CBSE Class 10 Physics Electricity Assignment

 Electricity Notes Class 10 Science

Charge: It is an inherent property of the body due to which the body feels attractive and repulsive forces. There are two types of electric charges: Positive and (ii) Negative
Like charges are repelling each other.
Unlike charges attract each other.

Conductors and insulators: Those substances through which electricity can flow are called conductors. All the metals like silver, copper, aluminum etc. are conductors.
Those substances through which electricity cannot flow are called insulators. Glass, ebonite, rubber, most plastics, paper, dry wood, etc., are insulators.

Potential Difference: The amount of work done in moving unit positive charge from one point to another in an electric field is known as potential difference. Potential difference = Work done/Quantity of charge transferred
If a W joule of work has to be done to transfer Q coulombs of charge from one point to another point, then the potential difference V between the two points is given by the formula:
Potential difference, V = W/Q
The SI unit of potential difference is volt (V) .

1 volt: One volt is defined as the potential difference between two points in a current carrying conductor when 1 joule of work is done to move a charge of 1 coulomb from one point to another. Therefore, 1 volt = 1joule/ 1 coulomb

Voltmeter: The potential difference is measured by means of an instrument called voltmeter. The voltmeter is connected in parallel across the points where the potential difference is measured. A voltmeter has high resistance.

Electric Current: The electric current is the rate of flow of electric charges (called electrons) in a conductor.
If a charge of Q coulombs flows through a conductor in time t seconds, then the magnitude I of the electric current flowing through it is given by
Current, I = Q/t
The SI unit of electric current is ampere and it is denoted by the letter A. Electric current is a scalar quantity.

Ammeter: Current is measured by an instrument called ammeter. The ammeter is connected in series with the circuit in which the current is to be measured. An ammeter should have very low internal resistance.

Ohm's Law: At constant temperature, the current flowing through a conductor is directly proportional to the potential difference across its ends. If 1 is the current flowing through a conductor and V is the potential difference across its ends. Then according to Ohm’s law
V α I
V = IR
Where R is a constant called 'resistance' of the conductor. The value of this constant depends on the nature, length, area of cross-section and temperature of the conductor.

Resistance of a Conductor: The property of a conductor due to which it opposes the flow of current through it is called resistance. The resistance of a conductor is numerically equal to ratio of potential difference across its ends to the current flowing through it. i.e.
Resistance = Potential difference/Current
R = V/I
The SI unit of resistance is ohm, which is denoted by symbol Ω
R = 1 volt/1 ampere= 1 ohm
Thus, the resistance of a conductor is said to be 1 ohm if 1 ampere current flows through the conductor when a potential difference of 1 volt is applied across it.

Factors affecting the Resistance of a Conductor: The resistance of the conductor depends:
on its length,on its area of cross-section,on the nature of its material.
The resistance of a given conductor is directly proportional to its length. R ∝ l
The resistance of a given conductor is inversely proportional to its area of cross-section.
R ∝1/A
By combining the equations (i) and (ii) , R ∝l/A
R = ρ (l/A)
Where ρ is called specific resistance or resistivity of the conductor. When l = 1m, A = 1m2, we have ρ = R
Thus, the resistivity of a conductor is the resistance of unit length and unit area of cross-section of the conductor.
The SI unit of resistivity is ohm metre (Ωm) .

Combination of Resistance: The resistance can be combined in two ways:
In series
In parallel
Resistance in series:
In series, the total potential difference, V = V₁ + V₂ + V₃
Applying Ohm’s law to the entire circuit V = IR

In parallel, the total current: I = I₁ + I₂ + I₃
Applying Ohm’s law to the entire circuit I = V/R
Applying Ohm’s law to each resistance separately, we have I₁ = V/R₁; I₂ = V/R₂; I₃ = V/R₃
From equations (i) , (ii) and (iii) , we have V/R = V/R₁ + V/R₂ + V/R₃
1/R = 1/R₁ + 1/R₂ + 1/R₃

Heating Effect of Current: When an electric current is passed through a high resistance wire, it becomes very hot and produces heat. This effect is called the heating effect of current.
When an electric charge Q moves against a potential difference V, the amount of work done is given by,
W=Q x V
But, current, I = Q/t, Q = I x t
From Ohm's law: V = I x R
Now, putting all these values in equation (i) , we have Work done, W = 1² x R x t
This work done is converted into heat energy for maintaining the flow of current I through the conductor for t second.
Heat produced, H = 1² x R x t joules.( It is also known as Joule’s law of Heating)

Applications Of Heating Effect of Current:

(i) In electrical heating appliances: All electrical heating appliances are based on heating effect of current. For example, appliances, such as electric iron, water heaters and geysers, room heaters, toaster, hot plates are fitted with heating coils made of high resistance wire such as nichrome wire.

(ii) Electric filament bulb: The use of electric filament bulbs (ordinary electric bulbs) is also based on the heating effect of current. Inside the glass shell of electric bulb there is a filament. This filament is made from a very thin high bulb resistance tungsten wire. When current flows through this filament, it gets heated up. Soon, it becomes white hot and starts emitting light.

Electric Power: The rate at which work is done by an electric current is known as electric power.
Power = Work done/Time
P = W/t = (V x Q) /t
The work done by current I when it flows for time t under a potential difference V is given by:
W = V x I x t joules [Because W = VQ and Q = It] Putting
P = (V x I x t) /t = VI
P = I²R [Because V = IR]
P = V²/R [Because I = V/R]
The unit of electric power is watt. Power = V x I
1 watt = 1 volt x 1 ampere

Electrical energy = Power x Time E = P x t
The electrical energy consumed by an electrical appliance depends upon
(i) Power rating of the appliance
(ii) Time for which it (appliance) is used. The SI unit of electrical energy is joule.
1joule is the amount of electrical energy consumed when an appliance of 1 watt is used for 1 second.

Commercial Unit of Electrical Energy: Kilowatt hour is the commercial unit of electrical energy. One kilowatt hour is the electrical energy consumed when an electrical appliance having 1kW power rating is used for 1 hour.
Energy used = Power x Time 1 kWh = 1 kW x 1h
= 1000 w x 60 x 60s
=1000Js-lX3600
= 3600000 J= 3.6 x 106 J

Very Short Answers

 

 

Question : The given figure shows three resistors

Find the combined resistance. 

Answer :

Let the three resistors are R₁,R₂ and R₃. Here R₁,R₂  are parallel to each other and R₃ is in series with them then equivalent resistance can be obtained by the given formula:

 

 

 

Question : Find the minimum resistance that can be made using five resistors each of . 

Answer :

For getting minimum resistance R we can connect five resistors in parallel Combination .We know that :

R = 1/5 , n = 5

R_eq = R/n = ^1/5 /₅   = 1/25 Ω

 

 

Question : Out of the two wires P and Q shown below which one has greater resistance? Justify it. 

Answer :  We know that resistance and cross-section area are inversly proportional to each other. So less area of cross section means more resistance .

[since lengths are same]

 

So, out of two, wire Q has greater resistance.

 

Question : Nichrome is used to make the element of electric heater. Why? 

Answer :  Nichrome is used to make the element of an electric heater because nicrome is an alloy which has high resistivity and high melting point. That's why nicrome is used to make the element of heater.

 

 

Short Answers

 

Question : An electric iron of resistance 20 Ω takes a current of 5A. Calculate the heat developed in 30s. 

Answer :  R = 20 Ω; I = 5A; t = 30 s.

H = I²Rt = (5)² (20) (30)

H = 15,000 J

 

Question : A metallic wire of resistance R is cut into ten parts of equal length. Two pieces each are joined in series and then five such combinations are joined in parallel. What will be the effective resistance of the combination? (3)

Answer :

 

 

 

Question : Which among iron and mercury a better conductor? 

Answer :  Resistivity of iron is 10 × 10^-8 ohm m and that of mercury 94 × 10^-8 ohm m, therefore iron is a better conductor as compared to mercury.

 

Long Answers

Question : Figure shows a 240 V A.C mains circuit to which a number of appliances are connected and switched on. 

 

i. Calculate the power supplied to the circuit.
ii. Calculate:
a. the current through the refrigerator,
b. the energy used by the fan in 3 hours,
c. the resistance of the filament of one lamp.

Answer : 

i. Power supplied to the circuit

= 1.2 1000 W + 200 W + 60 W + 60W

= 1200 W + 200 W + 60 W + 60 W

= 1520 W

= 1.52 kW

The power supplied to the circuit is 1.52kW

ii. a. Current in the refrigerator = Power / Voltage ( P = V x I )

= 240 W/240V

= 0.83 A

The current through the refrigerator is 0.83 A

b. Energy = Power x Time

= 1.2 kW x 3h

= 1.2 x 1000 x 3 x 60 x 60s

= 1200 x 3 x 3600 J

= 12960000 J

=  1.3 x 10⁷

The energy used by the fan in 3 hrs is

c. Current, I = P/V

= 60/240 = 0.25 A

Resistance (Filament) = V/I

= 240/0.25= 960 Ω

The resistance of the filament of the bulb is 960

 

Question : A circuit diagram is given as shown below:

Calculate

i. the total effective resistance of the circuit.

ii. the total current in the circuit.

iii. the current through each resistor. 

Answer : 

 

 

 

Question : Three equal resistors each equal to r and connected as shown in Fig. Calculate the equivalent resistance. 

Answer : 

Reducing the actual circuit to an equivalent circuit i.e. we find that the there resistors, each equal to r, are just placed parallel to each.

 

Question. When electric current is passed, electrons move from:
(a) high potential to low potential.
(b) low potential to high potential.
(c) in the direction of the current.
(d) against the direction of the current.
Answer. B

Question. Electrical resistivity of any given metallic wire depends upon
(a) its thickness
(b) its shape
(c) nature of the material
(d) its length
Answer. C

Question. What is the commercial unit of electrical energy?
(a) Joules
(b) Kilojoules
(c) Kilowatt-hour
(d) Watt-hour
Answer. C

Question. The heating element of an electric iron is made up of:
(a) copper
(b) nichrome
(c) aluminium
(d) iron
Answer. B

Question. A student carries out an experiment and plots the V-I graph of three samples of nichrome wire with resistances R1 , R2 and R3 respectively. Which of the following is true?

(a) R1 = R2 = R3
(b) R1 > R2 > R3
(c) R3 > R2 > R1
(d) R2 > R3 > R1
Answer. C

Question. In an electrical circuit two resistors of 2 Ω and 4 Ω respectively are connected in series to a 6 V battery. The heat dissipated by the 4 Ω resistor in 5 s will be
(a) 5 J
(b) 10 J
(c) 20 J
(d) 30 J
Answer. C

Question. Coulomb is the SI unit of:
(a) Charge
(b) current
(c) potential difference
(d) resistance
Answer. A

Question. Work of 14 J is done to move 2 C charge between two points on a conducting wire. What is the potential difference between the two points?
(a) 28 V
(b) 14 V
(c) 7 V
(d) 3.5 V
Answer. c

Question. Calculate the current flow through the 10 Ω resistor in the following circuit.

(a) 1.2 A
(b) 0.6 A
(c) 0.2 A
(d) 2.0 A
Answer. B

Question. If R1 and R2 be the resistance of the filament of 40 W and 60 W, respectively operating 220 V, then
(a) R1 < R2
(b) R2 < R1
(c) R1 = R2
(d) R1 ≥ R2
Answer. B

Question. Two resistors connected in series give an equivalent resistance of 10 Ω. When connected in parallel, give 2.4 Ω. Then the individual resistance is
(a) each of 5 Ω
(b) 6 Ω and 4 Ω
(c) 7 Ω and 4 Ω
(d) 8 Ω and 2 Ω
Answer. B

Question. The resistance of a wire of length 300 m and cross-section area, 1.0 mm² made of material of resistivity 1.0 x 10⁻⁷ Ω is:
(a) 2 Ω
(b) 3 Ω
(c) 20 Ω
(d) 30 Ω
Answer. D

Question. Which of the given statements is true regarding ammeter and voltmeter?
(a) Ammeter is connected in series with the required device, Voltmeter in parallel
(b) Both ammeter and voltmeter are connected in series with required device
(c) The voltmeter is connected in series with the device, Ammeter in parallel
(d) They can be connected in any way
Answer. A

Question. The obstruction offered by material of conductor to the passage of electric current is known as :
(a) Resistance
(b) Conductance
(c) Inductance
(d) None of these
Answer. A

Question. The unit of potential difference is :
(a) Volt
(b) Ohm
(c) Ampere
(d) Faraday
Answer. A

Question. The instrument used for measuring electric current is :
(a) Ammeter
(b) Galvanometer
(c) Voltmeter
(d) Potentiometer
Answer. A

Question. While a cell is being charged, energy is converted into energy.
(a) mechanical, electrical
(b) electrical, chemical
(c) heat, electrical
(d) chemical, heat
Answer. B

Question. Copper is not preferred to make fuse wire because it .
(a) is a good conductor of electricity
(b) has a low melting point
(c) has a high melting point
(d) is not easily available
Answer. B

Question. Identify the correct circuit diagram:

(a) i
(b) ii
(c) iii
(d) iv
Answer. D

Question. The unit of resistivity is:
(a) V A
(b) V A
(c) V m /A
(d) VA/m
Answer. C

ASSERTION-REASON TYPE QUESTIONS
Following questions consist of two statements – Assertion (a) and Reason (R). Answer these questions selecting the appropriate option given below:
(a) Both A and R are true and R is the correct explanation of (a)
(b) Both A and R are true but R is not the correct explanation of (a)
(c) A is true but R is false.
(d) A is false but R is true.

