CBSE Class 10 Science Electricity VBQs

CBSE Class 10 Science Electricity VBQs read and download in pdf. Value Based Questions come in exams for Science in Class 10 and are easy to learn and helpful in scoring good marks. You can refer to more chapter wise VBQs for Class 10 Science and also get latest topic wise very useful study material as per latest NCERT book for Class 10 Science and all other subjects for free on Studiestoday designed as per latest Class 10 CBSE, NCERT and KVS syllabus and examination pattern

VBQ for Class 10 Science Chapter 12 Electricity

Class 10 Science students should refer to the following value based questions with answers for Chapter 12 Electricity in Class 10. These VBQ questions with answers for Class 10 Science will come in exams and help you to score good marks

Chapter 12 Electricity VBQ Questions Class 10 Science with Answers

Fill in The Blank

Question. Kilowatt is the unit of electrical ......... but kilowatthour is the unit of electrical .........
Answer : Power, energy

Question. Current = Charge X
Answer : time

Question. Copper is a preferred material for making wire because of its low ..........
Answer : Resistivity

Question. The resistance of a conductor depends directly on its .........., inversely on its .........., and also on the .......... of the conductor.
Answer : Length, area of cross-section, material

Question. In the series combination of resistors, the current is the .......... in very part of the circuit.
Answer : same

Question. A fuse is connected in .......... to the .......... wire.
Answer : Series, live

Question. The .......... is always connected in parallel across the points between which the potential difference is to be measured.
Answer : voltmeter

Question. Energy converted per unit charge is measured with an instrument called a (n) ..........
Answer : Voltmeter

Question 1 volt X1 conductor.
Answer : Joule

Question. The S.I. unit of resistivity is .......... .
Answer : Ohm-meter

Question. Tungsten wire is used in the electrical bulb due to ...........
Answer : high

Question. The fuse is placed in .......... with the device.
Answer : series

Question. The unit of power is ..........
Answer : watt (W)

 

True/False

Question. Ohm’s law is a relation between the power used in a circuit to the current and the potential difference.
Answer : False

Question. The sun is visible two minutes before the actual sunrise due to atmospheric refraction.
Answer : True

Question. Direction of current is taken opposite to the direction of flow of electrons.
Answer : True

Question. The series arrangement is used for domestic circuits.
Answer : False

Question. The resistivity of all pure metals increases with the rise in temperature.
Answer : True

Question. The solar spectrum in general is an absorption spectrum.
Answer : False

Question. Clouds look white because water droplets of clouds scatter all colours of light equally.
Answer : False

Question. The focal length of a given lens depends on the surrounding medium.
Answer : True

Question. Fuse is a thin wire which melts and breaks the electric circuit due to only high voltage.
Answer : False

Question. Two wires of resistances 2 W and 4 W are connected in parallel. The combination is connected to a 220 V supply. The power dissipated in 2 W resistor is more.
Answer : True

 

Assertion and Reason

DIRECTION : In the following questions, a statement of assertion (A) is followed by a statement of reason (R). Mark the correct choice as:
(a) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A).
(b) Both assertion (A) and reason (R) are true but reason (R) is not the correct explanation of assertion (A).
(c) Assertion (A) is true but reason (R) is false.
(d) Assertion (A) is false but reason (R) is true.
(e) Both Assertion and Reason are false.

Question. Assertion : The product of resistivity and conductivity of a conductor depends on the material of the conductor.
Reason : Because each of resistivity and conductivity depends on the material of the conductor.

Answer : C

Question. Assertion : The connecting wires are made of copper.
Reason : The electrical conductivity of copper is high.
Answer : A

Question. Assertion : Positive charge inside the cell always goes from positive terminal to the negative terminal.
Reason : Positive charge inside the cell may go from negative terminal to the positive terminal.
Answer : D

Question. Assertion : 40 W tube light give more light in comparison to 40 w bulb.
Reason : Light produced is same from same power.

Answer : D

Question. Assertion : Heater wire must have high resistance will be melting point.
Reason : If resistance is high, the electric conductivity will be less.
Answer : A

Question. Assertion : Wire A is thin in comparison to wire B of same material same length then resistance of wire A is greater than resistance of wire B.
Reason : Resistivity of wire A is greater than resistance of wire B.
Answer : C

Question. Assertion : Electric appliances with metallic body have three connections, whereas an electric bulb has a two pin connection.
Reason : Three pin connections reduce heating of connecting wires.
Answer : C

Question. Assertion : If a graph is plotted between the potential difference and the current flowing, the graph is a straight line passing through the origin.
Reason : The current is directly proportional to the potential difference.
Answer : A

Question. Assertion : A conductor has +3.2 x 10-19C charge.
Reason : Conductor has gained 2 electrons.

