# NCERT Solutions for class 10th Science Chapter 12 Electricity

##### NCERT INTEXT QUESTIONS

Question 1. What does an electric circuit means?

Sol. A continuous and closed path of an electric current is called electric circuit. An electric circuit consists of electric devices, source of electricity and wires that are connected with the help of a switch.

Question 2. Define the unit of current.

Sol. S.I unit of current is ampere. If one coulomb of charge flows through a conductor in one second, then the amount of current flowing is said to be 1 ampere.

Question 3. Calculate the number of electrons constituting one coulomb of charge.

Question 4. Name a device that helps to maintain a potential difference across a conductor.

Sol. A source of electricity such as cell, battery etc. helps to maintain a potential difference across a conductor.

Question 5. What is meant by saying that the potential difference between two points is 1 V?

Sol. If 1 joule of work is done in carrying one coulomb charge between the two given points, then it is said that the potential difference between the two points is IV
Note: Work done in moving a charge Q through a potential difference V will be Q·V.

Question 6. How much energy is given to each coulomb of charge passing through a 6 V battery?

Sol. The energy given to each coulomb of charge is equal to the amount of work required to move it. The amount of work is given by the expression.

Question 7. On what factors does the resistance of a conductor depend?

Sol. Resistance of a conductor depends upon the following factors:

1. nature of material (resistivity p)
2. length of wire(£)
3. area of cross section of wire (A)
4. temperature of the conductor

Question 8. Will current flow more easily through a thick wire or a thin wire of the same material, when connected to the same source? Why?

Sol. The current will flow more easily through a thick wire than through a thin wire of the same material when connected to the same source. This is due to the fact that the resistance of a wire is inversely proportional to the square of its diameter. A thick wire has a greater diameter and hence lesser resistance making the current to flow through it more easily. On the other hand, a thin wire has smaller diameter and hence greater resistance to the flow of current through it.

Question 9. Let the resistance of an electrical component remain constant while the potential difference across the two ends of the component decreases to half of its former value. What change will occur in the current through it?

Sol. According to Ohm’s law, V = IR I=V/R
Now potential difference is decreased to half
:. New potential difference V’ = V/2 Resistance remains constant
So the new current I’ = V’/R
= (V/2)/R
= (1/2) (V/R)
= (1/2) I= 1/2
Therefore, the amount of current flowing through the electric component is reduced to half.

Question 10. Why are coils of electric toasters and electric irons made of an alloy rather thana pure metal?

Sol. Alloys have high resistivity in comparison to pure metals. Also alloys do not oxidise readily at high temperatures but pure metals do. Therefore, alloys are used for making coils of electric toasters and electric irons rather than a pure metal.

Question 11. Use the data in table 12.2 (given in NCERT text book) to answer the following­
(a) Which among iron and mercury is a better conductor?
(b) Which material is the best conductor?

Sol. (a) Iron is a better conductor than mercury because the resistivity of iron (=10 x 10-8Q m) is less than the resistivity of mercury (= 94 x 10-8Q m).

(b) It can be observed from table 12.2 (given in NCERT book) that the resistivity of silver is the lowest among the listed materials. Hence, silver is the best conductor.

Question 12. Draw a schematic diagram of a circuit consisting of a battery of three cells of 2V each. a 5 Q resistor, an 8 Q resistor, and a 12 Q resistor, and a plug key all connected in series.

Sol. The schematic diagram of a circuit:

Question 13. Redraw the circuit of question 12, putting in an ammeter to measure the current through the resistors and a voltmeter to measure the potential difference across the 12Q resistor. What would be the readings in the ammeter and the voltmeter?

According to Ohm’s law, V = IR
Now potential difference (V) = 6 V
Let current flowing through the circuit= I
Resistance of the circuit (in series combination), R = 5 + 8 + 12 = 25Q Since, I=V/R
= 6/25
=0.24A
Potential difference across 12Q resistor= V1 Current flowing through the 12Q resistor, I= 0.24 A Therefore, using Ohm’s law,
V1=IR
=0.24x 12=>2.88V
Therefore, the reading of the ammeter will be 0.24 and the reading of the voltmeter will be 2.88 V

Question 14. Judge the equivalent resistance, when the following are connected in parallel
(i) 1 Q and 106 Q (ii) 1 Q and 103 Q, and 106 Q ?

