CBSE Class 10 Science Chapter-12 Electricity – Free PDF Download
Free PDF download of Important Questions with Answers for CBSE Class 10 Science Chapter 12 – Electricity prepared by expert Science teachers from latest edition of CBSE(NCERT) books only by CoolGyan to score more marks in CBSE board examination.
CBSE Class 10 Science Chapter-12 Electricity Important Questions
CBSE Class 10 science Important Questions Chapter 12 – Electricity
3 Mark Questions
1. Two metallic wires A and B are connected in second wire A has length l and radius r, while wire B has length 2l and radius 2r. Find the ratio of total resistance of series combination and the resistance of wire A, if both the wires are of same material?
Ans. Resistance of wire 
Resistance of wire 
Total resistance in series
R=
Ratio of the total resistance is series to the resistance of A
2. Should the heating element of an electric iron be made of iron, silver or nichrome wire? Justify giving three reasons?
Ans. Heating element of an electric iron is made up of nichrome wire because of the following reasons.
(1) Resistivity is high because of which more heat is produced due to the passage of current.
(2) Melting point is high.
(3) It does not oxidize (or burn) easily at high temperature.
3. (a) Define electric resistance of a conductor?
(b)A wire of length L and resistance R is stretched so that its length is double and the area of cross section is halved. How will its
(i) resistance change
(ii) resistivity change?
Ans. (a) It is defined as the opposition offered by the resistor to the flow of current
i.e R= v/I its S.I. unit is ohm (
)
(b) (i) R= 
(1)
New length 
= 2L and 
(since volume remains same)
R1 = 
R1 = 
(2)
R1 = 
R1= 4R
Ie resistance of a wire becomes 4 times its original resistance.
(ii) since Resistivity does not depend on the dimensions of a wire. So it remains unchanged.
4. Two resistor of resistance R and 2R are connected in parallel in an electric circuit. Calculate the ratio of the electric power consumed by R and 2R?
Ans. power 
(consumption by R)
= 
(consumption by 2R)
Ratio 
= 
5. The length of different metallic wires but of same area of cross section and made of the same material are given below
(i) Out of these two wires which wire has higher resistance.
(ii) Which wire has higher electrical Resistivity? Justify your answer.
Ans. (i) Since R
l (length of the conductor)
Since length of wire C is more than A and B therefore wire C has higher resistance.
(ii) Electrical Resistivity of a wire depends on the nature of the material and not on dimensions of a wire hence Resistivity of all wires is same as material of all the wires is same.
6. Two resistors of resistances R and 2R are connected in series is an electrical circuit? Calculate the ratio of the electric power consumed by R and 2R?
Ans. Electric power consumed by
7. 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 parallel combinations would be
(a) 1:2
(b) 2:1
(c) 1:4
(d) 4:1
Ans. (c) Let resistance of each wire is R
In series, resistance is therefore=2R
Heat produced, 
In parallel total resistance = 
Heat produced 
8. Calculate
(i) effective resistance
(ii) current
(iii) Potential difference across 10 resistor of a circuit shown in the figure.
Ans. (i) 
(ii) 
(iii) Potential difference across 
9. Apiece of wire of resistance 
is drawn out so that its length is increased to twice its original length calculate the resistance of the wire is the new situation?
Ans. 
Since length of a wire is increased, its area of cross- section decreases, as volume of the wire remains constant.
10. A battery made of 5 cells each of 2 V and have internal resistance 
is connected across resistance. Draw circuit diagram and calculate the current flowing through 
Ans. Internal resistance 1.5 (0.1+0.2+0.3+0.4+0.5)
Total resistance = (1.5+10) = 11.5
I =V/R =10/11.5
I= 0.869 A
11. In the circuit diagram given here Calculate-
(a) The total effective resistance
(b) The total current
(c) The current through each resistor.
Ans. (a) since resistances are in parallel
(b) Total current 
(c) If 
be the current through 
Respectively
12. You have two circuits (i) a 6V battery is series with 
(ii) a 4V battery in parallel with 
Compare the power used in 
resistor in each case.
Ans. (i) 
Total Resistor 
(ii) 
13. How much energy is given to each coulomb of charge passing through a 6 volt battery?
Ans. Potential difference (V) = 6 V
Charge (Q) = 1 C
Energy = total work done (W) = Q x V = 1×6 = 6 joule.