Answer : C

Question. Assertion : Bending a wire does not affect electrical resistance.
Reason : Resistance of wire is proportional to resistivity of material.
Answer : A

 

Very Short Answer Type Questions

Question. Write the relation between resistance (R) of filament of a bulb, its power (P) and a constant voltage V applied across it.
Answer : P = V2/R

Question. Power of a lamp is 60 W. Find the energy in joules consumed by it in 1 s.
Answer : Here, power of lamp, P = 60 W time, t = 1 s
So, energy consumed = Power × time
= (60× 1)J = 60 J

Question. Why are filaments of incandescent lamps made of thin tungsten wire ?
Answer : Tungsten has low resistivity and as such only a thin wire filament will have high resistance necessary to produce large amount of heat to light the bulb brilliantly. Further, tungsten has very high melting point (3300 K) and as such the filament will not burn at the high temperature (2400 K) at which it becomes incandescent.

Question. State Ohm’s law.
Answer : It states that the potential difference V, across the ends of a given metallic wire in an electric circuit is directly proportional to the current flowing through it, provided its temperature remains the same. Mathematically,
V ∝ I
V = RI
where R is resistance of the conductor.

Question. Draw a schematic diagram of a circuit consisting of a battery of three cells of 2 V each, a 5 W resistor, a 8 W resistor, and a 12 W resistor, and a plug key, all connected in series.
Answer : The schematic circuit diagram is shown here: 

CBSE Class 10 Science Electricity VBQs

Question. Name a device that helps to maintain a potential difference across a conductor.
Answer : A battery consisting of one or more electric cells is used to maintain a potential difference across a conductor.

Question. Name the material with the least resistivity.
Answer : Silver metal is best conductor of electricity and has lowest resistivity.

Question. How is the resistivity of alloys compared with those of pure metals from which they may have been formed?
Answer : The resistivity of an alloy is generally higher than that of its constituent metals.

Question. Why is an ammeter placed in series of a conductor/resistor in a circuit?
Answer : Because ammeter is a low resistance instrument. So, the current flowing through the circuit remains virtually unaffected.

Question. Three resistors of 3 W, 6 W and 4 W are connected in series. Calculate the total resistance of the combination.
Answer : Resistance in a series circuit is given by the expression
R = R1 + R2 + R3
∴ Total resistance R = 3 W + 6 W + 4 W = 13 W.

Question. How are Maglev trains different from conventional trains?
Answer : 
Conventional trains use electric current or diesel as the source of energy while levitating trains use electricity to produce magnetic levitation. The former has wheels while the latter do not have wheels and use 30% less energy.

Question. Name three countries where Maglev system is operational
Answer : 
China, Japan and South Korea.

Question. Why are levitating trains called as Maglev trains?
Answer : 
They use magnetic levitation by using two sets of magnets, one to repel and lift the train off the track and other to push it forward.

Question. Are levitating trains more environmentally friendly? Suggest any two reasons to support your answer.
Answer : Yes. Levitating trains have no carbon emissions and no noise pollution, less maintenance as less wear and tear.

Question. The fastest commercially used levitating train is
Answer : 
Shangai Maglev train.

Question. The high speed of levitating train is due to
a) Less friction b) High friction c) No friction
Answer : C

 

Short Answer Type Questions

Question. In the given figure what is the ratio of current in A1,and A2
Electricity 1

Answer: V=IR V=const.
I ∞ 1/R     I1/I2 = R/2R     I1/I2 = 1/ 2

Question. Two wires of equal length, one of copper and the other of manganin (an alloy) have the same thickness. Which one can be used for i) electrical transmission ii) electrical heating devices? Why?
Answer : i) Copper wire can be used for electrical transmission lines because copper has very low resistivity and hence it is a very good conductor of electricity.
ii) Manganin can be used for electrical heating devices because it has very high resistivity and hence produces lot of heat when current passes through it,

