# Class 10 Science Chapter 12 Board Questions

Class 10 Science Chapter 12 Board Questions of Electricity. All the questions are divided into 1 mark, 2 marks, 3 marks and 5 marks questions. Answers are according to the marks of the questions. These questions are useful for CBSE Board Examination preparation and practicing to class tests. It includes last 10 years’ paper’s important questions from CBSE board and last 5 years’ questions answers from CBSE Sample papers and board examination papers. Just for the convenience of reading the questions are divided into sets. All the sets contain 5 questions of same categories. On the suggestions of students, we frequently update the questions, replacing the older questions with the latest one.

If you have any suggestion, you are welcome. Everything on Tiwari Academy website and Apps are free to use without any login or registration. For any inconvenience, please contact us for help.## Class 10 Science Chapter 12 Board Questions for Practice

Class: 10 | Science |

Chapter: 12 | Board Questions with Answers |

### Class 10 Science Chapter 12 Board Questions for Exams

Class 10 Science Chapter 12 Board Questions for exams practice are given below. Answers of all board papers questions are given just below the questions. Questions are taken from previous years CBSE Board papers but the answers are prepared from NCERT Textbook for class 10 Science.

#### 10th Science Chapter 12 Board Questions Set – 1 (1 Mark)

##### How is an ammeter connected in a circuit to measure current flowing through it? [CBSE 2011]

In series.

##### What happens to resistance of a conductor when its area of cross-section is increased? [CBSE 2011]

Resistance decreases as R ∝ 1/A

##### A given length of a wire is double on itself and this process is repeated once. By what factor does the resistance of the wire change? [CBSE 2011]

Length becomes one-fourth of the original length and area of cross-section becomes four times that of original.

i.e., I2 = 1/4 I1 and A2 = 4A1

R2/R1= I2/I1 × A1/A2 = 1/4×1/4 =1/16

R2 =1/16 R1

So, new resistance is (1/16)th of original resistance.

##### Two resistors of 10∩ and 15∩ are connected in series to a battery of 6 V. How can the values of current passing through them be compared? [CBSE 2009]

In series same current flows through each resistor, SO ratio of current 1:1.

##### A wire of resistance 20∩ is bend to form a closed square. What is the resistance across a diagonal of the square? [CBSE 2009]

Resistance of each side of a square = 20/4 = 5∩

1/Req= 1/(RAB+RBC) + 1/(RAD+RDC)

= (1 )/(5+5) + 1/(5+5)

= 1/10+1/10 = 1/5

#### 10th Science Chapter 12 Board Questions Set – 2 (1 Mark)

##### Why is a series arrangement not used after not used for connecting domestic electrical appliances in a circuit? [CBSE 2008]

If one stops working due to some reason, other will also stop working.

##### There are two electrical bulb (i) marked 60W, 220V and (ii) marked 100W, 220V. Which one of the two has a higher resistance? [CBSE 2006]

P = (V2 )/(R )

P ∝ 1/4 R ∝ 1/P

So, 60 W bulb has a higher resistance.

##### If four resistor each of 1∩ are connected in parallel, the effective resistance will be [CBSE 2011]

If four resistors each o each 1 ∩ are connected in parallel the effective resistance will be 0.25∩.

##### Resistivity of a given conductor depends on [CBSE 2013]

Resistivity of a given conductor depends only on its material and the temperature. Resistivity does not depend on the dimensions of the conductor.

##### Define electric current. [CBSE 2012, 2015]

Time rate of flow of charge through any cross-section of a conductor is called electric current.

#### Ohm’s Law

According Ohm’s law temperature remaining constant the current passing through a conductor is directly proportional to the potential difference across its ends.

V ∝ I or V = IR

Here constant R is known as the resistance of given conductor.

## 10th Science Chapter 12 Board Questions Set – 3 (1 Mark)

##### State the SI unit of electric current and define it. [CBSE 2012, 2015]

An ampere (A) which is defined as the rate of flow of 1 coulomb of charge per second.

##### In an electric circuit state, the relationship between the direction of conventional current and the direction of flow of electrons. [CBSE 2011, 2012, 2015]

In an electric circuit, the direction of conventional current is taken as opposite to the direction of flow of electrons.

##### Define potential difference between two points in a conductor. [CBSE 2012, 2014]

Potential difference between two points A and B in an electric field is defined as the amount of work done in order to move unit positive charge from point B to point A. thus,

VA – VB = (W^AB)/q

##### What is meant by the statement “potential difference between points A and B in an electric field is 1 volt”? [CBSE 2007, 2010, 2011, 2015]

Amount of work done to bring 1C charge from point B to point A in the electric field is 1 joule.

##### When do we say that the potential difference between two points of a circuit in 1 Volt? [CBSE 2019]

Potential difference between two points of an electric circuit is said to be 1 Volt, when a work of 1 J is to be done for moving a charge of 1 C between these two points.

#### 10th Science Chapter 12 Board Questions Set – 4 (1 Mark)

##### What is the name of physical quantity which is equal to V/I? [CBSE 2007]

Electrical resistance.

##### Define resistance. Give its SI unit. [CBSE 2012, 2013, 2019]

Resistance of a conductor is the measure of opposition offered by it for the flow of electrical charge through it. SI unit of resistance is ohm.