14. On what factor does the resistance of a conductor depend?
Ans. The resistance of a conductor depends on
i. length of conductor (l)
ii. Area of cross-section (A)
iii. Temperature
iv. Nature of material used to make conductor.
15. Will current flow more easily through a thick wire or a thin wire of the same material, when connected to the same source? Why?
Ans. The current flows more easily through a thick wire as compared to thin wire of the same material, when connected to the same source. It is due to the reason that resistance increases with decrease in thickness.
16. Let the resistance of an electric component remains 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?
Ans. It is given that resistance R of the electrical component remains constant but the potential difference across the ends of the component decreases to half of its value.
Hence, as per Ohm’s law, new current also decreases to half of its original value.
17. Why are coils of electric toasters and electric irons made of an alloy rather than a pure metal?
Ans. Coils of electric toasters and electric irons are made of an alloy due to the following reasons:
i. Resistivity of an alloy is generally higher than that of pure metal.
ii. At high temperature, an alloy does not oxidize readily. Hence, coil of an alloy has longer life.
18. Draw a schematic diagram of a circuit consisting of a battery of three cells of 2 V, each, a 5 Ώ resistor, 8 Ώ resistors and a 12 Ώ and a plug key, all connected in series.
Ans. The schematic diagram of circuit is as follows:
19. An electric lamp of 100 Ω, a toaster of resistance 50 Ω and a water filter of resistance 500 Ω are connected in parallel to a 220 V 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?
Ans. Here, voltage (V) = 220 V
1/R = 1/100+1/50 +1/500 = 16/500
R = 500/16 = 31.25 Ω
The resistance of electric iron, which draws as much current as all three appliances take together = R = 31.25 Ω. Current passing through electric iron (I) = V/R = 220/31.25 = 7.04 A.
20. What is (a) the highest, (b) the lowest total resistance that can be secured by combination of four resistance of 4 Ω, 8 Ω, 12 Ω and 24 Ω?
Ans. (a) To obtain highest resistance, all the four resistances must be connected in series
arrangement. In that case resultant 
= 4+8+12 48 Ω
(b) To obtain lowest resistance, all the four resistance must be connected in parallel arrangement. 
= 1/4 +1/8 +1/12 + 1/24 = 12/24 Ω
21. Why does the cord of an electric heater not glow while the heating element does?
Ans. Cord of heater and electric heater are joined in series and carry same current when joined to voltage source. As resistance of cord is extremely small as compared to that of heater element. hence, heat produced is extremely small in cord but much larger in heater element. So, the heating element begins to glow but cord does not glow.
22. A copper wire has diameter 0.5 mm and resistivity of 1.6 x 10-8 m. what will be the length of this wire to make its resistance 10? How much does the resistance change if the diameter is doubled?
Ans. Diameter of wire (d) = 0.5 mm, resistivity
, resistance (R) = 10 Ω.
R = ρL/A
= 122.5 m
If the diameter is doubled for given length of given material resistance is inversely proportional to the cross-section area of wire.
23. A battery of 9 V is connected in series with resistance of 0.2 Ω, 0.3 Ω, 0.4 Ω, 0.5 Ω and 12 Ω
respectively. How much current would flow through the 12 resistor?
Ans. Potential difference (V) = 9 V.
Total resistance 
= 0.2 +0.3 + 0.5 + 0.5 + 12 = 13.4 Ω
Current in the circuit (I) = V/R = 9 V / 13. 4 Ω = 0.67 A.
In series circuit same current flows through all the resistance, hence current of 0.67 A will flow through 12 Ω resistor.
24. How many 176 Ω resistors (in parallel) are required to carry 5 A on a 220 V line?
Ans. Let a resistor of 176 Ω are joined in parallel. Then their combined resistance (R)
1/R = 1/176 + 1/176 …… times = n/176 or R = 176/n Ω
It is given that V= 220 V and I = 5 A
R = V/I or 176/n = 220/5 = 44 Ω
n = 176/44 = 4, 4 resistors should be joined in parallel.
25. Show how you would connect three resistors, each of resistance 6 Ω so that the combination has resistance of (i) 9Ω (ii) 4Ω.
Ans. It is given here that
.