Question. In a household circuit an electric iron of 100 W is used for 10 hours and an electric oven of 1000 W is used for 2 hours every day. Calculate the cost of using them for 30 days if the cost of one unit of electrical energy is Rs 5.
Answer : electrical energy consumed by the electric iron = 100 / 1000 x 10 = 1kWh. electrical energy consumed by the electric oven = 1000/ 1000 x 2 = 2kWh.
Total energy consumed in 1 day = 1+2 = 3kWh.
Cost of electrical energy consumed 30 days = 3 x 30 x 5 = Rs 450

Question. In an electrical circuit two resistors of 2 Ω and 4 Ω respectively are connected in series to a 6 V battery. What will be the heat dissipated by the 4 Ω resistor in 5 s?
Answer : I = V / R = 6 / (2+4)
= 1A H = I2 R t = 1x 4 x5 = 20 J

Question. A wire of resistance R is bent in form of a closed circle, what is the resistance across a diameter of the circle?
Answer: 1/R’ =1/(R/2)+1/(R/2) R’=R/4.

Question. A charge of 6 C is moved between two points P and Q having , potential 10V and 5V respectively. Find the amount of work done.
Answer: W=q(V2-V1)=6(10-5)=30 joule

Question. Name the physical quantity whose SI unit is JC-1.
Answer: Potential

Question. Two wires of equal cross sectional area , one of copper and other of manganin have same resistance. Which one will be longer?
Answer: R=ρL/A (R,A=const .L=1/ρ)
ρ manganin  > ρ copper
L copper > L manganin

Question. A Rectangular block of iron has dimensions L X L X b. What is the resistance of the block measured between the two square ends ? Given ρ= resistivity.
Answer: R=ρ b/L2 

Question. Three equal resistances are connected in series then in parallel. What will be the ratio of their Resistances?
Answer: Rseries =3R.
Rparallel =R/3
Rseries / Rparallel =3R/(R/3)=9

Question. Jusitfy for any pair of resistance the equivalent resistance in series is greater equivalent resistance in parallel .
Electricity 2

Answer: Since, R=V/I
RA>RB
A=Series,B=Parallel

Question. How many bulbs of 8Ώ should be joined in parallel to draw a current of 2A from a battery of 4 V?
Answer: R=V/I=4/2=2 Ώ, let ‘n’ be the no of bulbs.
1/R=1/R1 + 1/R2 +……..1/Rn =n/8
½=n/8, n=4.

Question. Two cubes A and B are of the same material. The side of B is thrice as that of A. Find the ratio RA/RB.
Answer: RA = ρL/A RB=ρ3L/9A
RA : RB =3:1

Question. 3 X 1011 electrons are flowing through the filament of bulb for two minutes. Find the current flowing through the circuit. Charge on one electron=1.6X10-19 C.
Answer: q=ne=3x1011x1.6x10-19=4.8x108C
I=q/t=4.8x108/(2x60)=4x107

Question. A nichrome wire of resistivity 100X10-6ohm- m and copper wire of resistivity 1.62X10-8 ohm-m of same length and same area of cross section are connected in series , current is passed through them, why does the nichrome wire gets heated first?
Answer: Q=I2 Rt
Q= I2 { ρ L/A}t
Nichrome wire has higher resistivity than copper wire . Therefore, it is heated first

Question. In the given figure what is ratio of ammeter reading when J is connected to A and then to B
Electricity 3

Answer: when J is connected to A
I=V/R=3/5A=O.6A
When J is connected to B
V=1+2+3+4=10V
I=10/5=2A

Question. List the advantages of connecting electrical devices in parallel with an electrical source instead of connecting them is series.
Answer : (a) When a number of electrical devices are connected in parallel, each device gets the same potential difference as provided by the battery and it keeps on working even if other devices fail. This is not so in case the devices are connected in series because when one device fails, the circuit is broken and all devices stop working.
(b) Parallel circuit is helpful when each device has different resistance and requires different current for its operation as in this case the current divides itself through different
devices. This is not so in series circuit where same current flows through all the devices, irrespective of their resistances.

Question. A cylinder of a material is 10 cm long and has a cross section of 2 cm2. If its resistance along the length be 20 ohm. What will be its resistivity?
Answer : Here, l = 10 cm, A = 2 cm2
R = 20 ohm
Using, R = P ,1/A
P = RA/l  = 20 x 2 /10 = 4 ohm cm

Question. In the circuit shown below, the ammeter and the voltmeter readings are 3 A and 6 V respectively. Then find the value of resistance R. 
Answer : Case I : If ammeter A and voltmeter V are ideal, then
R= 6V/3A = 2Ω
Case II : If ammeter A and voltmeter V has some finite resistance, then R < 2 W.