##### When do you say that the resistance of a wire is 1∩? [CBSE 2010, 2011]

Resistance of a wire is said to be 1∩ if for flow of 1A current through the wire one has to maintain a potential difference of 1volt across its ends.

##### The potential difference across the terminals of a cell is 1.5 volt. It is connected with a resistance of 30 ohms. Calculate the current flowing through the circuit. [CBSE 2013]

Current

I = (Potentail difference V)/(Resistance R)

= 1.5V/(30∩)

= 0.05A

##### Which material is the best electrical conductor? [CBSE 2005]

Silver.

#### Electrical Resistance of a Conductor

The electrical resistance of a conductor is its property to oppose flow of electrical charge through it and is measured by the potential difference being applied across the conductor so as to maintain the flow of one ampere current through it.

#### 10th Science Chapter 12 Board Questions Set – 5 (1 Mark)

##### What is the shape of V-I graph for a metallic wire? Why? [CBSE 2010]

A straight inclined line passing through the original because for a metallic wire V∝I.

##### The resistance of a resistor is kept constant and the potential difference across its two ends is decreased to half of its former value. State the change that will occur in the current flowing through it. [CBSE 2011, 2014]

The current flowering through the conductor is reduced to one half of its previous value in accordance with the relation I=V/R.

##### Keeping the potential difference constant, the resistance of an electric circuit is double. State the change in the reading of an ammeter connected in the circuit. [CBSE 2011]

The reading of ammeter (current flowering in the circuit) is reduced to one half of its previous value because I = V/R.

##### The length of a wire is double and its cross-sectional area is also doubled. What is the change in its resistivity? [CBSE 2009, 2011]

There is no change because resistivity of a material depends only on its nature and is independent of its dimensions.

##### What is electrical resistivity? What is its SI unit? [CBSE 2010, 2012, 2014]

The electrical resistivity of a material is the resistance offered by unit cube of that material between its opposite faces. Its SI unit is ohm-meter.

#### 10th Science Chapter 12 Board Questions Set – 6 (1 Mark)

##### On what factor does the resistivity of a conductor depend? [CBSE 2010, 2011, 2012]

Resistance of a conductor (i) a directly proportional to its length, (ii) inversely proportional to its cross-section areas and depend on the material of the conductor. Resistance also dependence on the temperature.

##### List the factor on which resistance of a conductor in the shape of a wire depends? [CBSE 2019]

Resistance of a conductor (i) a directly proportional to its length, (ii) inversely proportional to its cross-section areas and depend on the material of the conductor. Resistance also dependence on the temperature.

##### How are two resistors with resistance R^1 ∩ and R^2 ∩ are to be connected to a battery of emf 3 volts to obtain maximum current flowing through it? [CBSE 2016]

For maximum current flow the net resistance of circuit must be least possible. Hence resistor R^1 and R^2 should be connected parallel.

##### What is potential difference is needed to send a current of 5A through the electrical appliance having a resistance of 18∩? [CBSE 2016]

Potential difference

V = RI

= 18 × 5

= 90 V.

##### You have two metallic wires of resistances 6Ω and 3Ω. How will you connected these wires to get the effective resistance 2Ω? [CBSE 2010]

In parallel because then 1/R = 1/(6 ) + 1/3 = (1+2)/6 = 1/2.

Hence, R = 2 Ω.

#### Voltmeter

Voltmeter is a measuring instrument used to measure potential difference between two points in an electric circuit.

#### 10th Science Chapter 12 Board Questions Set – 7 (1 Mark)

##### What happens to the resistance of a conductor when the light of the conductor is reduced to half? [CBSE 2010]

The resistance of conductor is reduced to one-half of its previous value because resistance R ∝ Length L.

##### What happens to resistance of a conductor when its temperature is increased? [CBSE 2010, 2014]

On increasing the temperature the resistance of given conductor also increases.

##### Name the instrument/device used to measure electric current in a circuit. [CBSE 2010, 2011, 2012]

An ammeter.

##### How is an ammeter connected in a circuit to measure current flowing through it? [CBSE 2011, 2012]

In series of electric current.

##### Name the instrument used to measure (i) electric current in a circuit, (ii) potential difference between two points in a circuit? [CBSE 2011, 2012, 2014]

(i) Ammeter

(ii) Voltmeter

#### 10th Science Chapter 12 Board Questions Set – 8 (1 Mark)

##### In a circuit if resistors of 5 Ω and 10 Ω are connected in series compare the current passing though the two resistors. [CBSE 2011]

In series arrangement of resistance same current flows through all the resistors.

##### Two resistors of 30 Ω and 60 Ω are connected in parallel in an electric circuit. How does the current passing through the two resistor compare? [CBSE 2010]

In parallel arrangement current flowing through different resistors are in the inverse ratio of there resistance. Hence current passing through 30 Ω resistor is double of current passing through 60 Ω resistor.

##### Write the relationship between electrical power in watt of a device with potential difference across it and current flowing through it. [CBSE 2013]

Power = Potential difference × Current.

