB:
Automatically breaks the circuit whenever there is an excess flow of current due to short circuits or overloading.

Kit Kat Fuse:
Protecting a circuit against excess current.

Switchboard:
To control the flow of power.

Meter:
Instrument used to measure the magnitude of a quantity.

Main switch:
The main power supply of the RC models.

Bulb holder:
Holding a light bulb or lamp.

Clamp Ammeter:
The vector sum of the currents flowing in all the conductors passing through the probe.

Multimeter:
Can measure voltage, current, and resistance.

AC Voltmeter:
To measure the AC voltage across any two points of electric circuit.

Wire stripper:
Used to strip the electrical insulation from electric wires.

Screwdriver (*):
Tool, usually hand-operated, for turning screws with shaped slotted heads.

Screwdriver (-):
Tool, usually hand-operated, for turning screws with shaped slotted heads.

Tester:
Touched to the conductor being tested.

Plier:
For gripping something round like a pipe or rod, some are used for twisting wires.

Gloves:
Leather protector gloves – Worn over rubber insulating gloves to help provide the mechanical protection needed against cuts, abrasions and punctures.

Insulation tape:
To insulate electrical wires and other materials that conduct electricity.

Wire ( cable):
To carry electrical currents.

PVC Pipe fittings:
Used for plumbing and drainage.

PVC channel:
In Control Panels and Electrical Cabinets.

PVC Pipe:
Used as the insulation on electric wires.

Text Book Page No. 72

Question 69.
Construct such a circuit in aplywood as shown in Fig 3.19. The list of materials required and the number of items are given in the table.

Answer:

• Main switch.
• Fuse.
• MCB.
• Main fuse.
• Switchboard.
• ELCB.
• Fan.
• Bulb.

Electro-Magnetic Induction Let Us Assess

Question 1.
Write down the names of some devices that work based on the principle of electromagnetic induction.
Answer:

• Electric generator,
• Transformer,
• Moving coil galvanometer,
• Induction cooker etc.

Question 2.
What are the components essential for proving electromagnetic induction experimentally?
Answer:
Magnet, soft iron core connecting wire and Galvanometer.

Question 3.
Which are the factors that affect the induced emf in electromagnetic induction?
Answer:

• Number of turns in the armature.
• Strength of magnetic field.
• Speed of armature or field magnet.

Question 4.
Take a used cell from a calculator or remote control and connect it to a galvanometer as shown in the figure. What do you observe?

Answer:
The galvanometer needle deflects to one direction because there is a slight current flow.

Question 5.
Write down the names of DC sources.
Answer:
Cell battery, Solar cell, Generator, Thermocouple, etc.

Question 6.

a. Write down the names of parts numbered.
b. State the working principle of this device.
Answer:
a. 1. Field magnet,
2.Armature,
3. Slip rings.

b.Generators convert mechanical energy into electrical energy. Flux is changed and current is produced in the coil when the armature rotates from an axis. So current is induced due to flux change.

Question 7.
Write down the special features of AC and DC.
Answer:
Direct current (DC): Direction constant, emf increases and decreases.
Alternating current (AC): Direction changes continuously, emf increases and decreases.

Question 8.
Analyze the given graph and find out the instances at which the emf is maximum and minimum.

Answer:
Maximum: T/4, 3T/4
Minimum : 0, T/2, T

Question 9.
There is only one type of generator AC generator. Write down your responses about this statement.
Answer:
According to the nature of output, generators can be classified into 2 types AC generator and DC generator. Though ac current is produced in a DC generator with the help of split-ring commutator ac is converted into dc current.

Question 10.
Line diagrams of a generator are given.

a. What is the specialty of the electricity reaching the galvanometer if the armatures of both the generators are made to rotate?
b. What is the speciality of the electricity reaching the galvanometer if the field magnets of both the generators are made to rotate?
c. Draw the graphical representation of electricity obtained in both.
Answer:
a. DC in first and AC in second.
b. AC in both.

Question 11.
Electromagnetic induction is
a. charging a substance.
b. process of developing a magnetic field around a coil by passing electricity through a coil.
c. process of rotating the armature of a generator.
d. process of making electricity by the relative motion of a magnet or a coiled conductor.
Answer:
d. process of making electricity by the relative motion of a magnet or a coiled conductor.

Question 12.
Which is the device used to generate electricity?
a. generator.
b. galvanometer.
c. motor.
d. ammeter.
Answer:
a. generator.

Question 13.
Write down the similarities and differences in the structure of an AC generator and a DC generator.
Answer:
Similarities:

• Field magnet and Armature are present.
• Works on the principle of electromagnetic induction.
• AC produced on armature.

Differences:

Question 14.
A conductor hung horizontally in the north-south direction is connected to a galvanometer. The conductor is situated in a magnetic field acting in the East-west direction, In which direction should you move the conductor if maximum current is to be induced in the conductor in the north-south direction? Justify your answer.
a. towards east.
b. downwards.
c. upwards.
d. towards north.
Answer:
b. downwards. According to Fleming’s Right-hand rule.

Question 15.
Copper wires of the same length and thickness are connected to points A and B in all the three circuits. In-circuit (a) copper wire is not coiled. In circuits (b) and (c), the copper wire is coiled. Observe the circuits and answer the following questions.

a. When circuit (a) is switched on, what do you observe?
b.When circuit (b) is switched on, what change do you observe in the intensity of light? Justify your answer.
c. When circuit (c) is switched on, what change do you observe in the intensity of light? Justify your answer.
Answer:
a. Bulb glows.
b. No change in the intensity of light. There is no back emf produced as DC current is used in the circuit.
c. Intensity of light decreases. Back emf is produced in the circuit as A.C current flows through it and current is decreased.

Question 16.
The current in the secondary coil of a transformer is 1A and that in the primary is 0.5 A.
a. What type of transformer is this?
b.If 200 V is available in the secondary coil of this transformer, what is the voltage in the primary?
c. Explain the working principle of a transformer.
Answer:
a. Step down Transformer.

c.Mutual Induction:
Keep two sets of coils side by side. When the direction of current flow or its intensity is changed in one, the magnetic flux around it also changes. An emf and current is induced in the secondary coil. This phenomenon is called mutual induction.

Question 17.
In connection with the working of a microphone, a few statements are given inboxes. Arrange them in the proper sequence.

Answer:
Sound is produced → diaphragm vibrates → Voice coil vibrates → electrical signals are produced in the voice coil.

Question 18.
Thick insulated copper wires are used in the primary coil of a step-up transformer and in the secondary of a step-down transformer. What is the necessity of this?
Answer:
The power in the primary and secondary.of a transformer is equal. Current is more where voltage is less because P = VI. Thick wires are needed to flow more quantity of current. In the primary of step up and in the secondary of step down, quantity of current is more.

Question 19.
Which situation causes short circuit?
Answer:
When a phase and neutral comes into direct contact in a household circuit or when two-phase come in contact or when a phase and neutral comes into direct contact in a distribution line.

Question 20.
What is the role of earthing wire in a household circuit?
Answer:
The pin E of a three-pin plug comes into contact with the earth line. This pin is now connected to the body of the appliance. If at all the body comes into contact with an electric connection, electricity flows to the; earth through the earth wire.

The flow of current to the earth through a circuit of low resistance increases the current. As a result heat generated in the fuse wire increases and the circuit gets broken. This ensures the safety of instrument and the person handling it.

Question 21.
Why do we say that metallic devices should be earthed?
Answer:
Chances of electric shock are” more when metallic devices are used. When metallic devices are earthed, chances of electric shock can be prevented.

Question 22.
An electric heater calibrated 1.5 kW, 230 V is connected to a household branch circuit having 5 Afuse wire and is, made to work. What will happen?
Answer:
Current produced the circuit when the device works is
I = P/V = 1500/230 = 6.52 A
So 5 Amperage fuse bums.

Question 23.
Which are the devices connected in series in a household circuit?
Answer:
Fan, Switch, Regulator, etc.

Question 24.
What can be done to save electrical energy in schools and houses?
Answer:

• Use devices of good efficiency.
• Use electricity only when required.
• Reduce cooking using electricity.
• Reduce the use of air conditioners and refrigerators.
• Use LED bulbs instead of other bulbs.

Question 25.
Why do some mobile phones use three-pin plugs?
Answer:
In mobile phones, instead of earth pin, a plastic pin is used. For safety purpose, the phone neutral socket will have caps. Only if the pin enters the earth socket the other two sockets will open. It will be difficult to connect two-pin in such sockets. To solve this problem in mobile charges three-pin plug can be used.

Electro Magnetic Induction Extended Activities

Question 1.
Michael Faraday, the Father of electricity, did not even get elementary education. Are j you not inspired by the achievements of Faraday in the field of science? Conduct a seminar on “Contributions df Faraday and the hard work behind it.”
Answer:
Michael Faraday, an eminent scientist in the fields of physics and chemistry. He made his first invention in 1821. He proved that when a wire is kept in the magnetic field and current is passed through it, the wire moves.

By the series of experiments conducted in 1831 by using magnetic power Faraday found the important theory of production of current. He was called the father of electric current. He made valuable contributions in the field of chemistry also. It is wonderful to know that he didn’t get any formal education.

Question 2.
Energy is precious, especially electrical energy. Society must be convinced of the necessity of reducing the consumption of electrical energy. Prepare and propagate posters for this purpose.
Answer:

• Control the consumption of current, or else we will be in darkness.
• Next-generation also deserves current.
• Electricity is only for essential purpose.
• Use only efficient machinery to use minimum current.

Question 3.
Compare the induced current obtained when the armature coil rotates once in between the poles of a magnet, and the induced current obtained when the experiment using a magnet and coil was performed.
Answer:

Question 4.
Exhibit a model of electrical distribution network.
Answer:

Question 5.
Draw an electrical circuit containing the electrical appliances required for your classroom.
Answer:

Question 6.
How can the earthing be done in order to ensure safety in electrical circuits? Discuss and prepare a note.
Answer:
Overloading or short-circuiting may occur during the working of electrical devices due to some problems in the mains or within the circuit. The excess current produced is discharged to the ground through the earth wire. This is known as earthing.

Thus earthing provides safety in the electrical circuits. In household electrical circuits for the effective functioning of the earth wire, the earthing system should be proper.

• The pit in which the earth wire is placed should be filled with salt and coal.
• During summers it is required to wet the places where it is earthed.

Question 7.
Observe and record the meter reading in your house for 10 consecutive days. Based on this, find out the average consumption per day. Find out methods to reduce consumption and record them. Present your findings in the Energy Club.
Answer:
Average usage = Total usage/no.of days
Methods to reduce the usage of electricity.

• Use devices of good efficiency.
• Use electricity only when required.
• Reduce cooking using electricity.
• Reduce the use of air conditioners and refrigerators.

Electro Magnetic Induction Orukkam Questions and Answers

Question 1.
Observe the figure.

a. Identify the device in figure 1?
b.What is Rj and Bj indicate in the figure?
c. What is the working principle of that device?
d. Identify the graph which indicate the emf produced by that device?
e. Suggest a way of method to increase the value of emf produced in armature coil of the device?
f. Which law is related with the direction of the induced emf in the armature? Explain?
g. Which part is used as rotor?
h.Which part is used as stator?
i. What are the advantages of armature used as stator?
j. Field magnets in power generator are electromagnet. Why?
k. Identify the device in figure 2?
l. Identify the graph given above of emf which is produced by that device?
m.What is the angle between the armature coils of a three-phase generator?
n. Some statements are given below which related to single-phase generator and three-phase generator. Classify them.
a. For each field magnet, there is only one armature.
b. There are three sets of armature coils for each field magnet.
c. AC generated in all the armature coils will be of the same phase at the same time.
d. AC generated in all the three armature coils will be of different phase at the same time.
Answer:
a. AC generator.
b. R1 – Slip Rings,
B1 – Brushes.
c. Electromagnetic Induction.
d. Graph 1.
e. Increase the number of turns.
f. Refer page no: 39
g. Field magnet.
h. Armature.
i. We can avoid rings and graphite brushes when the armature is used as stator. So there is no possibility of producing sparks.
j. Refer page no: 72 Question No: 1
k. Three-phase generator.
l. Graph 2.
m. 120°.
n. Single-phase generator – a, c.
Three phase-generator – b,d.

Question 2.
Observe the graph and answer the following.

a. In which angles are the rate of changes of flux is maximum?
b. In which angles are the induced emf maximum?
c. What is meant by period of AC?
d. What is meant by frequency of AC?
e. What is the aim of increasing number of magnetic pole and armature coil in a power generator?
Answer:
a. 90° and 270°
b. 90° and 270°
c. The time taken by the armature coil for a full rotation is called Period.
d. The number of cycles per second is the frequency of AC.
e. To increase the amount of electricity produced.

Question 3.
Observe the figures.

a. What kind of transformer is in figure 1?
b. What kind of transformer is in figure 2?
c. What is the working principle of a transformer?
d. Some statements are given below related to step-up transformer and step down transformer. Classify them. Number of turns in primary is less than secondary.

• Thick wires are used in secondary.
• Thick wires are used in primary.
• Voltage in secondary is high.
• Voltage in secondary is low.

e. Some relations are given about step-up transformer and stepdown transformer. Classify them.
Answer:
a. Step-up transformer.
b. Step down transformer.
C. Mutual induction.

Question 4.
Construct the circuit as in the figure. Observe the intensity of light when switch in the circuit is turn on.

a. What change is observed when 6V battery is connected instead of 6V AC in the circuit?
b. What change will occur in magnetic flux when 6V battery is connected to the circuit and 6V AC connected to the circuit?
c. What is the peculiarity of the induced emf when AC source is connected in the circuit instead of DC.
d. What is the name of this induced emf?
e. Name the phenomenon caused to reduce the intensity of light in the circuit.
f. What is the change in intensity of light will be occurred when soft iron core is inserted into the inductor?
g. What is the reason for the change in the intensity of light?
h. What is the phenomenon which caused to glow the LED?
i. Explain the phenomenon.
Answer:
a. Intensity of light decreases
b. When 6V AC is given, there is a change in the magnetic flux linked with the solenoid. Due to the change in flux, an induced emf is developed.
c. Emf will be in the opposite direction.
d. Back emf.
e. Self Induction.
f. Intensity of light decreases.
g. Change in flux increased.
h. Mutual Induction.
i. When the strength or direction of the current in one coil changes, the magnetic flux around it changes. As a result, an emf is induced in the secondary coil. This phenomenon is called Mutual induction.

Question 5.
Match the following related to ideas of different kind of power stations are given below.
Pallivasal, coal, Nuclear power station, Flowing water, Thermal power station. Kalpakam, Hydroelectric power station, Kayamkulam, Nuclear energy.
Answer:
Pallivasal :
Flowing water – Hydroelectric power station.

Kalpakam:
Nuclear energy – Nuclear power station

Kayamkulam:
Coal – Thermal power station.

Question 6.
There are some terms in connection with various power stations given. Fill in the blanks suitably.

Answer:
a. Nuclear Power station.
b. Kinetic energy.
c. Electrical energy.
d. Chemical energy.
e. Mechanical energy.
f. Heat energy.
g. Electrical energy.

Question 7.
Power stations are centers that generate and distribute large quantities of electricity. Power transmission is the process of sending electricity to distant places through wires from the power stations.
a. What is the voltage at which electricity is generated at the power station?
b. What is the voltage of electricity supplied for domestic consumption?
c. In which stage of power transmission is the step-up transformer used?
d. Which are the stages in which step down transformers are used in power transmission?
e. What are the problems arises during distant energy transmission?
f. What are the factors which depend when electricity passes through the conductor?
g. What are the methods to reduce the resistance?
h. What change will occur in power when intensity of current is reduced in the circuit?
i. How is it possible to reduce the intensity of current without power loss?
j. What will be the change in secondary current when secondary voltage of a step-up transformer is increased 10 times?
Answer:
a. 11 kV or 11000 V.
b. 230 V.
c. Phase I
d. Phase ii, iii, iv.
e. Voltage drop, Power loss.
f. Current, Resistance, Time.
g. Increase the thickness of the conductor.
h. Power decreases
i. Increase the voltage.
j. Become less by 1/10.

Question 8.
Step-up or Step down transformer in the power distribution system erected for the purpose of household distribution.
a. How many lines reach the input of the distribution transformer?
b.What is the potential of each phase line?
c. How many lines go out of the distribution transformer? Which are they?
d.Which method is adopted for connecting secondary coil in the distribution transformer?
e. Three-phase lines from output of distribution transformer are connected into a common point. What is the name of the common point?
f. What is the potential difference between a phase line and neutral line?
g. What is the potential difference between any two-phase line?
h.What is the potential difference between the earth and the neutral line?
i. What is the potential difference between a phase line and the earth line?
Answer:
a. 3 lines.
b. 400 V.
c. 4 lines come out, 3 phase and one neutral.
d.Star connection.
e. Neutral.
f. 230 V.
g. 400 V.
h.0 V.
i. 0 V.

Question 9.

a. How are the bulbs arranged in figure 1?
b. How are the bulbs arranged in figure 2?
c. Which circuits has more resistance?
d. All the switches in both the circuits may be turned on, in which circuit do the bulb give brighter light?
e. In which circuit current flowing through each bulb is maximum?
f. Which mode of connecting devices in your household electrical circuit is advisable?
Answer:
a. Series.
b. Parallel.
c. Series Connection.
d. Parallel Connection.
e. Parallel Connection.
f. Parallel Connection.

Electro Magnetic Induction Some more activities

Question 1.
Three-pin plug is used in the circuit of power devices like electric iron.
a. In three pin plug, what each pin indicate?
b. Which part of the device is connected to the earth pin?
Answer:
a. E-Earth line, N-Neutral line, P-Phase line.
b. To the body of the device.

Question 2.
What is the reason for massaging the body of a person who got electric shock?
Answer:
To increase the body temperature.

Electro Magnetic Induction SCERT Questions and Answers

Question 1.
Four identical solenoids and magnets are, depicted in the diagram. Magnets are shown as moving in the figures 1 and 2 and solenoids are shown as moving in figures 3 and 4.

a. Which are the figures showing inducing of emf? How?
b. Pair the pictures in which the galvanometer needle deflects in the same direction. What does the deflection indicate?
c. Draw the picture of a device that can produce continuous electricity using the same principle as shown above for producing electricity.
Explain its working.
Answer:
a. Emf will be induced in all the 4 coils in the diagrams; flux change happens in all the four activities.
b. Pairs (1) and (4), (2) and (3). deflection of galvanometer needle indicates the direction of the flow of current.

Question 2.
Certain factors that influence the induced emf in electromagnetic induction are listed below:
i. Increasing the speed of magnet.
ii. Decreasing the strength of magnet.
iii. Increasing the number of turns.
iv. Reducing the speed of rotation of the magnet.
v. Increasing the strength of magnetic field.
vi. Using solenoid with less number of turns.
Which of the above ideas can be used to get maximum emf by electromagnetic induction?
Answer:
i, iii, v.

Question 3.
The diagram shown below is an arrangement for producing 10V AC using electromagnetic induction. Observe the diagram carefully and answer the questions.

a. Which law is used to find out the direction of induced emf when the armature coil ABCD starts rotating?
b. In which direction the side CD should move (up/down) to produce a flow of current from x to y? What should be the direction of movement of the side AB (up/down) to produce a flow of current in the same direction?
C. Find out the frequency by drawing a time emf graph if the armature completes 10 cycles in 5 seconds.
Answer:
a. Fleming s Right Hand Rule.
b.Upward, side AB moves down.

Question 4.
Parts of an AC generator are given below. Field magnet, Armature, Slip rings, Brush.
a. Explain the position of these parts in an AC generator.
b. Write down the functions of any two.
Answer:
a. Armature rotates in the magnetic field of the field magnet. Springs are soldered to the ends of armature coil, Graphite Brush is connected in such a way as to have constant contact with springs.
b. Armature – Generate emf.
Field magnet – creates magnetic flux.

Question 5.
The various stages of rotation of an armatures coil while completing one rotation in a magnetic field and the graph of the emf produced by the coil are shown below:

a. Select the appropriate points from the pictures m,n,o,p corresponding to the positions 1,2,3,4 and 5 in the graph.
b. Explain the scientific basis for the pairing.
c. Which two pictures show the maximum flux changes? Point out one difference between the two.
Answer:
a. 1-m, 2-0, 3-n, 4-p, 5m OR 1-n, 2-p, 3-m,4-0, 5-n
b. The change in direction and intensity of magnetic flux in armature file.
c. n and o. Magnetic flux change takes place in the opposite directions in these two.

Question 6.
Maximum voltage produced in an AC generator completing 60 cycles in 30 seconds is 250 V.
a. What is the period of the armature?
b. How many cycles are completed in T/2 seconds?
c. What is the maximum emf produced when the armature completes 1800 rotation?
Answer:
a. Period = Time taken to complete one full rotation.
Period = Time taken for rotation / Number of rotations = 30/60 = 12 = 0.5s
When the Armature ABCD rotates, change in the magnetic flux depends on its direction and intensity.
b. only 1/4 of a rotation. T 1/2,T/2,1/4
c. 250V

Question 7.
In single-phase generators armature functions as rotor whereas in power generators armature functions as stator.
a. List out the other differences between single-phase and three-phase generators?
b. What are the advantages of using armature as stator in power generators?
Answer:
a.

b.

• Can increase the no.of turns of the armature coil.
• Can avoid slip rings and also the subsequent sparks.

Question 8.
If a rotor in a power generator completes 3000 rotations in a minute.
a. Which type of AC is produced here, single-phase or three-phase?
b. What will be the frequency of the AC produced if a pair of magnetic poles are ) used in the generator?
c. What would be the time interval to produce the same emf in the two phases if j the rotor completes 20 rotations in a minute?
Answer:
a. Three phase AC
b. Frequency = 3000/1×60 = 50Hz
c. 20 rotations in a minute -3 seconds for completing one rotation. So the time interval to produce same emf is 1 second.

Question 9.

a. Observe the picture and identify the parts labeled as a, b, and c in the picture shown.
b. What is the energy change taking place in a moving coil microphone? What are the roles of the parts a,b, and c?
c. Why is the output signal of a moving coil microphone sent to the loudspeaker only after feeding it directly into an amplifier?
Answer:
a.1. Diaphragm.
2. Permanent magnet.
3. Voice coil.
b.
Mechanical energy to electrical energy Diaphragm – Vibrates in accordance with the sound falling on it.
Permanant magnet – Creates a magnetic field.
Voice coil – Creates electric signals corresponding to the sound.
c. The weak signals obtained from the microphone are strengthened by an amplifier.

Question 10.
Coils wound around a soft iron core connects two bulbs, B1 and B2 of 6 V. Analyse the figure and answer the questions.

a. If 6 V, DC is given in the coil A and the switch is on, which of the bulbs, B, & B2 will glow? Why?
b. If AC is given to the coil A instead of DC, which of the bulbs will glow B1 or B2?
c. Was there any variation in the brightness of Bulb B1 when AC and DC were supplied? Why?
d. Write down the names of two instruments that use the principle of mutual induction.
Answer:
a. Only B1 because electricity flows only through B1
b. B1 and B2 because of mutual induction.
c. Light of B1 will be reduced when AC is passed through it. This is due to the back emf produced due to self-induction.
d. Transformer, Relay switch.

Question 11.

Two solenoids A and B made of insulated copper wire are shown in the diagram. Observe the diagram and answer the questions below:
a. Write down different methods to light up the bulb connected in the coil B using the principle of electromagnetic induction.
b. What is the scientific principle applied to light up the bulb B?
Answer:
a.i. Switching S1 on and off by giving a DC supply.
ii. Keeping switched on after giving AC supply to the coil A.
iii.When coil A is moved relatively after supplying DC.
b. Mutual induction.

Question 12.
A transformer without power loss carries 4000 turns in the primary and 2000 turns in the secondary. If the primary voltage is 120 V and the current is 0.2 A.
a. What is the output voltage of transformer?
b. What is the current in the secondary?
c. What is the maximum output power?
d. Where are the thick wires to be used in this transformer in the primary or secondary? Why?
Answer:

d. In the secondary. Since P = V1 current is more in the secondary.

Question 13.
In a step up transformer, primary power is 50 W and output voltage is 100 V. If the current in the primary is 1 A.
a. What is the primary voltage?
b. What will be the current in the second ary?
c. Secondary power / Primary power =
Answer:
a. P = V_p × I_p
500 W = V_p × 1A, V_p = 500 V
b. I_s = \(\frac { I_p × V_p }{ V_s }\) = 1×500/100 = 5A
c. Primary and secondary powers are equal
Secondary power / Primary power = 1

Question 14.
Power Station A :
Potential energy → Kinetic energy → Mechanical energy → Electrical energy.
Power Station B :
Chemical energy → Heat energy → Mechanical energy → Electrical energy.
Power Station C :
Nuclear energy → Heat energy → Mechanical energy → Electrical energy.
a. Identify the above power stations A, B, and C based on the energy change taking place in them.
b. Write down the merits and demerits of the power stations A, B, and C.
c. Name other three energy sources that can be used instead of these.
Answer:
a. A- Hydroelectric power station.
B- Thermal power station.
C- Nuclear power station.

b.1. Merits: No pollution, Energy of waterfall is used.
Demerits: Deforestation, Extinction of rare species of flora and fauna.

2. Merits: less area needed, danger possibility is less.
Demerits: Pollution, High consumption of fossil fuels.

3. Merits: Energy production on large scale, Less area needed.
Demerits: Pollution, Problems of radiation.

c. (1) Solar energy.
(2) Tidal energy.
(3) Windmills.

Question 15.
Names of power stations and statements related to them are given below in two groups. Match them suitably.
A. (1) Uses the energy of water stored at a height.
(2) Uses the heat generated by the burning of fuels.
(3) Uses nuclear energy.
B. (a) Pallivasal Power Station.
(b) Ramagundam Power Station.
(c) Kalpaklcam Power Station.
Answer:
a – 1,
b – 2,
c – 3.

Question 16.
Different stages of power production and transmission are given. Arrange them in the proper sequence.
a. Distribution transformer reduces the voltage from 11 kV to 230 V.
b. Voltage is reduced from 220 kV to 66 \ kV to supply electricity to large scale industries.
c. Turbine is rotated to operate the generator.
d. Transmission of electricity from the Power Station starts at 220 kV.
e. Electricity is produced at 11 kV.
f. Domestic consumers get electricity.
Answer:
c,
e,
d,
b,
a,
f.

Question 17.
Different stages of power transmission and distribution are given in a flow chart.

i. Choose the suitable terms from the bracket given below and arrange them in the given boxes in the correct order. (Power generator, substation, power transformer, star connection, distribution transformer, household connection, power grid)
ii. What do you mean by power grid? Give two advantages of power grid.
Answer:
i.
a. Power generator.
b. Power transformer.
c. Substation.
d. distribution transformer.
e. Star connection.
f. household connection.
ii. Power grid is the network that connects different power generating center and distribution systems to one another. Due to this arrangement, if any defect occurs either at the generator or transmission lines, electricity can be taken from any other power generating center through another set of transmission lines.

Question 18.
We use 230 V for our household purpose.
But the power produced at 11 kV is transmitted to distant places after increasing the voltage.
a. Why is electricity transmitted to distant places at high voltage?
b. Write down one disadvantage of transmitting electricity at high voltage.
c. What are the use of substations? Which type of transformer is used here?
Answer:
a. To reduce energy loss in the form of heat high voltage is used for long-distance transmission.
b. Transmission through populated areas is difficult. Transmission lines must be at a great height to ensure safety.
c. To reduce high transmission line voltage as and when required. Step down transformers are used for this purpose.

Question 19.
Power stations are centers where electricity is produced and distributed on a large scale.
a. What is the voltage of power production in our country?
b. Calculate the current if the power produced here in 11 MW.
Answer:
a. 11kV
b. p = VI
I = P/V = 11MV/11kV

Question 20.
Identify the relation and fill up the blank.
Step up transformer: power transformer
Step down transformer:
Answer:
Distribution transformer.

Question 21.
Given diagram shows the star connection drawn by a child in his science diary.

a Can you draw a neutral line from the diagram? Why?
b.Draw the correct diagram of star connection.
Answer:
a. No: Three-phase lines do not meet at a single point.