Question. For a heater rated at 4 kW and 220 V, calculate (a) the current, (b) the resistance of the heater, (c) the energy consumed in 2 hours, (d) the cost, if 1 kW h is priced at 50 paise. 
Answer : Power, P = 4 kW
Voltage, V = 220 V
Time, t = 2 h
(a) Current, I = P/V = 4000 W/220 V = 18.2 A
(b) Resistance, R = V//l = 220 V/18.2 V 12.1 Ω
(c) Energy consumed = VIt
= 220 V × 18.2 A × 2 h = 8008 W h = 8 kW h
(d) Cost = 8 kW h × ` 0.50/kW h = ` 4.00

Question. Which uses more energy, a 250 W TV set in 1 h or a 1200 W toaster in 10 minutes?  
Answer : Energy consumed by TV set
= 250 W × 1 h = 250 J s–1 × 60 × 60 s
= 900,000 J
Energy consumed by toaster
= 1200 W × 10 min = 1200 J s–1 × 10 × 60 s
= 720,000 J.
Thus, the TV Set will use more energy.

Question. 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?
Answer : Since both the bulbs are connected in parallel and to a 220 V supply, the voltage across each bulb is 220 V. Then Current drawn by 100 W bulb, 20

Question. Three 2 W resistors; A, B and C, are connected as shown in figure. Each of them dissipates energy and can withstand a maximum power of 18 W without melting. Find the maximum current that can flow through the three resistors?  
Answer : As P = I 2R, I = √P/R
Thus, maximum current that can flow through 2 W resistor
(rating 18 W), I= √18W/2Ω = 30A
Since resistor B (= 2 W) and resistor C(= 2 W) are in parallel, and the current through their combination , which is in series with resistor is in series with resistor A(= 2 W) is also 3 A,
current through B and C (having equal resistance and in parallel) = 3 A/2 = 1.5A

Question. Name a device that you can use to maintain a potential difference between the ends of a conductor. Explain the process by which this device does so.
Answer : A cell or a battery can be used to maintain a potential difference between the ends of a conductor. The chemical reaction within a cell generates the potential difference across the terminals of the cell, even when no current is drawn from it. When it is connected to a conductor, it produces electric current and maintain the potential difference across the ends of the conductor.

Question. (i) List three factors on which the resistance of a conductor depends. (ii) Write the SI unit of resistivity.
Answer : (i) Resistance of a conductor depends upon the following factors:
(1) Length of the conductor : Greater the length (l) of the
conductor more will be the resistance (R).
R ∝ l
(2) Area of cross-section of the conductor: Greater the crosssectional area of the conductor, less will be the resistance.
R ∝ 1/A
(3) Nature of conductor.
(ii) SI unit of resistivity is W m.

Question. (a) List the factors on which the resistance of a conductor in the shape of a wire depends. (b) Why are metals good conductors of electricity whereas glass is a bad conductor of electricity ? Give reason. (c) Why are alloys commonly used in electrical heating devices ? Give reason.
Answer : (a) Resistance of a conductor depends upon the following factors:
(1) Length of the conductor : Greater the length (l) of the
conductor more will be the resistance (R).
R ∝ l
(2) Area of cross-section of the conductor: Greater the crosssectional area of the conductor, less will be the resistance.
R ∝ 1/A
(3) Nature of conductor.
(b) Metal have very low resistivity and hence they are good conductors of electricity.
Whereas glass has very high resistivity so glass is a bad conductor of electricity.
(c) Alloys are commonly used in electrical heating devices due to the following reasons
(i) Alloys have high melting point
(ii) Alloys have higher resistivity than metals
(iii) Alloys do not get oxidised or burn readily.

Question. Three resistors of 3 W each are connected to a battery of 3 V as shown. Calculate the current drawn from the battery.  
Answer : As given in circuit diagram, two 3 W resistors are
connected in series to form R1, so
R1 = 3 W + 3 W = 6 W
And, R1 and R2 are in parallel combination,
Hence, equivalent resistance of circuit (Req) is given by 
We get,
3V = I × 2 W
or I = 3/2 A 1.5 A
Current drawn from the battery is 1.5 A.

Question. An electric iron has a rating of 750 W, 200 V. Calculate:
(i) the current required.
(ii) the resistance of its heating element.
(iii) energy consumed by the iron in 2 hours.
Answer : Here, P = 750 W, V = 200 V
(i) As P = VI
I = P/V = (750/200) A = 3.75 A
(ii) By Ohm’s law V = IR or R = V/I
∴ R = 200/3.75Ω = 53.3 Ω
(iii) Energy consumed by the iron in 2 hours
= P × t = 750 W × 2 h = 1.5 kWh
or E = (750 × 2 × 3600) J = 5.4 × 106J

Question. Two identical resistors are first connected in series and then in parallel. Find the ratio of equivalent resistance in two cases. 
Answer : Let resistance of each resistor be R. For series combination,
1/RP = 1/R1 + 1/R2 or RP = R1R2/R1 + R2
So, RP = R X R/R + R = R/2
Required ratio = RS /RP = 2R/R/2 = 4/1 ⇒ RS : RP = 4 : 1

Question. Define resistance of a conductor. State the factors on which resistance of a conductor depends. Name the device which is often used to change the resistance without changing the voltage source in an electric circuit. Calculate the resistance of 50 cm length of wire of cross sectional area 0.01 square mm and of resistivity 5 × 10–8 W m.
Answer : Resistance is the property of a conductor to resist the flow of charges through it.
Resistance of a conductor depends upon the following factors:
(1) Length of the conductor : Greater the length (l) of the conductor more will be the resistance (R).
R ∝ l
(2) Area of cross-section of the conductor: Greater the crosssectional area of the conductor, less will be the resistance.
R ∝ l/A
(3) Nature of conductor.
Rheostat is the device which is often used to change the resistance without changing the voltage source in an electric circuit.
We are given, length of wire, l = 50 cm = 50 × 10–2m
cross-sectional area, A = 0.01 mm2
= 0.01 × 10–6m2
and resistivity, r = 5 × 10–8 W m.
As, resistance, R = P 1/A
R = (5 X 10 -8 X 50 X 10 -2/0.01 X 10-6) Ω

Question. Calculate the total resistance between A and B as shown in the figure. G stands for a galvanometer whose resistance is 110 W. It was noticed that the galvanometer did not show any deflection.  
Answer : Since the galvanometer did not show any deflection, i.e., no current is passing through the galvanometer. Therefore potential at Q is equal to potential at S and hence
galvanometer resistance becomes ineffective. The equivalent circuit diagram as shown in the figure. 

CBSE Class 10 Science Electricity VBQs

Question. Calculate the resistance of a metal wire of length 2 m and area of cross section 1.55 × 10–6 m2, if the resistivity of the metal be 2.8 × 10–8 W m.
Answer : For the given metal wire,
length, l = 2 m
area of cross-section, A = 1.55 × 10–6 m2
resistivity of the metal, r = 2.8 × 10–8 W m
Since, resistance, R = P 1/A   
So , R = (2.8 X 108- X 2/1.55 X 10-6)Ω
= 5.6 /1.55 X 10-2 Ω = or R = 0.036 W

Question. Study the V-I graph for a resistor as shown in the figure and prepare a table showing the values of I (in amperes) corresponding to four different values V (in volts). Find the value of current for V = 10 volts. How can we determine the resistance of the resistor from this graph?  
Answer : Since, the graph is straight line so we can either extrapolate the data or simply mark the value from graph as shown in figure. 
Hence, the value of current for V = 10 volts is 5 amperes
(or 5 A).
From Ohm’s law, V = IR,
We can write, R = V/l
At any point on the graph, resistance is the ratio of values of V and I. Since, the given graph is straight line (ohmic conductor) so, the slope of graph will also give the resistance
of the resistor
R= 10V/5 A = 2Ω 
Alternately, R = (8 -2 ) V /(4 -1)A= 6 V/3 A = 2Ω

Question. V-I graph for a conductor is as shown in the figure  
(i) What do you infer from this graph?
(ii) State the law expressed here.
Answer : (i) As graph is a straight line, so it is clear from the graph that V ∝ I
(ii) It states that the potential difference V, across the ends of a given metallic wire in an electric circuit is directly proportional to the current flowing through it, provided its temperature remains the same. Mathematically,
V ∝ I
V = RI
where R is resistance of the conductor.