##### What is the heating effect of electric current? [CBSE 2012]

When electric current is pass through a resistor electrical energy is dissipated and appear as heat energy. This is known as the heating effect of electric current.

##### How many Joules are equal to kWh? [CBSE 2015]

3.6 × 10^6 J = 1 kWh.

#### Kilowatt Hour – kWh

A kilowatt hour is the commercial unit of electrical energy. It is the energy consume when 1 Kw power is used for 1 hour.

#### 10th Science Chapter 12 Board Questions Set – 9 (1 Mark)

##### Out of 60 W and 40 W lamp which one has a higher electrical resistance when in use? [CBSE 2008]

40 W lamp has a higher electrical resistance because R = V²/P.

##### Power of a lamp is 60W. Find the energy in SI unit consumed by it in 1 sec. [CBSE 2016]

Energy consumed

E = Pt = 1 W × 1 s

= 1 J.

##### Out of the two a toaster of 1kWh and an electric heater of 2kWh , which has a greater resistance? [CBSE 2006]

Toaster has a greater resistance because its power is less and R=(V²)/P

##### Name any two appliances/devices based on heating effect of electrical current? [CBSE 2012]

Electric iron, electric toaster, electrical oven etc.

##### Nichrome is used to make the element of an electric heater. Why? [CBSE 2010, 2017]

Nichrome is an alloy of high resistivity and high melting point and does not oxidise.

#### Joules Law of Heating

As per Joules law the heat produced in a resistor is H = I²RT.

- (i) directly proportional to square of current flowing through
- (ii) directly proportional to resistance and
- (iii) directly proportional to time.

#### 10th Science Chapter 12 Board Questions Set – 10 (2 Marks)

##### How much current will an electric bulb draw from 220V source if the resistance of the bulb is 1200Ω? If in place of bulb, a heater of resistance 100Ω is connected to the source calculate the current drawn by it. [CBSE 2012]

Given V: 220V, R^1 =1200Ω, I^1 =?

R^2 = 100Ω, I^2 =?

Using Ohms law,

V= I^1R^2

I^1 = V/(R^1) = 220/1200 = 0.18A

And, I^2 = V/(R^2) = 220/100 = 2.2A

##### A 9Ω resistance is cut into three equal part and connected it into parallel. Find the equivalent resistance of the combustion. [CBSE 2011]

Resistance of each part = R/3 = 9/3 = 3Ω

R^1 = R^2 = R^3 = 3Ω

In parallel combination,

1/(R^p) = 1/(R^1) + 1/(R^2) + 1/(R^3) = 1/3 + 1/3 + 1/3 = 3/3 = 1

R^p = 1Ω

##### An electric iron has a rating of 750 W, 220V Calculate the (i) current flowing through it and (ii) its resistance when in use [CBSE 2007, 2011]

Given: P = 750 W, V = 220V

(i) P = VI

750 = 220 × I

I = 750/220 = 3.40A

(ii) P = (V^2)/R

R = (V^2)/R = (220^2)/R , R= 64.53Ω

##### The charge possessed by an electron is 1.6 × 10^-19 the coulombs. Find the number of electrons that will flow per second to constitute a current of 1 ampere. [CBSE 2011]

Given: q= 1.6 × 10^-19 C, I= 1A, n=?, t=1s

We know, q=It and q =ne

ne = It

n = (It )/(e ) = (1×1)/(1.6×10-19)

##### Explain the role of fuse in series with any electrical appliances in an electric circuit. Why should a fuse with defined rating for an electrical circuit no be placed by one with larger rating? [CBSE 2011]

Fuse wire is a safety devices connected in series with the live wire of circuit. It has high resistivity and low melting point. It melts when a sudden urge of large current passes through it and disconnects the entire circuit from the electrical supply. But in disconnects the entire circuit from the electrical supply. But in case if we use a larger rating instead of a defined rating, then it will not protect the circuit as high current will easily pass through it and it will not melt.

Question: Would you connect a fuse in series or in parallel to an electrical circuit? [CBSE 2011]

Answer: In series of electric circuit before appliances present in the circuit.

##### 10th Science Chapter 12 Board Questions Set – 11 (2 Marks)

##### The wattage of a bulb its effective wattage if it operates a 12V battery. Calculate its effective wattage if it operates on a 6V battery (Neglect the change in resistance due to unequal heating of the filament in the two cases). [CBSE 2016]

Given P^1 = 24 W, V^1 = 12V, P^2 =?, V^2 = 6V

Using P = (V^2)/R

(P^1)/(P^2) = (V^1)/(V^2)

P^2 = [ (V^2)/(V^1)] × P^1

[6/12]^2 × 24 = 1/4× 24 = 6W

##### Two metallic wire A and B of the same material are connected in parallel. Wire A has length L and radius r wire B has a length 2L and radius 2r. Calculate the ratio of the equivalent resistance in parallel combination and the resistance of wire A. [CBSE 2009]

For wire A:

I^1 = l, r^1 = r, p^ = p

So R^A = p^1 (l^1)/(A^1) = p^1(l^1)/(πr^2) =pl/(πr^2)

For wire B:

l^2 =2l, r^2 =2r, p2 =p

R^B = P^2(l^2)/(A^2) = p2(l^1)/(πr^2) = (p(2l))/(π(2r)^2)

1/2 = {pl/(πr^2)} = (R^A)/2

In parallel combination of R^A and R^B

1/(R^P) = 1/(R^A) + 1/(R^B) = 1/(R^A) + 1/(R^Al2) = 3/(R^A)

(R^P)/(R^A) = 1/3 = 1:3

##### A piece of wire of resistance 20Ω is drawn out so that its length increased too twice its original length. Calculate the resistance of the wire in the new situation. [CBSE 2009]

Using R= pl/A

We have, (R^1)/(R^2) = (I^1)/(A^1). (A^2)/(I^2)

Given I^2 =2l^1

Volume of material will be conserved. So A^1l^1 = A^2l^2

(A^1)/(A^2) = (l^2)/(l^1) = 2

(R^1)/(R^2) = (l^1)/(l^2). (l^2)/(l^1) = (l^2/l^1 )^2 = 1/4

R^2 = 4R^1 = 80Ω

##### An electric lamp is marked 25 W, 220V. It is used for 10 hours daily. Calculate (i) Its resistance while glowing (ii) energy consumed in kWh per day. [CBSE 2007]

(i) R= (V^2)/P = (220^2)/25 = (220×220)/25 = 1936 Ω

(ii) Energy consumed = Power × Time

= 25 × 10 = 250 Wh = 0.25 kWh

##### Two wires of equal lengths one copper and the other of Manganin have the same thickness. Which one can be used for (i) electrical transmission lines (ii) electrical heating devices? Why? [CBSE 2008]

(i) Copper as it has the lesser resistivity.

(ii) Manganin as it has the comparatively higher resistivity, less oxidation even at high temperature.

Question: What is an electric circuit? Distinguish between an open and closed circuit? [CBSE 2009]

Answer: An arrangement for maintaining the continuous flow of electric current by the electrical energy source through various electrical components connected with each other by conducting wires is termed as electric circuit.

##### 10th Science Chapter 12 Board Questions Set – 12 (3 Marks)

##### (a) n electrons each carry a charge –e is flowing across a unit cross—section of a metallic wire in unit time east to west. Write an expression for electric current and also give its direction of flow. Give reason for your answer. (b) The charge possessed by an electron is 1.6 × 10^-19 coulomb. Find the number of electrons that will flow per second to constitute a current of 1 ampere. [CBSE 2011, 2012, 2013, 2015]

(a) Electric current in a circuit is defined as the time rate of flow of electric charge through any cross-section and its direction is opposite to that of flow of electrons. Hence in present case

Electric current I = q/t = ne/1 = ne

As electron are flowing from east to west, the direction of electric current is from west to east.

(b) Here current I= 1A, time t= 1s and charge on each electron e = 1.6 × 10^-19 C.

Hence number of electrons flowing n= (It )/e = (1 ×1)/(1.6 × 10^-19) = 6.25 × 10^18

##### (a) Define the term Volt. (b) State the relation between work, charge and potential difference for an electric circuit. Calculate the potential difference between the two terminals of a battery if 100 joules of work is required to transfer 20 coulombs of charge from one terminals of the battery to the other. [CBSE 2009]

(a) The term ‘volt’ is the SI unit of potential difference. Potential difference is said to be 1 volt if one joules work is to be done to carry 1 coulomb charge from one point to another.

(b) The relation between work (W), charge (Q) and potential difference (v) for an electric circuit is

V = (W )/Q

Given that W = 100J and Q = 20 C

Potential difference V = (W )/Q = (100J )/(20 C ) = 5V

##### (a) Mention the condition under which charges can move in a conductor. Name the device which is used to maintain this condition in an electric circuit. (b) A current of 2A passes through a circuit for 1 minutes. If potential difference between the terminals of the circuit is 3V what is the work done in transferring the charge? [CBSE 2012]

(a) Electric charges can move in a conductor when potential differences is maintained across its ends. An electric cell is used to maintain potential difference across the conductor and thus can be used for flow of charges.

(b) Here I = 2A, time t = 1 minutes = 60s and V = 3V

Total charge flowing q = It = 2 × 60 = 120C

Work done W = qV =120 × 3 = 360 J

##### State Ohms law. How can it be verified experimentally? Does it hold good in all conditions? Comment. [CBSE 2011, 2014]

George Simon ohm established a relationship between the electric current flowing through a conductor and the potential difference across its ends due to which current flows. According to Ohms law temperature remaining constant the current passing through a conductor is directly proportional to the potential difference across its ends

V ∝ I or V = IR

Here constant R is known as the resistance of given conductor. For a given conductor its resistance is constant at a given temperature. XY is a resistance wire, An ammeter and V a voltmeter. A battery of four cells is being used as a current source and K is a plug key. Initially use one cell only. Put plug in key K and note current and voltage by nothing ammeter and voltmeter reading respectively. Let those be I^1 and V^1. Then connect two cells in the circuit and note current I^2 and potential difference V^2 across the resistance. Similarly take reading with three cells and four cells in the circuit. From our observations, we find that

(V^1 )/V^2 = (V^2 )/I^2 = (V^3 )/I^3 = (V^4 )/I^4 = a constant = R.