(i) To get net resistance of 9 Ω we should join three resistors as below:
(ii) To get 4Ω net resistance we should join three resistors as below:
26. A hot plate of an electric oven connected to a 220 V line has two resistance coils A and B. Each of 24 Ω resistances, which may be used separately, in series or in parallel. What are the currents in the three cases?
Ans. It is given that potential difference (V) = 220 V.
Resistance of coil (A) = Resistance of coil (B) = 24 Ω
(i)When either coil is used separately, the circuit (I) = V/R = 220 V/ 24 Ω
= 9.2 A.
(ii)When two coils are used in series total resistance (R)
= = 24 +24 = 48 Ω
Current flowing (I) = V/ R = 220 V/ 48 Ω = 4.6 A.
(ii)When two coils are joined in parallel. Total resistance (R) = 1/24 + 1/24
= 2/24, R = 12 Ω.
Current (I) = V/R = 220 V / 12 Ω = 18.3 A.
27. Compare the power used in the 2 Ω resistor in each of the following circuits:
(i) a 6 volt battery in series with 1 Ω and 2 Ω resistors and,
(ii) a 4 V battery in parallel with 12Ω and Ω resistors.
Ans. (i) When a 2Ω resistor is joined t a 6 V battery in series with 1Ω and 2Ω resistors. Total
resistance (R) = 2 + 1 + 2 = 5Ω.
Current (I) = 6V/5Ω = 1.2 A
Power used in 2 A resistor = 
= 2.88 W
(ii) When 2Ω resistor is joined to a 4 V battery in parallel with 12Ω resistor and 2Ω resistors, the current flowing in 2Ω = 4 V/ 2Ω = 2 A/.
Power used in 2Ω resistor = I2R = 8 W
Ratio = 2.88/8 = 0.36: 1.
28. In the given figure what is ratio of ammeter reading when J is connected to A and then to B
Ans. when J is connected to A
I=V/R=3/5A=O.6A
When J is connected to B
V=1+2+3+4=1OV
I=1O/5=2A
29. Given a resistor each of resistors R. How will you combine them to get the (i) maximum and (ii) minimum effective resistance? What is the ratio of the maximum to minimum resistance?
Ans. for maximum resistance Rs= nr (Equivalent of series combination) for minimum resistance Rs=nr (Equivalent of parallel combination) 
30. A wire of length L and resistance R is stretched so that its length its doubled. How will its (a) Resistance change (b) Resistively change?
Ans. Resistance of a wire is related to its length, area of cross-section and resistivity as 
Hence, if the length is doubled and area is halved, then we have
Hence, resistance of the wire becomes four times the original value.
Resistivity of the wire is the property of the material from which wire is made.
Hence, by changing the dimensions of the wire, its resistivity does not change.
31. Two students perform the experiments on series and parallel combinations of two given resistors (R) and 
and plot the following V-I graphs.
Ans. Both are correct because AV/A1= resistance(R) and A1/AV=l/R Series means high resistance and parallel means low resistance. Which of the graphs is (are) correctly labelled in terms of the words ‘series’ and parallel’ Justify your answer.
32. A household uses the following electric appliances
(i) Refrigerator of rating 4 for ten hours each thy.
(ii) Two electric fans of rating 8 each for twelve hours each day.
(iii) Six electric tubes of rating 18 W each for 6 hours each day.
Calculate the electricity bill of the household for the month of June if the cost
per unit of electric energy is Rs. 3.00.
Ans. Month of June has 30 days.
i. Refrigerator of 400 W is run 2 hours each day.
Total hours it is run in 30 days = 2 × 30 = 60 h
Energy consumed in kWh is = 400 × 60/1000 = 24 kWh
ii. Two electric fans of 80 W are run 12 hours each day.
Total hours they are run in 30 days = 12 × 30 = 360 h
Energy consumed in kWh is = 2 × 80 × 360/1000 = 57.6 kWh
iii. Six electric tubes each of 18 W are run 6 hours daily.
Total hours it is run in 30 days = 6 × 30 = 180 h
Energy consumed in kWh is = 6 × 18 × 180/1000 = 19.44 kWh
Net energy consumed in the month of June is = 24 + 57.6 + 19.44 = 101.04 kWh
Thus, the electric bill is = 3 × 101.04 = Rs 303.12