 

Long Answer Type Questions
 

Question. Define Ohm’s law. Draw a labelled circuit diagram to verify this law in the laboratory. If you draw a graph between the potential difference and current flowing through a metallic conductor, what kind of curve will you get? Explain how would you use this graph to determine the resistance of the conductor.
Answer : Ohm’s law : It states that the potential difference V,
across the ends of a given metallic wire in an electric circuit is
directly proportional to the current flowing through it, provided
its temperature remains the same. Mathematically,
V ∝ I
V = RI
where R is resistance of the conductor. 
CBSE Class 10 Science Electricity VBQs
The shape of the graph obtained by plotting potential difference applied across conductor against the current flowing through it will be a straight line. 
According to Ohm’s law,
V = IR or R = V/1
So, the slope of V-I graph at any point
represents the resistance of the given conductor.

Question. An electric lamp of resistance 20 Ω and a conductor of resistance 4 W are connected to a 6 V battery as shown in the circuit. Calculate. 
(i) the total resistance of the circuit
(ii) the current through the circuit,
(iii) the potential difference across the (a) electric lamp and (b) conductor, and
(iv) power of the lamp.
Answer : Resistance of the lamp = 20 Ω
External resistance = 4Ω
(i) As both the lamp and external resistance are connected
in series, therefore the total resistance,
R = 20 + 4 = 24 Ω   

CBSE Class 10 Science Electricity VBQs

Question. Two wires A and B are of equal length and have equal resistances. If the resistivity of A is more than that of B, which wire is thicker and why ?  
For the electric circuit given below, calculate
(i) current in each resistor,
(ii) total current drawn from the battery, and
(iii) equivalent resistance of the circuit.
Answer : Let lA, aA and RA be the length, area of cross-section
and resistance of wire A and lB, aB and RB are that of wire B.
Here, lA = lB and RA = RB
If rA and rB are the resistivities of wire A and B respectively then  
CBSE Class 10 Science Electricity VBQs
Since rA > rB therefore aA > aB
Hence, wire A is thicker than wire B.
For parallel combination,
V1 = V2 = V3 = 6 V
(i) Using Ohm’s law
I1 = V1/R1 = 6/30 = 0.2 A
I2 = V2/R2 = 6/10 = 0.6 A
I3 = V3/R3 = 6/5 = 1.2 A
(ii) Total current drawn from battery,
I = I1 + I2 + I3 = 0.2 + 0.6 + 1.2 = 2 A
(iii) Equivalent resistance of the circuit, Req can be obtained
by Ohm’s law
V = I Req
So, 6 V = 2 A × Req or, Req = 6/2 = 3Ω

Question. State ohm’s law and represent it graphically. In the given circuit diagram calculate  
CBSE Class 10 Science Electricity VBQs

(i) the total effective resistance of the circuit.
(ii) the current through each resistor.
Answer : Ohm’s law: It states that the potential difference V, across the ends of a given metallic wire in an electric circuit is directly proportional to the current flowing through it, provided its temperature remains the same. Mathematically, 
V ∝ I
V = RI
where R is resistance of the conductor.
Graphical representation of Ohm’s law

CBSE Class 10 Science Electricity VBQs

Question. Derive the expression for the heat produced due to a current ‘I’ flowing for a time interval ‘t’ through a resistor ‘R’ having a potential difference ‘V’ across its ends. With which name is the relation known? How much heat will an instrument of 12 W produce in one minute if it is connected to a battery of 12 V?  
Answer : As we know, from the definition of potential difference (V), V = W/Q
here, W is work done in moving charge from one point to another, Q is the amount of charge.
W = V × Q
W = V X Q/t x t (On multiplying and dividing by time ‘t’)
∴ W = VIt
Since this work done is converted into heat energy, so, we can write
H = VIt ...(ii)
Where H is heat energy produced by electrons.
From Ohm’s Law,
V = IR, here, R is the resistance of the resistor.
Putting this in equation (ii), we get
H = I2Rt
This relation is also known as Joule’s law of heating.
Since, heat developed = power × time
∴ H = P × t
Given, P = 12 W, t = 1 min = 60 s
So, heat developed in 1 min = 12 × 60 = 720 J

 

Case Based Questions :

 

Read the passage given below and answer the following questions from Two or more resistances are connected in series or in parallel or both, depending upon whether we want to increase or decrease the circuit resistance.  
CBSE Class 10 Science Electricity VBQs

The two or more resistances are said to be connected in series if the current flowing through each resistor is same. The equivalent resistance in the series combination is given by RS = R1 + R2 + R3

Question. In the following circuit, find the equivalent resistance between A and B is (R = 2 W) 
CBSE Class 10 Science Electricity VBQs