##### (a) What do you mean by resistance of a conductor? Define its unit. (b) In an electric circuit with a resistance wire and a cell the current flowing is I. What would happen to this current if wire is replaced by another thicker wire of same material and same length? Give reason. [CBSE 2010, 2011, 2016]

(a) The resistance of a conductor is a property of the conductor which affects the flow of current through it on maintaining a potential difference across its ends.

Unit of resistance is ohm. Resistance of a conductor is said to be 1 ohm if a potential difference of 1V is to be applied across its ends for maintaining flow of 1A current.

(b) If given resistance wire is replaced by another thicker wire of same material and same length then cross-section area wire is increased and consequently its resistance decreases. So the current flowing in the circuit increases.

Question: Define electric current. An electric circuit comprising of a cell, a resistor, an ammeter, a voltmeter and a closed switch. Distinguish between an open and a closed circuit. [CBSE 2014, 2015]

Answer: A continuous and closed path of an electric current is called an electric circuit. An electric circuit showing a cell E a resistor R, an ammeter A, a voltmeter V and a closed switch S. An electric circuit is said to be an open circuit when the switch is in off mode and no current flows in the circuit. The circuit is said to be a closed circuit when the switch is in on mode and a current flow in the circuit.

##### 10th Science Chapter 12 Board Questions Set – 13 (3 Marks)

##### (a) List the factor on which the resistance of a cylindrical conductor depends and hence write an expression for its resistance. (b) How will the resistivity of a conductor change when its length is triple by stretching it? [CBSE 2013, 2015, 2019]

(a) The resistance of a cylindrical conductor a wire (R) is (i) directly proportional to its length L, (ii) inversely proportional to its cross-section area A and (iii) depend on nature of material.

R ∝ (L)/A

Here p is known as the resistivity of given material. It is defined as the resistance offered by a unit cube of given material when current flows perpendicular to the opposites faces.

(b) The resistivity of the conductor remains unchanged.

##### Describe a simple expression to demonstrate variation of resistance on (i) length, (ii) cross-section area and (iii) material of the conductor. What are the conclusion drawn? [CBSE 2016]

To demonstrate dependence of resistance on length, cross-section area and material of the conductor we complete an electric circuit.

(i) Effect of length: Take the resistance wire number 1 of length l and connect in the circuit. Put plug in key K and note ammeter reading I^1. Now instead of wire number 1, connected wire number 2 of double length but of same material and same thickness. As before note ammeter reading I^2. Experimentally it is observed I^2 = (I^1 )/2 It shows the resistance of 2nd wire is double of 1st wire thus, we conclude that

(R^2 )/R^1 = (l^2 )/I^1 = (2l )/l = 2

(ii) Effect of cross-section area: Replace the wire by resistance wire number 3 of length l but thickness more than wire number 1. Put plug in key and note ammeter reading I^3. It is more than I^1. It means for a thicker wire the resistance is less. Extract calculation shows that resistance is inversely proportional to cross-section area of the conductor.

Thus R∝ (1 )/A

(iii) Effect of material: Now take a wire number of 4 of some other material but of length l and same thickness as wire number 1. Connect the wire in the circuit and again note ammeter reading I^4 is difference from I^1. It shows that resistance depends on material of the conductor.

##### (a) Write the relationship between resistance and electrical resistivity of the material of a conductor in the shape of a cylinder of length l and area of cross-section A Hence derive the SI unit of electrical resistivity. (b) Resistance of a metal wire of length 5m is 100∩. If the area of cross-section of the wire is 3 × 10^7 m^2, calculate the resistivity of the material. [CBSE 2019]

(a) The resistance R of a conductor, the shape of a cylinder, of length l and area of cross-section A is given as:

R = pl/A

Where p is a constant which is known as the electrical resistivity of the material of conductor.

Thus resistivity p= (RA )/(l )

SI unit of resistivity p shall be (ohm × m^2 )/m = ohm m= ∩m

(b) Here l = 5m, R = 100∩ A = 3 × 10^7m^2

Resistivity of the material of wire p = (RA )/(l ) = (100 ×3×10-^7)/(5 )

= 6.0 × 10^-6 m^2

##### Derive the relation R = R1 + R2 + R3, when resistance are joint in series. [CBSE 2005, 2012, 2019]

In series combination, the same current flows in all the resistance but the potential difference across each of the resistance is different.

According to Ohm’s law we have

V1 = IR1, V2 = IR2, V3 = IR3

If the total potential difference between A and B is V, then

V = V1 + V2 + V3

= IR1 + IR2 + IR3

= I(R1 + R2 + R3)

Let the equivalent resistance be R then

V = IR

IR = I(R1 + R2 + R3)

R = R1 + R2 + R3.

##### Derive the relation (1 )/R = (1 )/R1 + (1 )/R2 + (1 )/R3, when the resistors are joined in parallel. [CBSE 2013, 2014, 2015, 2017, 2019]

In parallel combination of three resistance R^1, R^2 and R^3 the current in each of the resistance is different. If I is the current drawn from the cell then it is divide into branches I1, I2 and I3. Thus,

I = I1 + I2 + I3.

Thus from Ohm’s law I1 = (V )/R1, I2 = (V )/R2, I3 =(V )/R3

If R is the equivalent resistance then,

I= (V )/R

(V )/R = (V )/R^1 + (V)/R^2 + (V)/R^3

(1 )/R = (1)/R^1 + (1)/R^2 + (1)/R^3

Question: Why do electrician wear rubber hand gloves while working? [CBSE 2016]

Answer: Rubber is an electrical insulator. Hence electrician can work safely while working on an electric circuit without a risk of getting any electrical shock.

##### 10th Science Chapter 12 Board Questions Set – 14 (3 Marks)

##### (a) A bulb is rated 40W, 220V. Find the current drawn by it when it is connected to a 220V supply. Also find its resistance. (b) If the given bulb is replaced by a bulb of rating 25W, 220V will there be any change in the value of current and resistance? Justify your answer and determine the change. [CBSE 2019]

(a) Here power of bulb P = 40W and voltage V= 220V

Current drawn by the bulb I = (P)/V = (40)/220 = (2)/11A

And resistance of the bulb R = (V)/I = (220)/((2/11)) = (220×11)/2 = 1210Ω

(b) On taking another bulb power P= 25V and voltage V= 220V. there is a change in the value of current and resistance because their values depend on the power of the bulb.

New Current I’ = (P’)/V = (25)/220 = (5)/44A

And new resistance R’ = (V)/I’ = (220)/((5/44)) = 1936Ω

##### An electric kettle of 2kW works for 2h daily. Calculate the (i) energy consumed in SI and commercial unit, (ii) cost of running it in the month of June at the rate of ₹3.00 per unit. [CBSE 2013]

Here power P = 2kW, time of use t= 2h daily

Total time of use in the month of June t = 2 × 30 = 60h

(i) Energy consumed E = P × t = 2kW × 60h = 120kWh = 120 units

E = 120 × 3.6 × 10^6 J = 4.32 × 10^8J

(ii) Cost ₹ 3.00 per unit = 3.6 × 120 = ₹ 360

##### An electric iron consumer energy at a rate of 840W when heating is at the maximum rate and 360W when the heating is at the minimum. The voltage is 220V. What are the current and resistance in each case? [CBSE 2016]

(a) When heating is at maximum rate the power rating P = 840W and supply voltage V= 220V

Current flowing I= (P)/V = (840W)/220V = 3.82A

and resistance of electric iron R= (V )/I = (220 V)/3.82A =57.6Ω

(b) When heating is at maximum rate the power rating P = 360W

Current flowing I = (P)/V = (360W )/220V = 1.64A

And resistance of electric iron R = (V )/I = (220V )/1.64A = 134.2Ω

##### The potential difference between the terminals of an electric heater is 110V. When it draws a current of 5A from the source. What current will the heater draw and what will be its wastage if the potential difference is increased to 220V. Considered that the resistance of the heater element does not change with temperature. [CBSE 2012]

As here V= and I=5A, hence resistance of the electric heater

R = (V)/I = (110)/5 = 22∩

When potential difference is increased to V’ = 220V then

(a) the current I’ = (V’)/R = (220)/22 = 10A.

(b) the power P’ = V’I’ = 220×10 = 2200W = 2.2Kw

##### How many 40W; 220V lamps can be safely connected to a 220V, 5A line? Justify your answer. [CBSE 2014, 2015]

The current drawn by a 40W, 220V electric lamp I^1 = (P^1)/V = (40)/220 = (2)/11A

As the electric line of rating 220V, 5A hence we can connect n lamps in parallel where

n= (I)/I^1 = (5)/((2/11)) = 27.5

Thus we can safely have connected 27 lamps of 40W, 220V rating to a 220V, 5A line.

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##### 10th Science Chapter 12 Board Questions Set – 15 (3 Marks)

##### Calculate the amount of heat generated while transferring 90000 coulombs of charge between the two terminals of a battery of 40V in one hour. Also determine the power expanded in the process. [CBSE 2014, 2015]

Here charge transferred Q = 90000C, potential difference between the terminals of battery V= 40 and time t = 1h =3600s.

Current I = (Q)/t (9000)/3600 = 25A

Amount of heat generated H = VIt = 40×25 × 3600 = 3600000J = 3.6×10-^6

And power expanded P = (H)/t = VI = 40 × 25 = 1000W

##### Two identical resistors each of resistance 10∩ are connected (i) in series and then (ii) in parallel in line to a battery of 6volts. Calculate the ratio of power consumed in the combination of resistors in the two cases. [CBSE 2013]

Here R^1 = R^2 = 10∩ and V= 6 volt

(i) In series net resistance R^s = R^1 + R^2= 10 + 10 + 20∩

And power P^s = (V^2)/R^s = (6^2)/20 = 1.8W

(ii) In parallel net resistance (1)/R^p

= (1)/R^1 + (1)/R^2 = (1)/10 + (1)/10 = (1)/5

and power Pp = (V^2)/R^p = (6^2)/5 =7.2W

(P^s)/P^p = (1.8)/7.2 = (1)/4

##### A torch bulb is rated 5V and 500mA. Calculate (i) its power (ii) its resistance (iii) the energy consumed if this bulb is lighted for 4 hours. [CBSE 2013]

Here rating of torch bulb is V= 5V and I = 500 mA = 0.5A

(i) Power of torch bulb P = VI = 5 × 0.5 = 2.5 W

(ii) Resistance of torch bulb R = (V)/I = (5)/0.5 = 10∩

(iii) Energy consumed by bulb in 4 hours

E = Pt = 2.5 × 4 = 10Wh.

##### Derive the expression for power P consumed by a device having resistance R and potential difference V. [CBSE 2010, 2012, 2019]

Amount of work done in carrying a charge Q through a potential difference V is

W = QV

Q = It

As power is defined as the rate of doing work hence

Power P = W/t = (VIt )/t = VI.

If R be the value of resistance of the conductor, then V= RI and hence

P = VI = (RI)I = I^2R

Again P= VI = V(V/R) = V^2/R

Thus in general we can say that electric power is given by

P = VI = I^2R = V^2/R

##### Two identical resistors are first connected in series and then in parallel. Find the ratio of equivalent resistance in two cases. [CBSE 2013]

Let two resistors of value R each are first joined in series when equivalent resistance

R^s = R + R = 2R

On joining these very resistors in parallel let equivalent resistance be R^p where

1/R^P = 1/R + 1/R = 2/R

R^s/R^p = 2R/(R/2) = 4/1

Question: Why are metals good conductors of electricity whereas glass is a bad conductor of electricity? Give reason. [CBSE 2018]

Answer: Metals are good conductor of electricity because they have a large number of free electrons. Thus conduction of charge is not easily possible in glass and so glass is a bad conductor of electricity.

###### 10th Science Chapter 12 Board Questions Set – 16 (5 Marks)

##### (a) State Ohm’s law. Express it mathematically. (b) An electric bulb is rated 220V and 100W. When it is operated on 110V, what will be the power consumed? [CBSE 2011]

(a) According to Ohm’s law, physical conditions like temperature remaining constant, the current passing through a conductor is directly proportional to the potential difference maintained across the ends of the conductor i.e.,

V∝ I or V = IR

Hence, R is known as the resistance of given conductor. It is a property of the conductor which affects the flow of current. For a given conductor value of resistance R is a constant at a given temperature.

(b) As rating of bulb is 220V, 100 W, hence P = V1 = V^2/R

Resistance of bulb filament R=V^2/P = (220 ×220)/100 = 484 Ω.

When the bulb is operated at a voltage V’=110V, the power consumed is

P= V^2/R = (110 × 110)/484 = 25.W]

The power will be New power = 25 W

##### (a) Define electric resistance of a conductor. (b) List two factors on which resistance of a conductor depends. [CBSE 2013]

(a) Electrical resistance of a conductor may be considered as a measure of the opposition offered by it for flow of electric charge through it.

Mathematically, resistance R = V/I ((potential difference))/((current))

SI unit of electrical resistance is ohm (Ω), where 1Ω =1V/1A

(b) At a given temperature resistance of a conductor depends on its (i) length L, (ii) cross-section area A, and (iii) nature of the material of conductor. It is found that R∝ L and R∝ 1/A

Mathematically, ρL/A

Where ρ is a constant known as the resistivity of the material of conductor. Its value depends only on the nature of material of the conductor and the temperature and is independent of the dimensions (i.e. length and cross-section area) of the conductor.

The resistivity of a given material is defined as the resistance offered by a cube of that material of side 1m, when current flows perpendicular to the opposite faces. Its unit is ohm-meter (Ωm).

##### What is meant by electric current? Name and define its SI unit. In a conductor electrons are flowing from B to A. What is the direction of conventional current? Give justification for your answer. A steady current of 1 ampere flows through a conductor. Calculate the number of electrons that flows through any section of the conductor in 1 second. [CBSE 2015]

Electric current is defined as the rate of flow of electric charge through a cross-section of a conductor. If Q charge passes through a section of a conductor in time t then current I= Q/t

SI unit of electric current is an ampere (A). current is said to be one ampere. If rate of flow of electric charge through a cross section of conductor be 1 coulomb per second. Direction of conventional current is taken as the direction of flow of positive charges or opposite to the charges of flow of negative charge. If negatively charged in a conductor flow from B to A then the direction of conventional current will be from A to B. Here current I = 1A time t= 1s and charged on electron e = 1.6 × 10^-19C. let one n electron flow through a section of conductor so that charge passing through the section is Q= ne.

I= Q/t = (ne )/t , n = (It )/e (1×1 )/(1.6 × 〖10〗^(-19) ) = 6.25 × 10^18

##### Establish a relationship to determine the equivalent resistance R of a combination of three resistor having resistance R^1, R^2, and R^3 connected in parallel. [CBE 2016, 2013]

In parallel combination of three resistance is R^1, R^2 and R^3 the current in each of the resistance is different. If I is the current drawn from the cell then it divide into branches I^1, I^2 and I^3Thus I = I^1 + I^2 + I^3

The potential difference across each of these resistance is same.

Thus from Ohm’s law I^1 = V/R^1 , I^2= V/R^2 , I^3= V/R^3

I= V/R

V/R = V/R^1 + V/R^2 + V/R^3

1/R = 1/R^1 + 1/R^2 + 1/R^3

###### 10th Science Chapter 12 Board Questions Set – 17 (5 Marks)

##### A wire of resistance 8 ∩ is bent in the form of a closed circle. What is the effective resistance between the point A and B at the end of a diameter of the circles? What is ammeter reading? [CBSE 2014]

As resistance of entire wire bent in the form of a circle is 8Ω, hence resistance of each half part (semi-circle) will be 8/2 = 4Ω.

Now, these two resistances of 4∩ each are joined in parallel between points A and B, hence effective resistance of the circuit R is given as

1/R = 1/4 + 1/4 = 1/2 = R = 2Ω

The ammeter reading= current flowing in the circuit I= V/R = 2V/(2Ω) = 1A

##### Establish a relationship to determine the equivalent resistance R of a combination of three resistor having resistance R^1, R^2, and R^3 connected in series. (b) Calculate the equivalent resistance R of a combination of three resistors of 2Ω, 3Ω and 6Ω are joint in parallel. [CBSE 2016]

(a) In series combination, the same current flows in all the resistance but the potential difference across each of the resistance is different.

According to Ohm’s law we have

V^1 = IR^1, V^2 = IR^2, V^3 = IR^3

If the total potential difference between A and B is V, then

V = V^1 + V^2 + V^3

= IR^1 + IR^2 + IR^3

= I(R^1 + R^2 + R^3)

Let the equivalent resistance be R then

V = IR

IR = I(R^1 + R^2 + R^3)

R = R^1 + R^2 + R^3.

(b) Here R^1= 2Ω, R^2 = 3Ω, R^3 = 6Ω

The equivalent resistance R in parallel combination of resistor will be

1/R = 1/R^1 + 1/R^2 + 1/R^3 = 1/2 + 1/3 + 1/6 = 1/1 R = 1Ω

##### Three resistor of resistance R^1, R^2 and R^3 are connected in (i) series, and (ii) parallel. Write expression for the resistance of the combinations in each case. (b) Two identical resistance of 12Ω each are connected to a battery of 3V. Calculate the ratio of the power consumed by the resulting combination with minimum resistance and maximum resistance. [CBSE 2019]

(a) (i) In series arrangement equivalent resistance R^s = R^1 + R^2 + R^3

(ii) In parallel arrangement equivalent resistance R^p is given as:

1/R^p = 1/R^1 + 1/R^2 + 1/R^3

(b) Here R^1 = R^2 = 12Ω and V= 3V.

For minimum resistance two resistor must be connected in parallel so that

1/R^p = 1/R^1 + 1/R^2 = 1/12 + 1/12 = 1/6 R= 6Ω

Hence power P^p = V^2/R^p = ((3)^2)/6 = 24Ω

For maximum resistance two resistance must be connected in series so that

R^s = R^1 + R^2 = 12 + 12 = 24Ω

So the power P^s = V^2/R^s = (3)^2/24 = 0.375W

P^p/(Ps ) = (1.5 )/(0.375 ) = 4/1

##### Experimentally prove that in series combination of three resistances: (a) current flowing through each resistance is same and (b) total potential differences is equal to the sum of potential differences across individual resistors. [CBSE 2016]

Series combination of resistor: We take three resistor R^1, R^2 and R^3 and join them in series between the points X and Y in an electric circuit.

(a) Plug the key and note the ammeter reading Then change the position of ammeter to anywhere in between the resistor and again note the ammeter reading. We find that ammeter reading remains unchanged. It shows that in series arrangement same current flow through each resistor.

(b) Inert a voltmeter across the ends X and Y of the series combination of resistors. Plug the key so as to complete the circuit and note the voltmeter reading V across the series combination of resistors.

Take out plug from key K and disconnected the voltmeter. Now inert the voltmeter across the ends of first resistor R^1. Plug the key and note the voltmeter reading V^1. Similarly measures the potential difference across the other two resistors R^2 and R^3 separately. Let these potential differences be V^2 and V^3 respectively. Experimentally we find that V= V^1 + V^2 + V^3 It shows that in series arrangement of resistors total potential difference equal to the sum of potential differences is equal to the sum of potential differences across individual resistors.

##### An electric lamp of resistance 20Ω and a conductor and a conductor of resistance 4Ω are connected to a 6V battery as shown in the circuit. Calculate (a) the total resistance of the circuit., (b) the current through the circuit, (c) the potential differences across the (i) electric, lamp and (ii) conductor and [CBSE 2019]

Here voltage of battery V = 6V resistance of electric lamp R^1 = 20 Ω and resistance of conductor R^2 = 4Ω

(a) Since R^1 and R^2 are connected in series the total resistance of the circuit.

R= R^1 + R^2 = 20 + 4 = 24Ω

(b) The current through the circuit I = V/R = (6 )/24 =0.25A

(c) (i) Potential differences across the electric lamp V^1= IR^1 = 0.25× 20 = 5V

(ii) Potential differences across the conductor V^2 = IR^2 = 0.25× 4 = 1V

(d) Power of the lamp P= I^2R^1 = (O.25)^2 × 20 = 1.25 W

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