Chapter 4 - Effects of electric current
1. Tell the odd one out. Give a proper explanation.
1. Fuse wire, bad conductor, rubber gloves, generator.
Ans: Fuse wire.
Explanation – Fuse wire protects the electric circuit from short circuit whereas other types of wires are used in home electrical connections.
2. Voltmeter, Ammeter, galvanometer, thermometer.
Ans: Thermometer.
Explanation – Others work on the magnetic effect of electric current while the thermometer does not work on magnetic effect of electric current.
3. Loud speaker, microphone, electric motor, magnet.
Ans: Magnet.
Explanation – Others are devices which convert energy from one form to another while magnet is not used as a device for energy conversion.
2. Explain the construction and working of the following. Draw a neat diagram and label it.
a. Electric motor
Ans: It is a device which converts electrical energy into mechanical energy.
Diagram:
Construction:
Electric motor consists of following parts:
(a) Rectangular coil: A large number of turns of insulated copper wire wound on iron core in rectangular shape forms a coil ABCD as shown in figure.
(b) Strong magnet: The coil is placed in between two pole pieces (N and S) of a strong horse shoe magnet which provides a strong magnetic field. The sides AB and CD are perpendicular to the direction of the magnetic field.
(c) Split ring: It consists of two halves of a metallic ring ( R₁ and R₂ ) The ends of the coil are connected to these rings. The outer conducting surfaces of split rings R₁ and R₂ are in contact with the two stationary carbon brushes. Split rings reverse the direction of current in the coil.
(d) Brushes: Two carbon brushes B₁ and B₂ are used to press the split rings.
(e) Axle: The two halves of the rings have resistive coating in their inner surfaces and are tightly fitted on the axle. The main function of the axle is to rotate with the coil.
(f) Battery: The battery supplies the current to the coil.
Working:
(a) When the circuit is completed, the current passes through the brush B, from A to B in branch AB.
(b) As the magnetic field acts from north pole to south pole, according to Fleming’s left hand rule, the force experienced by arm AB is in the downward direction and arm CD is in the upward direction.
(c) Both these forces are equal in magnitude and opposite in direction and hence, the coil rotates in anticlockwise direction.
(d) After half rotation, the split rings R₂ and R₁ come in contact with brushes B₁ and B₂ respectively.
(e) Again the current starts passing through the coil in the opposite direction.
(f) The force acting on branch BA is upward and DC is downward, this force again moves the coil in anticlockwise direction.
(g) This shows that the current in the loop ABCD is reversed after every half rotation, and the loop continues to rotate in the same direction.
(h) Thus, the coil rotates with the help of electrical energy.
b. Electric Generator (AC)
Ans: A generator which converts mechanical energy into electrical energy in the form of alternating current is called A.C. generator.
Diagram:
Construction:
The main components of A.C. generator are:
(a) Rectangular coil: A large number of turns of insulated copper wire wound on iron core in rectangular shape forms a coil ABCD as shown in figure.
(b) Strong magnets: The coil is placed in between two pole pieces (N and S) of a strong magnet. This provides a strong magnetic field.
(c) Rings: The two ends of the coil are connected to two rings R₁ and R₂ These rings rotate along with the coil.
(d) Brushes: Two carbon brushes B₁ and B₂ are used to press the rings.
(e) Axle: The two rings have resistive coating in their inner surfaces and are tightly fitted on the axle and the function of the axle is to rotate with the coil.
Working:
(a) When the axle is rotated with the help of a machine from outside, the coil ABCD starts rotating.
(b) On rotating the axle, the branch AB moves upwards and branch CD moves downwards hence coil ABCD rotates clockwise.
(c) According to Fleming’s right hand rule, electric current flows in the direction A → B → C → D Therefore, current flows from B₂ to B₁ in the external circuit through galvanometer.
(d) After half rotation, the branch AB and CD exchange their position and the induced current flows as D → C → B → A.
(e) Since, branch BA is in contact with brush B₁ and branch DC is in contact with B₂ current flow from B₁ to B₂ in the external circuit i.e., in the direction opposite to the previous half rotation.
(f) This repeats after every half rotation and in this way, alternate current is produced using an AC generator.
3. Electromagnetic induction means –
a. Charging of an electric conductor.
b. Production of magnetic field due to a current flowing through a coil.
c. Generation of a current in a coil due to relative motion between the coil and the magnet.
d. Motion of the coil around the axle in an electric motor.
Ans: Option (c) – Generation of a current in a coil due to relative motion between the coil and the magnet.
4. Explain the difference :
AC generator and DC generator.