(a) 10 W
(b) 5 W
(c) 2 W
(d) 4 W

Answer : A

Question. In the given circuit, the current in each resistor is  
CBSE Class 10 Science Electricity VBQs

(a) 3 A
(b) 6 A
(c) 9 A
(d) 18 A

Answer : A

Question. When three resistors are connected in series with a battery of voltage V and voltage drop across resistors is V1, V2 and V3, which of the relation is correct?  
(a) V = V1 = V2 = V3
(b) V = V1 + V2 + V3
(c) V1 + V2 + V3 = 3V
(d) V > V1 + V2 + V3

Answer : B

Question. When the three resistors each of resistance R ohm, connected in series, the equivalent resistance is  
(a) R/2
(b) > R
(c) < R/2
(d) < R

Answer : B

Question. There is a wire of length 20 cm and having resistance 20 W cut into 4 equal pieces and then joined in series. The equivalent resistance is  
(a) 20 W
(b) 4 W
(c) 5 W
(d) 10 W

Answer : A

Read the passage given below and answer the following questions from The heating effect of current is obtained by transformation of electrical energy in heat energy. Just as mechanical energy used to overcome friction is converted into heat, in the same way, electrical energy is converted into heat energy when an electric current flows through a resistance wire. The heat produced in a conductor, when a current flows through it is found to depend directly on (a) strength of current (b) resistance of the conductor (c) time for which the current flows. The mathematical expression is given by H = I2Rt.
The electrical fuse, electrical heater, electric iron, electric geyser etc. all are based on the heating effect of current.

Question. A fuse wire melts at 5 A. It is is desired that the fuse wire of same material melt at 10 A. The new radius of the wire is  
(a) 4 times
(b) 2 times
(c) 1/2 times
(d) 1/4 times

Answer : B

Question. What are the properties of heating element?  
(a) High resistance, high melting point
(b) Low resistance, high melting point
(c) High resistance, low melting point
(d) Low resistance, low melting point.

Answer : A

Question. What are the properties of electric fuse?  
(a) Low resistance, low melting point
(b) High resistance, high melting point.
(c) High resistance, low melting point
(d) Low resistance, high melting point

Answer : C

Question. When the current is doubled in a heating device and time is halved, the heat energy produced is  
(a) doubled
(b) halved
(c) four times
(d) one fourth times

Answer : A

Read the passage carefully and answer the following questions from Ohm’s law is the relationship between potential difference and current in a circuit which was first established by George Simon Ohm. The law states that the current passing through a metallic conductor is directly proportional to the potential difference applied between its ends. V ∝ I i.e., V = kI where k is the resistance offered by the conductor and is constant for a given conductor. Although a large class of materials is known to follow Ohm’s law, there do exist materials used in circuits that do not follow the direct relationship between V & I.

Question. The slope of the V-I graph shall give: 
(a) resistance
(b) reciprocal of resistance
(c) power
(d) charge

Answer: C

Question. Four students have plotted the graph between V-I for a conductor. Which one is correct? 

CBSE Class 10 Science Electricity VBQs

CBSE Class 10 Science Electricity VBQs

Answer: C

Question. If in a circuit both the potential difference and resistance are doubled, then 
(a) current is doubled.
(b) current is halved.
(c) current remains same.
(d) current is four times.

Answer: C

Question. When a battery of 9 V is connected across a conductor and the current flowing is 0.1 A, the resistance is: 
(a) 90 W
(b) 0.9 W
(c) 9 W
(d) 900 W

Answer: A

Question. By increasing voltage across a conductor: 
(a) current will increase.
(b) current will decrease.
(c) resistance will decrease.
(d) resistance will increase.

Answer: D

Read the passage carefully and answer the following questions from In a circuit, several resistors may be combined to form a network. The combination must have two endpoints to connect it with a battery or other elements of the circuit. When the resistors are connected in series then the current flowing in each remains the same but potential differences across each resistor will vary. When the resistances are connected in parallel, the potential difference across each resistor will be the same though a different amount of current will flow in each resistor.

Question. Two resistances 10 Ω and 3 Ω are connected in parallel across a battery. If there is a current of 0.2 A in 10 resistors, the voltage supplied by the battery is: 
(a) 2 V
(b) 1 V
(c) 4 V
(d) 8 V

Answer: A

Question. Two wires each having a resistance value equal to R are first connected in series and then connected in parallel. The plot shows the graphical representation of resistances in both cases. 
CBSE Class 10 Science Electricity VBQs

(a) (a) A denotes parallel combination
(a) (b) B denotes series combination
(a) (c) A denotes series combination and B denotes parallel combination
(a) (d) None of the above

Answer: C

Question. The equivalent resistance (in Ω) of the network across A and B is: 
CBSE Class 10 Science Electricity VBQs

(a) 2
(b) 1.5
(c) 2.5
(d) 3

Answer: A

Question. The household circuits are connected in: 
(a) series
(b) parallel
(c) both series and parallel
(d) neither series nor parallel

Answer: B

Question. The equivalent resistance of two resistors x and y is Z when connected in series and M when connected in parallel. Z:M is: 
(a) xy
(b) x + y × y
(c) (x + y)2/xy
(d) xy (2x + 2y)

Answer: C

Read the passage carefully and answer the following questions from Resistivity or electric resistivity is the inverse of the electrical conductivity. Resistivity is a fundamental property of a material and it demonstrates how strongly the material resists or conducts electric current. A low resistivity is a clear indication of a material which readily allows electric current. The common representation of resistivity is by the Greek letter ρ. Also, the SI unit of electrical resistivity is ohm-meter (Ω-m). Resistivity refers to the electrical resistance of a conductor of a particular unit cross-sectional area and unit length. 
CBSE Class 10 Science Electricity VBQs

Experts can use resistivity for comparing different materials on the basis of their ability to conduct electric currents. High resistivity is the designation of poor conductors.

Question. The value of resistivity depends upon: 
(a) length of wire
(b) area of cross section
(c) nature of conductor
(d) radius of wire

Answer: C

Question. A student plotted the graphs as shown below to study the variation of resistances R of a wire with its length l and radius r: 
(I) The resistance of a wire is inversely related to the length of 1/l the wire, i.e., R ∝ .
(II) The resistance of a wire is directly related to the length of the wire, i.e., R ∝ l.
(III) The resistance of a wire is inversely related to the radius of 1/r the wire, i.e., R ∝ .
(IV) The resistance of a wire is inversely related to the square of 1/r2 the radius of the wire, i.e., R ∝
(a) Both (I) and (III)
(b) Both (II) and (III)
(c) Both (I) and (IV)
(d) Both (II) and (IV)

Answer: D

Question. A wire of length l and of radius of cross-section r has a resistance of R Ω. Another wire of same material and of radius of crosssection 2r will have the same R if the length is: 
(a) l/4 
(b) 2l
(c) 4l , 
(d) l/2

Answer: C

Question. A wire has the same resistance as the one given in the figure. Calculate its resistivity if the length of the wire is 10 m and its area of cross section is 2 m.
CBSE Class 10 Science Electricity VBQs

(a) 16 Ω – m
(b) 8 Ω – m
(c) 16 kΩ – m
(d) 8 kΩ – m

Answer: B

Question.The resistivity of alloys is: 
(a) very low
(b) very high
(c) generally lower than its constituent metals
(d) more than resistivity of insulators

Answer: A

 

Chapter 01 Chemical Reactions and Equations
CBSE Class 10 Science Chemical Reactions and Equations VBQs
Chapter 02 Acids Bases and Salts
CBSE Class 10 Science Acids Bases and Salts VBQs
Chapter 03 Metals and Non metals
CBSE Class 10 Science Metals and Non metals VBQs
Chapter 04 Carbon and its Compounds
CBSE Class 10 Science Carbon and its Compounds VBQs
Chapter 05 Periodic Classification of Elements
CBSE Class 10 Science Periodic Table VBQs
Chapter 07 Control and Coordination
CBSE Class 10 Science Control and Coordination VBQs
Chapter 08 How do the Organisms Reproduce
CBSE Class 10 Science How Do Organisms Reproduce VBQs
Chapter 09 Heredity and Evolution
CBSE Class 10 Science Heredity and Evolution VBQs
Chapter 10 Light Reflection and Refraction
CBSE Class 10 Science Light Reflection and Refraction VBQs
Chapter 11 Human Eye and Colourful World
CBSE Class 10 Science Human Eye and Colourful World VBQs
Chapter 13 Magnetic Effects of Electric Current
CBSE Class 10 Science Magnetic Effects of Electric Current VBQs
Chapter 16 Sustainable Management of Natural Resources
CBSE Class 10 Science Sustainable Management of Natural Resources VBQs

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