Sol. (a) Since l Q and l0 Q are connected in parallel then the equivalent resistance R will be

Question 15. An electric lamp of 100 Q, a toaster of resistance 50 Q, and a water filter of resistance 500 Q are connected in parallel to a 220V source. What is the resistance of an electric iron connected to the same source that takes as much current as all three appliances, and what is the current through it?

Therefore, the current flowing through an electric iron is 7.04A.

Question 16. What are the advantages of connecting electrical devices in parallel with the battery instead of connecting them in series ?

Sol. The advantages of connecting electrical devices in parallel with the battery instead of connecting them in series are as follows:

1. In a parallel circuit, if one electrical appliance stops working due to some defect, then all other appliances keep working normally. In a series circuit, if one electrical appliance stops working due to some defect, then all other appliances also stop working.
2. In parallel circuits, each electrical appliance gets the same voltage as that of the power supply line. In a series circuit, appliances do not get the same voltage as that of the power supply line.
3. In the parallel connection of electrical appliances, the overall resistance of the household circuit is reduced, due to which the current from the power supply is high. In the series connection, the overall resistance of the circuit increases too much, due to which the current from the power supply is low.

Note: It has the advantage over series combination that if one of the devices fails the other will continue working without being affected.

Question 17. How can three resistors of resistances 2 Q, 3 Q and 6 Q be connected to give a total resistance of
(a) 4 Q, (b) 1 Q

Question 18. What is (a) the highest and (b) the lowest total resistance that can be secured by combinations of four coils of resistance 4 Q, 8 Q, 12 Q and 24 Q?

Question 19. Why does the cord of an electric heater not glow while the heating element does?

Sol. According to the Joule’s law of heating effect,
H = I2RT and H oc R
Because the resistance of heating element is very high. So, more heat is developed, hence bulb glows. But the cord has very low resistance, so it does not glow.

Question 20. Compute the heat generated while transferring 96000 coulomb of charge in one hour through a potential difference of 50 V.

Question 21. An electric iron of resistance 20 Q takes a current of SA. Calculate the heat developed in 30 sec.

Question 22. What determines the rate at which energy is delivered by a current?

Sol. The rate at which energy is delivered by a current is called the power which is given by
P=VI
Thus, the potential difference (V) determine the power delivered by a current.

Question 23. An electric motor takes SA from a 220 V line. Determine the power of the motor and the energy consumed in 2 h.

Exercise

Question 1. A piece of wire of resistance R is cut into five equal parts. These parts are then connected in parallel. If the equivalent resistance of this combination is R’ then the ratio R/R’ is
(a) 1/25 (b) 1/5 (c) 5

Note: Equivalent resistance in parallel combination decreases, but the power consump­tion increases.

Question 2. Which of the following terms does not represent electrical power in a circuit?
(a) I2R (b) IR2 (c) VI (d) V2/R

Sol. (b) We know that electric power is given by P= VI … (i) So, the option (c) is correct.
According to Ohm’s law, V = IR … (ii)
Now putting the value of V from (ii) in (i), Power P = (IR) x I= I2RSo, the option (a) is correct.
Now putting the value of l from (ii) in (i), Power P = V (V/R) = V2/R So, the option (d) is correct. Hence, the option (b) does not represent electrical power in a circuit.

Question 3. An electric bulb is rated 220V and 100W. When it is operated on 110V, the power consumed will be-
(a) 100 W (b) 75 W (c) 50 W (d) 25 W

Question 4. Two conducting wires of the same material and of equal lengths and equal diameters are first connected in series and then in parallel in an electric circuit. The ratio of heat produced in series and in parallel combinations would be-
(a) 1 : 2 (b) 2 : 1 (c) 1 : 4 (d) 4 : 1

Question 5. How is a voltmeter connected in the circuit to measure the potential difference between two points?

Sol. Voltmeter is connected in parallel to the appliance across which the potential difference is to be measured.

Note: In a voltmeter, a coil is connected in series with a resistor having a high resistance.

Question 6. A copper wire has diameter 0.5 mm and resistivity of l.6 x 10-8 Q m. What will be the length of this wire to make its resistance 10 Q? How much does the resistance change if the diameter is doubled?

Question 7. The values of current I flowing in a given resistor for the corresponding values of potential difference ‘V’ across the resistor are given below –

Plot a graph between V and I and calculate the resistance of that resistor.

Question 8. When a 12V battery is connected across an unknown resistor, there is a current of 2.5 mA in the circuit? Find the value of the resistance of the resistor.