Question 22.
Identical five bulbs of 230V and 40W are connected using star connection in the diagram given below.

a. What do you call the point N which is earthed in the circuit?
b. State the voltage differences between two phases and between a phase and neutral line in a star connection.
c. Which of the connected bulbs will glow? Why?
d. Two houses have electric connection from the same distribution transformer, but unfortunately one of the houses suffers power failure frequently; Explain it on the basis of household distribution.
Answer:
a. Neutral point.
b. 400 V in between two phases.
230 V in between a phase and neutral.
c. B4 glows because it is connected between the lines phase and neutral B, and B2 glows with low brightness. Each bulb gets 200 V as it is connected in series with 400 V.
d. Three phases in a transformer have separate fuses. So power supply problem in a line need not affect the other phases.

Question 23.
Examine the given circuits and answer the questions.

a. Identify series and parallel connection from the given circuits,
b. Parallel mode of connected devices is advisable for household electrification? Why?
Answer:
a. Figure A in series connection.
Figure B in parallel connection.
b. If the appliances are connected in parallel it will help to control each appliance with separate switches. It gives the needed voltage and current to appliances.

Question 24.
Four 60W, 230V bulbs are connected between a phase and a neutral as shown in the diagram. (Line voltage is 230 V)

a. How are the bulbs B, and B, are connected? What about B1 and B2?
b. Which bulb will glow at 60W power?
c. Which bulbs will glow with low power? Why?
d. Which bulbs in the circuit can be controlled using separate switches.
Answer:
a. B, and B, in series B₁ and B₂ in parallel.
b. B, and B₂.
c. B₃ and B₄ because each bulb gets less voltage as they are in series. So the bulb’s intensity will be less.
d. B₁ and B₂.

Question 25.
A flowchart of art household wiring is given in the diagram below. Fix the suitable devices in the space provided by picking from the given list.

(ELCB, MCB, Main switch, Watt-hour meter, Three-pin plug)
Answer:
(A) Main switch.
(B) ELCB.
(C) MCB.
(D) Three-pin plug.

Question 26.
We know that earth is an electron bank and that it has zero potential. But quite often we get electric shock from computer despite proper earthing.
a. Can electric shock from electric devices be avoided by earthing alone?
b. Proper earthing will give us safety from electric shock. What is meant by proper earthing? How can the earthing arrangement by done properly to ensure safety?
Answer:
a. No
b. Earth wire should carry all charge from the metal body of an electrical device to the earth effectively.

• Wire of suitable thickness must be used.
• Pits must be prepared to ensure effective earthing.
• Metal plates having large area as in plate earthing can be used.

Question 27.
Examine the diagram.

a. How would you connect the wires A, B and C from the electric iron to the wires of three-pin top correctly?
b.What is your response to the opinion that two pin top can be used instead of three-pin top in an electric iron?
Answer:
a. X wire is to be connected’ to earth pin. So connection should be made to be B. A and C must be connected to the wires Y and Z respectively.
b. Do not agree.
Insulation damage will carry electricity to the metallic part of electric iron. If one touches the metallic part he will get a electric shock. If the metallic part is earthed the charge will flow to the earth through the less resistance path.

Due to low resistance path the electric flow increases and the fuse burns out, thus safety is ensured or Agree. Special techniques are employed in new type iron boxes for avoiding shocks. Two pins shall be used only in such type appliances.

Question 28.
Which is the electric line to be connected to the point marked E in the diagram?

Answer:
The pin marked E must be connected to the earth line.

Question 29.
A bulb of 60W in a classroom remains switched on for a long time from 9 am to 5 pm due to the negligence of students in the class.
a. Calculate in joule the electric energy consumed by the bulb?
b. If similar negligence had happened in 3 classrooms, how many units of electric energy would have been wasted?
c. List out the different ways that can be adopted to save energy.
Answer:
a. Energy consumed =
= 60 x 8 x 60 x 60
= 1,728,000 J = 1728 KJ

b. Energy loss = 60×8×3/1000 = 2.4 unit.

c.i. Avoid unnecessary use of lights and bulbs.
ii. Promote maximum use of natural light and wind.
iii. Use energy-efficient electrical appliances.
iv. Promote use of LED bulbs instead of filament lamps and CFL.

Question 30.
In a house 5 CF lamps each of 24 W work for a period of 5 hours and fans of 80W work for 5 hours.
a. What will be the electric consumption in kWh for a month?
b. Calculate the bill amount for one month by collecting the rate from the given table.

c. Find out the change in the monthly bill amount if CF lamps are replaced by LED lamps of 3W.
Answer:
a. One month’s consumption of energy by CFL
= 24×5×5/1000 × 30 = 18 Unit
One month’s consumption of energy by
= 80×5×5/1000 × 30 = 24 Unit
One month’s total consumtion of energy
= 18 + 24 = 42 Unit.
One months total bill amount
= 2.80 × 42 = Rs.118
c. One month’s total energy consumption by LED
= 3×5×5/1000 × 30 = 2.25 Unit
2.25 + 24 = 26.25 Unit
One months bill amount
= 26.25 × 1.50
=Rs. 39.375 ≈ 40
Difference in bill amount
= 118 – 40 = 78

Electro Magnetic Induction Exam Oriented Questions and Answers

Very Short Answer Type Questions (Score 1)

Question 1.
Instead of permanent magnets, electromagnets are used as field magnets in power generators. Find out reasons from the following and write them down.
a. The strength of permanent magnets goes on decreasing.
b. The strength of electromagnets goes on decreasing.
c. Electromagnets can retain the required magnetic strength.
Answer:
a & c

Question 2.
Using the relation from the first pair, complete the other.
AC generator: slip rings
DC generator:
Answer:
Split rings.

Question 3.
Find the odd one in the group and write the reason.
[Voice coil, permanent magnet, slip rings, diaphragm]
Answer:
Slip rings. Others are parts of microphone.

Question 4.

Name the type of transformer.
Answer:
Step-up transformer.

Question 5.
What is the basic principle of working of a transformer?
Answer:
Mutual induction

Question 6.
A solenoid is connected to a galvanometer. A magnet is moved into and out of the solenoid frequently.

Which of the following will be most similar to the time emf graph of the electricity available from the solenoid.

Answer:
Figure 1

Question 7.
Using the relation from the first pair, complete the other.
AC generator: armature.
Moving coil galvanometer:
Answer:
Voice coil

Question 8.
Name the solenoid which is used to control emf.
Answer:
Inductor.

Question 9.
What is the number of turns 6f a secondary coil of a step-down transformer compared to primary?
(increasing, decreasing, doesn’t change)
Answer:
Increasing.

Question 10.
Which is used as rotor in the power generators?
Answer:
Field magnet.

Question 11.
Name the network which is connected in different power generating centers and distribution systems.
Answer:
Grid.

Question 12.
Using the relation from the first pair, complete the other.
Moolamattom – Hydroelectric power station
Kota –
Answer:
Nuclear power station

Question 13.
Find the odd one in the group and write the reason.
[Nuclear energy, Thermal energy, Kinetic energy, Electrical energy]
Answer:
Kinetic energy. Others are energy changes related to nuclear power stations.

Question 14.
What is the potential of the point where all the 3 phase lines are connected together?
Answer:
Zero potential.

Question 15.
What is the voltage of electricity supplied for domestic consumption?
a. 400 V
b. 250V
c. 230V
d. 50V
Answer:
230V

Question 16.
Using the relation from the first pair, complete the other.
Armature – Stator
Field magnet –
Answer:
Rotor.

Question 17.
The following diagram shows a three-pin Plug.

To which line is the fuse connected.
Answer:
The fuse is connected to the phase line.

Question 18.
In a household electric circuit. Where is the MCB (Miniature Circuit Breaker) connected?
Answer:
MCB is connected to the main fuse board.

Question 19.
Using the relation from the first pair, complete the other.
Hydroelectric power station flowing water Thermal power station –
Answer:
Steam at high pressure and temperature.

Question 20.
From given below, which device is used to measure electrical energy.
a. Transformer.
b. Watt-hour meter.
c. Three-pin plug.
d. Generator.
Answer:
b. Watt-hour meter.

Very Short Answer Type Questions (Score 2)

Question 21.
The figure shows a conductor PQ situated in a magnetic field, connected to circuit.

a. If the conductor is moved in the direction A, what is the direction of induced current?
b. State the rule used to find out the direction of induced current.
Answer:
a. from Q to P
b. Fleming’s right-hand rule. It states that hold the thumb, forefinger and middle finger of the right hand in mutually perpendicular direction, forefinger represents the direction of the magnetic field, middle finger represents the direction of current and thumb finger represents the direction of motion of the conductor.”

Question 22.
There are 10000 turns in the primary and 500 turns in the secondary of a transformer. The primary voltage is 240V and intensity of current in the primary is 0.1 A. Find the intensity of current if the secondary voltage is 12V?
Answer:

Question 23.
Match the following suitably.

Answer:

Question 24.
Arun decided to construct a transformer to change 240V to 12V.
a. Where does it need more number of turns in the primary or in the secondary?
b. Find the number of turns in the primary if 200 turns are used in the secondary.
Answer:
a. More number of turns are used in the primary.
B. V_s/V_p = N_s/N_p,
V_p = 240 V,
N_s = 200,
V_s = 12 V
N_p = V_p×N_s/N_s
= 240×200/12
= 4000 Turns

Question 25.
Back e.m.f reduces the effective, voltage in the circuit. What is your opinion about this statement? Substantiate.
Answer:
When an AC passes through a solenoid, a changing magnetic field is generated around it. Due to this, an induced emf is generated inside the solenoid. This induced emf is in a direction opposite to the emf applied on the coil. Hence this is called back emf. Back e.m.f reduces the effective voltage in the circuit.

Question 26.
Arrange the following in given order.
Distribution transformer → Power generator → Substation → Power transformer
Answer:
Power generator → Power transformer → Substation → Distribution transformer.

Question 27.
In a household electric circuit
a. Where is the MCB (Miniature Circuit Breaker) connected?
b. What is the function of ELCB (Earth Leakage Circuit Breaker)
Answer:
a. MCB is connected to the main fuse board.
b. Prevents loss of. electricity through the circuits.

Question 28.
Write the voltages of the labeled parts a, b, c and d

Answer:
a. 11 KV
b. 220 KV
c. 11 KV
d. 230 V

Question 29.
Given is a schematic diagram of a household circuit which includes bulbs and a three-pin socket.

i. What does the dotted line represent?
ii. Which three parts are marked wrong in the diagram?
iii. Suggest methods to correct this.
Answer:
i. Earth line.
ii. 1. fuse is connected to the neutral
2. The switches S₁ and S₂ are connected to the neutral.
3. Three-pin socket is connected to the earth.

Very Short Answer Type Questions (Score 3)

Question 30.

Identify the device shown in the figure.

b. Suggest two methods to increase the emf obtained when such device work.
Answer:
a. ac generator
b. Increase the number of turns in the coil Use high strength magnets. Increase the speed of the motion of magnet.

Question 31.

a. Where is thick wire used in the primary or secondary?
b. Find the number of turns in the primary if there are 50 turns in the secondary coil?
Answer:
a. Secondary coil.
b. \(\frac { V_s }{ V_p }\) = \(\frac { N_s }{ N_p }\)
Primary Voltage V_p = 100 V
Secondary Voltage V_s = 10 V
No. of turns in the secondary (Ns) = 50
= 10_s/100_p = \(\frac { 50 }{ N_p }\)
N_p = \(\frac { V_p × N_s }{ V_s }\)
= 100×50/10
= 500 turns

Question 32.
Secondary coil is arranged near the primary coil both made up of insulated copper wire without touching on an iron core. A bulb is connected in the secondary coil. When an A.C was given in the primary it is found that the bulb glows.

i. How is current reached into the secondary coil from the primary even though they do not touch?
ii. Name the phenomenon.
Answer:
i. When an A.C flows through the primary coil a changing magnetic field is produced in it. This magnetic field induces a current in the secondary coil. So bulb in the secondary coil glows.,

ii. Mutual induction.

Question 33.
i. Which part produces magnetic field for the working of a generator?
ii. What is the function of the armature coil in the generator?
iii. Which parts in the A.C generator helps to reach current in the external circuit from the moving armature?
Answer:
i. Field magnet
ii. The armature coil cuts the magnetic flux lines and an induced emf is produced in the armature coil.
iii.Brushes and slip rings.

Question 34.
Convert 1 Kilowatt-hour of electric energy into Joules.
Answer:
1k Wh = 1000W × 1 hr
= 1000 W × 60 × 60 s = 3600000 Ws
= 3600000 J

Question 35.

Answer:
a. Hydroelectric power station.
b. Hydroelectric power station.
c. Water from dams.
d. Thermal power station.
e. Thermal power station.
f. Lignite.

Question 36.
Answer the following questions related to a three-pin plug.
a. How is the earth pin different from other pins?
b. In which of the three connecting wires should a switch be connected?
Answer:
a. The length and thickness of the earth pin is more than that of other pins. Since the length is more, when the three-pin is introduced into the socket, the earth pin comes into contact with the circuit first.
b. Phase wire.

Question 37.
a. Some energy sources are given below.
From them find out which source produces green energy.
i. Energy from fossil fuel.
ii. Nuclear energy.
iii.Solar energy.
iv.Energy from wind.
b. Write any two applications of green energy at the time of making of a house.
Answer:
a. Solar energy, Energy from wind
b. Avoid solar panels.
Use biogas plants.

Question 38.
In a household, there are 4 CFL’s of 15W each working for 5 hrs a day, 2 fans of 60W working for 5 hrs daily, 500W iron box working for 15 minutes a day. How many units of electricity is used in this house. What is the cost of electricity costing Rs.3 per unit?
Answer:
CFL Power =15 × 4 = 60 W,
time = 5hrs.
Fan Power = 60 × 2 = 120 W,
time = 5 hrs.
Iron box Power = 500 W,
time = 15 min = 1/4 hrs .
Energy in k Wh = (Power used in house/1000 ) × time
= 1.025
Electricity used per day= 1.025 kWh
Electricity used per month = 1.025 × 30
= 30.75 kWh
Cost of electricity at Rs.3 per unit = 30.75 × 3
= Rs.92.25

Very Short Answer Type Questions (Score 4)

Question 39.
i. What is the basic principle of working of a transformer?
ii. Write the function of a step up and a step-down transformer.
iii.Transformers work only in AC. They do not work in DC. why?
Answer:
i. Mutual induction
ii. A step-up transformer raises AC voltage. A step-down transformer decreases AC voltage.
iii. The direction of current flow is changed in AC. When AC flows through the primary, current is induced in the secondary due to mutual induction. There is no mutual induction in DC as its direction does not change. So a transformer cannot work in DC.

Question 40.

a. Write the observation when the switch is kept on and off continuously.
b. What can we do for getting light continuously?
c. Explain the process.
Answer:
a. Bulb will light only alternatively. When switch is in the ON position bulb will not light.
b. When an AC current is given in primary the primary coil its primary coil its magnetic flux around will continuously change and electricity is induced in the secondary coil.
c. Mutual Induction
If two coils are set apart and the electric current through one of the coil changes by its intensity or direction, the magnetic flux around it changes and an electricity is induced in the second coil. This phenomenon is called Mutual Induction.

Question 41.
An experiment is performed using a magnet and a coil of wire as shown in the figure.

a. From the following activities, find out the instants in which the galvanometer deflects.
i. The magnet is placed at rest inside the solenoid.
ii. Magnet is quickly inserted into the solenoid.
iii. The magnet is placed in the solenoid and then solenoid and the magnet are moved with the same speed in the same direction.
iv. The magnet is placed within the solenoid and the solenoid alone is moved to one side.
b.Write down the principle of the above observations.
Answer:
a. (ii) and (iv).
b. Electromagnetic induction.

Question 42.
Given below are the readings obtained from a transformer.

a. What type of transformer is this?
b. What is the principle of a transformer?
c. If power loss occurs, how much is it?
Answer:
a. Step-up transformer
b. Electromagnetic Induction
c. Power in the primary P = VI
= 150 × 1 = 150 W
Power in the secondary = VI
= 1000 × 1/10 = 100 W
Power loss = 150 – 100
= 50 W

Question 43.
a. A transformer working on a 240V AC supplies 12 V. The number of turns in the primary coil is 1000. Calculate the number of turns in the secondary coil.
b. In this transformer thick wire is used in which coil primary or secondary and what is the reason for it?
Answer:
a. \(\frac { V_s }{ V_p }\) = \(\frac { N_s }{ N_p }\)
V_s × V_p = N_s × N_p
V_s = 12 V
N_p = 1000
V_p = 240 V
N_s = V_s × \(\frac { V_s }{ V_p }\)
= 12 V × 1000/240V
= 50 turns
b. For maintaining same power.
Power = Voltage × Current Therefore if voltage is lower, thickness of the coil must be increased. Therefore thick wire is used in the secondary coil.

Question 44.
Kuttiadi, Kayamkulam, Kaiga, Kalpakkam, Mooiamattom
Some power stations in India shown in above.
a. Tabulate the above power stations according to the energy changes.
b. How does the electricity produced from Moolamattam power station?
c. Industrially electricity produced in the form of a.c. Why?
Answer:
a.

b.Water stored at a height is allowed to flow down a penstock pipe. The energy of flowing water is used to rotate the turbine and generate electricity.
c. It helps the transformer for rising and decreasing voltage.

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