Maharashtra Board Class 10 Science Part 1 Solution Chapter 4 – Effect Of Electric Current
Balbharati Maharashtra Board Class 10 Science Part – 1 Solution Chapter 4: Effect Of Electric Current. Marathi or English Medium Students of Class 10 get here Effect Of Electric Current full Exercise Solution.
|
Std |
Maharashtra Class 10 |
|
Subject |
Science Part – 1 Solution |
| Chapter |
Effect Of Electric Current |
Chapter: -4
(Effect Of Electric Current)
Exercise: –
1) Tell the odd one out. Give proper explanation.
a) Fuse wire, bad conductor, rubber gloves, generator.
b) Voltmeter, Ammeter, galvanometer, thermometer.
c) Loud speaker, microphone, electric motor, magnet.
Ans: – a) Generator (As all except generator restricted the flow of current)
b)Thermometer (Voltammeter, ammeter, galvanometer is use in measuring the current and voltage.)
c) Except magnet all other instrument is use in electromagnetism.
2) Explain the construction and working of the following. Draw a neat diagram and label it.
a) Electric motor
b) Electric Generator (AC)
Ans: – a) Electric motor: – Electric motor convert electrical energy to mechanical energy. Electric motor is consisting of different equipment like rectangular coils, strong magnet, split ring, brushes, axle, battery. In an electrical motor a coil is placed between two magnets an d then current passes through the coil. when current passes through the coil it rotates and energy converted to mechanical energy.
b) Electric Generator: -Electric generator convert mechanical energy into electrical energy. In a generator a coil is placed between magnet and it rotates and electric current induced. In electric generator the current direction is according to the Fleming right hand rule.
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: –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 |
| Alternating current is produced. | Direct current is produced. |
| Two ring commutator are present. | A single slip ring commutator are present. |
6) How does the short circuit form? What is its effect?
Ans: – In household there are mainly three electric wire present such as live, neutral and earth wire. So by any chance when the neutral and live wire came in contact with each other we see a large current is flowing which is known as short circuit.
When short circuit happened the various equipment catch fire.
7) Give Scientific reasons.
a) Tungsten metal is used to make a solenoid type coil in an electric bulb.
Ans: – In an electric bulb there we see some heating effect. To overcome this problem, we need a wire which is more resistivity and high melting point so we use tungsten meatal as in a solenoid.
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: – In electric equipment we use Nichrome weir to flow current as there needed a high melting point like weir. And Nichrome has also high resistivity and doesn’t get oxidized.
c.) For electric power transmission, copper or aluminium wire is used.
Ans: – For power transmission we need a weir which has large electron flow. As copper weir or aluminum has more free electron in them they are used.
d) In practice the unit kWh is used for the measurement of electrical energy, rather than joule.
Ans: – The different electrical equipment consumed high electricity which needed high measuring system. As Julie is law as comparison to KWH (1kwh = 3.6×10^6j). So we use kwh for measuring electricity.
8) Which of the statement 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. The magnetic lines of force are perpendicular to the conductor going radially outward.
c) The magnetic lines of force are in concentric circles with the wire as the centre, in a plane perpendicular to the conductor.
Ans: – The statement which described the magnetic field near a long straight current carrying conductor is – d) The magnetic lines of force are in concentric circles with the wire as the centre, 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: – A solenoid is material which has high resistivity and high melting point and for this benefit we uses it to flow current.
If we compare both the magnetic field produced by solenoid and bar we will see they are identical. After passing current through solenoid it acts as bar magnet.
10) Name the following diagrams and explain the concept behind them.
Ans:-
11) Identify the figures and explain their use.
Ans: –a) This is image of fuse which is use in ac, tv, household circuit board, geyser etc.
b) This is MCB which act as a circuit breaker, it cuts the flow of current when high electricity is flowing suddenly.
c) The image is showing is a DC generator which convert mechanical energy to electrical energy.
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?
Ans: – We know that power of electric circuit is P= I^2R.
It is given current = I= A; power =P= 100w, R=?
Then, R=P/R^2;
Or, R= 100/9 = 11.1ohm.
So the resistance of the circuit is 11.1 ohm.
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 ?
Ans: – From the question we know that power of first bulb (p1) is 100w and the second bulb (p2) is 60w.
So the resistance of first bulb (R1) = V^2/P1 = 220^2/100 = 484 ohm.
And the resistance of second bulb (R2) = 220^2/60 = 806.7ohm.
When two resistances in parallel the equivalent resistance is (R1×R2/R2 + R1).
So R(eq) = (484×806.7)/ (484 + 806.7) = 302.5ohm.
Now the current I (i=v/a = 220/302.5 = 0.72) 0.72A.
c) Who will spend more electrical energy? 500 W TV Set in 30 mins,
or 600 W heater in 20 mins?
Ans: – We know energy = power × time.
So the energy of TV is (500 × 30 × 60= 900×10^3 j-s).
And the energy of heater is (600×20×60= 720×10^3 j-s).
Then TV will spend more electrical energy than 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).
Ans: – The power of electric iron (p)l 1100w.
Daily electrical consumption time is (2 × 60 × 60= 7200) 7200s.
The electrical consumption in a day by iron is (7200 × 1100 = 7290000) 7290000 j.
So electrical consumption in April month is ( 7290000 × 30 = 237600000) 2376 × 10^5 j.
We know one unite is 3.6×10^5 j. So the total unite consumed by the iron is (2376×10^5/3.6×10^6= 66) 66 unite.
The expenses of the month April is (66×5 = 330) 330 RS.
Here is your solution of Maharashtra Board Class 10 Science Part – 1 Chapter 4 Effect Of Electric Current.
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