Ans:
| AC generator | DC generator |
|---|---|
(i) It is used to produce alternating current. | (i) It is used to produce direct current. |
(ii) In the A.C. generator, the flow of current in the circuit reverses the direction after every half rotation of the coil. | (ii) In the D.C. generator, the direction of current is the same as long as the coil rotates in the magnetic field. |
(iii) Rings used in an A.C. generator are not split. | (iii) Split rings are used in a D.C. generator. |
5. Which device is used to produce electricity? Describe with a neat diagram.
a. Electric motor
b. Galvanometer
c. Electric Generator (DC)
d. Voltmeter
Ans: Option (c) – Electric Generator (DC)
A generator which converts mechanical into electrical energy in the form of direct current is called a D.C. generator.
Diagram:
Construction:
The main components of D.C. energy generator are:
(i) Rectangular coil (ABCD): It consists of a large number of turns of insulated copper wire wound on an iron core in rectangular shape.
(ii) Strong magnets: The coil is placed between the two pole pieces (N and S) of a strong magnet that produces a strong magnetic field.
(iii) Split ring: It consists of two halves of a metallic ring to which the ends of the coil are connected.
(iv) Brushes: Carbon brushes (B₁ and B₂) are used to press the rings.
(v) Axle: The function of the axle is to rotate with the coil.
Working:
(i) When the axle is rotated with the help of a machine from outside, the coil ABCD starts rotating.
(ii) On rotating the axle, the branch AB moves upwards and branch CD moves downwards and coil ABCD rotates clockwise.
(iii) According to Fleming’s right hand rule, electric current flows in the direction A → B → C → D. Therefore, current flows in the external circuit through the galvanometer.
(iv) After half rotation, the branch AB and CD exchange their position and the induced current flows as D → C → B → A.
(v) Since, branch BA is always in contact with brush B₁ and branch DC is always in contact with brush B₂, current flows in the same direction in the external circuit, thereby producing direct current.
6. How does the short circuit form? What is its effect?
Ans:
(i) The household electrical connections, mainly 51, consist of live, neutral, and earth wires.
(ii) The potential difference between live and neutral wires is around 220 V.
(iii) Sometimes, due to faults in the electrical appliances or damage to the insulating coating around the two wires, i.e., neutral and live, the two wires come into contact with each other and large current flows through them, causing a short circuit.
(iv) A large amount of heat is produced due to the flow of this large current. Any inflammable material like wood, plastic, cloth, etc. present around the short-circuited place can catch fire.
(v) If the household connection has fuse wire connected to it, then the short circuit results in the melting of the fuse wire and the possible damage to the house can be avoided.
7. Give Scientific reasons.
a. Tungsten metal is used to make a solenoid type coil in an electric bulb.
Ans:
(i) An electric bulb works on the principle of the heating effect of current.
(ii) When current is passed through the bulb, the solenoid-type coil of the bulb gets heated to a high temperature (up to 3400 °C) and starts glowing.
(iii) To avoid getting melted at high temperatures, the filament of the solenoid type of coil of the bulb needs to have a high resistivity and a very high melting point.
(iv) Tungsten has a high resistivity (5.6 × 10⁻⁸ Ω) and a high melting point (3422 °C).
(v) Hence, tungsten metal is used to make a solenoid-type coil in an electric bulb.
b. In the electric equipment producing heat e.g. iron, electric heater, boiler, toaster etc, an alloy such as Nichrome is used, not pure metals.
Ans:
(i) Heating devices such as irons, electric heaters, toasters, boilers, etc. work on the principle of the heating effect of electric current.
(ii) Alloys such as nichrome, as compared to metals, have higher resistivity.
(iii) Unlike metals, they can be heated to a higher temperature.
(iv) Hence, heating devices like electric heaters, boilers, toasters, etc. are made of alloys rather than pure metal.
c. For electric power transmission, copper or aluminium wire is used.
Ans:
(i) Copper and aluminium contain large numbers of free electrons.
(ii) These free electrons can move through the conductor easily.
(iii) This results in copper and aluminium having low values of resistivity.
(iv) Thus, copper and aluminium are good conductors of electricity and offer low resistance to the flow of current.
(v) Hence, aluminium and copper are used for electric power transmission.
d. In practice the unit kWh is used for the measurement of electrical energy, rather than joule.
Ans:
(i) We know that, Electrical energy = Electrical power × time.
(ii) For practical applications, devices consuming high electrical power are used over a period of time.
(iii) This leads to a heavy expenditure of electrical energy. To measure such a value of energy per unit joule is significantly difficult.
(iv) Hence, the unit kWh is used in practice. 1 kWh represents the power of 1 kW used for 1 hour and equals 3.6 × 10⁶ J.
8. Which of the statements given below correctly describes the magnetic field near a long, straight current carrying conductor?
a. The magnetic lines of force are in a plane, perpendicular to the conductor in the form of straight lines.
b. The magnetic lines of force are parallel to the conductor on all the sides of conductor.
c. The magnetic lines of force are perpendicular to the conductor going radially outward.
d. The magnetic lines of force are in concentric circles with the wire as the center, in a plane perpendicular to the conductor.