Sol. Potential difference (V) across a resistor= 12V Electric current (I)= 2.5 mA = 2.4 x 10-3 A According to Ohm’s Law (V) = IR
Or R=V/I

Question 9. A battery of 9V is connected in series with resistors of 0.2 Q, 0.3 Q, 0.4 Q,
0.5 Q and 12 Q respectively. How much current would flow through the 12Q resistor?

Sol. There is no current division occurring in a series circuit. So, the current flow through the component is the same. Resistances are connected in series. Hence, the sum of the resistances will give the value of R.
R = 0.2 + 0.3 + 0.4 + 0.5 + 12 = 13.4Q ( In series combination, R= R1 + R2 + R3… )
Potential difference, V = 9 V I=V/R 9/13.4
=0.67A
:. the current that would flow through the l 2Q resistor will be 0.671 A.

Note: In series combination, current through all the resistors is same

Question 10. How many 176 Q resistors (in parallel) are required to carry SA on a 220 V line?

Question 11. Show how you would connect three resistors, each of resistance 6Q, so that the combination has a resistance of (i) 9 Q, (ii) 4 Q

Sol. (i) For getting a resistance of 9 Q, the two resistors should be connected in parallel and one in series with them.
The equivalent resistance of the two resistors in parallel, Rp is given by

(ii) To get a resistance of 4 Q, two resistors should be connected in series and one is parallel to them.
The equivalent resistance of the two resistors in series is given by Rs =6Q+6Q=l2Q
Now, the equivalent resistance of Rs and 6Q in parallel is given by

Question 12. Several electric bulbs designed to be used on a 220V electric supply line, are rated lO W. How many lamps can be connected in parallel with each other across the two wires of 220 V line if the maximum current is SA?

Sol. Resistance of each bulb= R (say)

Question 13. A hot plate of an electric oven connected to a 220 V line has two resistance coils A and B, each of 24 Q resistance, which may be used separately, in series or in parallel. What are the currents in the three cases?

Question 14. Compare the power used in the 2 Q resistor in each of the following circuit.
(i) a 6V battery in series with 1 Q and 2 Q resistors (ii) a 4V battery in parallel with 12 Q and 2 Q resistors.

Question 15. Two lamps, one rated 100 W at 220 V, and the other 60 W at 220 V, are connected in parallel to electric mains supply. What current is drawn from the line if the supply voltage is 220 V?

Question 16. Which uses more energy, a 250 W TV set in 1hr, or a 1200 W toaster in 10 minutes?

Note: We pay the electric company for this energy that we use in our household appliances.

Question 17. An electric heater of resistance 8 Q draws ISA from the service mains in 2hrs. Calculate the rate at which heat is developed in the heater.

Sol. Resistance of the electric heater (R) = 8 Q
Current drawn (I) = 15 A Power, P= I2R
= 15 X 15 X 8
= 1800 W or J/s
:. Heat is produced by the heater at the rate of 1800 J/s.

Question 18. Explain the following.
(a) Why is the tungsten used almost exclusively for filament of electric lamps?
(b) Why are the conductors of electric heating devices, such as bread-toasters and electric irons made of an alloy rather than a pure metal?
(c) Why is the series arrangement not used for domestic circuits?
(d) How does the resistance of a wire vary with its area of cross-section?
(e) Why are copper and aluminium wires usually employed for electricity transmission?

Sol. (a)Tungsten has very high melting point (3380°C). So it cannot melt up to a high temperature. This is the reason why tungsten is used almost exclusively for filament of electric lamps.

(b) Resistivity of an alloy is generally higher as compared to pure metals. When current is passed through an element made up valloy, a large amount of heat is produced.

(c) In series arrangement, the potential drop is divided into different appliances. So, some appliances can not get sufficient potential drop for its proper working. This is possible only in parallel connection. Hence, series arrangement is not used for domestic circuits.

(d) The resistance of a wire is inversely proportional to its area of cross-
section, i.e., R oc i which means larger will be the area, lower will be A
resistance and vice-versa.

(e) Copper and aluminium have low resistivity. Therefore, there will be less energy loss. Hence, they are usually employed for electricity transmission

Note: It wires of copper and aluminium have equal length, then for same resistance aluminium wire is lighter than the copper wire. So, aluminium wires are preferred over copper wire for overhead power cables.

Related Articles: