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
5 Mark Questions
1. Two wires A and B are of equal length, different cross sectional areas and made of same metal.
(a)(i) Name the property which is same for both the wires,
(ii) Name the property which is different for both the wires.
(b)If the resistance of wire A is four times the resistance of wire B, calculate
(i) the ratio of the cross sectional areas of the wires and
(ii) The ratio of the radii of the wire.
Ans. (a) (i) Resistivity – since the Resistivity is a property of a substance hence it remains the same for both the wires.
(ii) Resistances – As both the wires are of different cross sectional areas, so both wires are
considered as different objects.
(b) (i) Since R=
For wire A 
For wire B 
Since 
(ii) 
2. (a) State ohm’s law?
(b) The value of (I) current following through a conductor for the corresponding valves of (V)potential difference are given below
Plot a graph between V and I and also calculate resistance.
Ans. (a) At constant temperature the electric current flowing in a conductor is directly proportional to the potential difference across the end of a conductor provided the temperature and other physical conditions of the conductor remain the same.
(b) Along x-axis IV=1 cm
Y-axis IA=1cm
3. (a) Define electrical energy with S.I. unit?
(b) A house hold uses the following electric appliance;
(i) Refrigerator of rating 400w for ten hour each day.
(ii) Two electric fans of rating 80w each for twelve hours each day.
(iii) Six electric tubes of rating 18w each for 6hours 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. (a) The work done by a source of electricity to maintain current in a circuit is known as electrical energy. Its S.I. unit is joule.
(b) (i) Electricity consumed by refrigerator in one day
= power 
time
=400 W 10 h
= 4000 Wh=4 kwh
(ii) Electricity consumed by 2 electric fans in 1 day
= power Time
= 2 80 W 12 h
= 1920 Wh = 1.92 kwh
(iii) Electricity consumed by 6 electric tubes in 1 day
= 618 W 6 h
= 648 wh = 0.648 kwh
Total energy consumed in one day
=4+1.92+0.648=6.548kwh
Total energy consumed in one month
=6.56830 = 197.04 kwh
Cost of 1 unit (kwh) = Rs 3.00
Cost of 197.04 kwh = 197.043
Electricity bill =Rs 591.12
4. Redraw the circuit of question 1, putting in an ammeter to measure the current through the resistors and a voltmeter to measure the potential difference across the 12 Ώ resistors. What would be the reading in the ammeter and voltmeter?
Ans.
Here ammeter A has been joined in series of circuit and voltmeter V is joined in parallel to 12 ohms resistor.
Total voltage of battery V = 3×2 = 6 V.
Total resistance R = R1+ R2+ R3 = 5 Ω +8 Ω +12 Ω = 25 Ω
Ammeter reading (current) = I = V/R = 6/25 = 0.24 A.
Voltmeter reading = IR = 0.24 x 12 = 2.88 V.
5. What are the advantages of connecting electrical devices in parallel with the battery instead of connecting them in series?
Ans. Advantage of connecting electrical devices in parallel with the battery are as follows:
(i)Voltage across each connecting electrical device is same and device take current as per its
resistance.
(ii) Separate on/off switches can be applied across each device.
(iii) Total resistance in parallel circuit decreases, hence, a great current may be drawn from cell.
(iv) If one electrical device is damaged, then other devices continue to work properly.
6. How can three resistors of resistance 2Ω, 3Ω and 6Ω be connected to give a total resistance of (a) 4 Ω (b) 9 Ω?
Ans. (a) If we connect resistance of 3 Ω and 6Ω in parallel and then resistance of 2Ω is connected in series of the combination, then total resistance of combination is 4 Ω.
(b) If all the three resistance are joined in parallel the resultant resistance will be 3Ω.
7. The value 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.
Ans. From the given data the I-V graph is a straight line as shown below:
Resistance of resistor (R) = VA-VB/1A-1B = 12 V – 6 V/ 3.6 A – 1.8 A
= 6V/ 1.8 A = 3.3 Ω
8. 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 wire vary with its area of cross-section?
(d) Why are copper and aluminum wires usually employed for electric transmission?
Ans. (a) For filament of electric lamp we require a strong metal with high melting point.
Tungsten is used exclusively for filament of electric lamps because its melting point is extremely
high.
(b) Conductors of electric heating devices are made of an alloy rather than a pure metal due to high resistivity than pure metal and high melting point to avoid getting oxidized at high temperature.
(c) Series arrangement is not used for domestic circuits as current to all appliances remain same in spite of different resistance and every appliance can not be switched on/ off independently.
(d) Resistance of a wire is inversely proportional to its cross-section area.
(e) Copper and aluminum wires are usually employed for electricity transmission because they are good conductor with low resistivity. They are ductile also to be drawn into thin wires.