Ans: Option (d) – The magnetic lines of force are in concentric circles with the wire as the center, in a plane perpendicular to the conductor.
9. What is a solenoid? Compare the magnetic field produced by a solenoid with the magnetic field of a bar magnet. Draw neat figures and name various components.
Ans:
(i) When a copper wire with a resistive coating is wound in a chain of loops (like a spring), it is called a solenoid.
(ii) Whenever an electric current passes through a solenoid, magnetic lines of force are produced in a pattern as shown in the figure.
(iii) The magnetic field produced by a solenoid is similar to the magnetic field of a bar magnet. One end of the coil acts as south pole and the other end acts as north pole whereas the field lines are parallel inside the solenoid.
10. Name the following diagrams and explain the concept behind them.
Ans:
Fleming’s right hand rule:
Stretch the thumb, index finger and middle finger of the right hand so that they are perpendicular to each other. If the index finger indicates the direction of magnetic field and the thumb shows the direction of motion of the conductor, then the middle finger will show the direction of induced current.
Ans:
Fleming’s left hand rule:
Stretch the index finger, the middle finger and the thumb of the left hand mutually perpendicular to each other. If the index finger is in the direction of the magnetic field and the middle finger points in the direction of the current, then the thumb will point towards the direction of the force on the conductor.
11. Identify the figures and explain their use.
Ans: Fuse
Use: A fuse is used in electric circuits to protect electrical appliances from damage due to a short circuit or overloading.
Ans: Miniature circuit breakers (MCB) switches
Use: MCB switches are used in households. When the current in the circuit suddenly increases, this switch opens and stops the current.
Ans: DC generator
Use: A DC generator is used to convert mechanical energy into electrical energy in the form of direct current.
12. Solve the following example.
a. Heat energy is being produced in a resistance in a circuit at the rate of 100 W. The current of 3 A is flowing in the circuit. What must be the value of the resistance?
Given:
Power (P) = 100W
Current (I) = 3 A
To find:
Resistance (R)
Solution:
We know that,
Power (P) = V × I
∴ V = \(\large \frac {P}{I}\)
∴ V = \(\large \frac {100}{3}\)
∴ V = 33.33 V
Now,
According to ohm’s law,
V = IR
∴ R = \(\large \frac {V}{I}\)
∴ R = \(\large \frac {33.33}{3}\)
∴ R = 11.11 Ω
Ans: The value of resistance in the circuit is 11 Ω.
b. Two tungsten bulbs of wattage 100 W and 60 W power work on 220 V potential difference. If they are connected in parallel, how much current will flow in the main conductor?
Given:
Potential difference (V) = 220V
Power of bulb 1 (P₁) = 100W
Power of bulb 2 (P₂) = 60 W
To find:
Total current (I)
Solution:
We know that,
Power (P) = V × I
∴ I₁ = \(\large \frac {P₁}{V}\)
∴ I₁ = \(\large \frac {100}{220}\)
∴ I₁ = 0.4545 A
∴ I₂ = \(\large \frac {P₂}{V}\)
∴ I₂ = \(\large \frac {60}{220}\)
∴ I₂ = 0.2727 A
Now,
I = I₁ + I₂ [∵ Both the bulbs are connected in parallel, the potential difference will be same]
∴ I = 0.4545 + 0.2727
∴ I = 0.7272 A
Ans: The total current flowing in the main conductor is 0.7272 A.
c. Who will spend more electrical energy? 500 W TV Set in 30 mins, or 600 W heater in 20 mins?
Solution:
(i) For T.V. set:
Power = 500 W,
Time = 30 min = 30 × 60 s = 1800 s
We know that,
Energy = Power × time
E₁ = 500 × 1800
∴ E₁ = 900 × 10³ J
(ii) For Heater:
Power = 600 W,
Time = 20 min = 20 × 60 s = 1200 s
∴ E₂ = 600 × 1200
∴ E₂ = 720 × 10³ J
Comparing E₁ and E₂,
E₁ > E₂
Energy spent by TV set E₁ is greater.
Ans: The energy consumed by the TV set is more than that consumed by the heater.
d. An electric iron of 1100 W is operated for 2 hrs daily. What will be the electrical consumption expenses for that in the month of April? (The electric company charges Rs 5 per unit of energy).
Given:
Power (P) = 1100
W = 1.1 kW
Time (t) = 2 hour
To find:
Expense of using electric iron in April month
Solution:
We know that,
Energy = Power × time
∴ Energy consumed in one day = 1.1 × 2
∴ Energy consumed in one day = 2.2 kWh
Energy consumed in the month of April for 30 days will be,
∴ Energy = 2.2 × 30
∴ Energy = 66 kWh
∴ Energy = 66 unit …[∵ 1 kWh = 1 unit]
Now,
Electrical consumption expenses = 66 × 5
∴ Electrical consumption expenses = ₹ 330
Ans: The expense of using iron for 2 hr daily in the month of April is ₹ 330.