Question. Assertion (a) : Longer wires have greater resistance and the smaller wires have lesser resistance.
Reason (R) : Resistance is inversely proportional to the length of the wire.-
Answer. C

Question. Assertion (a) : Tungsten metal is used for making filaments of incandescent lamps.
Reason (R) : The melting point of tungsten is very low.
Answer. C

Question. Assertion (a) : Alloys are commonly used in electrical heating devices, like electrical iron, toasters et(c)
Reason (R) : Alloys do not oxidise (burn) readily at high temperatures.
Answer. A

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

CASE STUDY BASED QUESTIONS

1.Electrical resistivities of some substances, at 20°C are given below in the table. Study the table and answer the given questions.

Question. Which is a better conductor of electric current ?
(a) Silver
(b) Copper
(c) Tungsten
(d) Mercury
Answer. A

Question. Which element will be used for electrical transmission lines ?
(a) Iron
(b) Copper
(c) Tungsten
(d) mercury 
Answer. B

Question. Nichrome is used in the heating elements of electric heating device because:
(a) It has high resistivity
(b) It does not oxidise readily at high temperature
(c) Both of the above
(d) None of the above 
Answer. C

Question. Series arrangement is not used for domestic circuits because:
(a) Current drawn is less
(b) Current drawn is more
(c) Neither of the above
(d) Both of the above
Answer. A
 

2. In the given circuit, three identical bulbs B1, B2 and B3 are connected in parallel with a battery of 4.5 V. Study the diagram and answer the questions given below:

Question. What will happen to the other two bulbs if the bulb B3 gets fused ?
(a) They will also stop glowing.
(b) Other bulbs will glow with same brightness.
(c) They will glow with low brightness.
(d) They glow with more brightness.
Answer. B

Question. If the wattage of each bulb is 1.5 W, how much readings will the ammeter A show when all the three bulbs glow simultaneously?
(a) 1.1 A
(b) 2.1 A
(c) 1.5 A
(d) None of the above
Answer. A

Question. Find the total resistance of the circuit.
(a) 1.0 Ω
(b) 4.1 Ω
(c) 1.5 Ω
(d) 2.0 Ω
Answer. B

Question. How many resistors of 88 Ω are connected in parallel to carry 10 A current on a 220 V line ?
(a) 2 resistors
(b) 1 resistors
(c) 3 resistors
(d) 4 resistors
Answer. D

(a): Which among iron and mercury is a better conductor?

Answer- Iron.

 

(b): Which material is the best conductor?

Answer- Silver.

 

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

Answer-

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

Answer-

R₁ = 5 Ω, R₂ = 8 Ω, R₃ = 12 Ω

As resistors are combines in series, So total resultant resistance is:

R = R₁ + R₂ + R₃ = 5 + 8 + 12 = 25 Ω

Now total current or reading of ammeter = I =  V/R =6/25 = 0.24 A

Now for 12 Ω Resistor, Reading of voltmeter = V = IR = 0.24 x 12 = 2.88 V.

Now equivalent resistance (R) for the circuit is:

Question : Four different measuring instruments are shown below. Out of these, the instrument that can be used for measuring current is/are the instruments labelled as 

 

 

Question : Why closed path is required for the flow of current? 

Answer : A closed path is required for the flow of current so that charges can move in a particular direction in a given circuit. But if path is not closed means circuit is open, then there is air between the gap, and we know that air is an insulator for flow of charges, so the current stop flowing in the circuit.

 

 

Question : What is resistance of dry air? 

Answer :  The resistance of dry air is infinity.

 

 

Question :`Draw a circuit diagram using a battery of two cells, two resistors of 3 each connected in series, a plug key and a rheostat. 

Answer :   When we connect rheostat, key, battery of two cells and two resistors of three ohm each then required circuit diagram will be:

 

 

Short Answers

 

Answer : Combined resistnace of 100Ω, 50Ω and 500Ω in parallel i.e. R_p is given by

Answer :  

 Resistors 1Ω and 2 Ω are in parallel.

 

 

Long Answers

Question : Two metallic wires A and B are connected in series. 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 the same material? 

Answer :  

 

Question : Redraw the circuit putting an ammeter to measure the current through the resistors and voltmeter to measure the potential difference across 12 Ω resistor. what would be the reading in the ammeter and the voltmeter? 

Answer :  

Modified circuit is as shown. Since 5Ω, 8 Ω and 12 Ω are in series, therefore the total resistance in series.

Rs = R₁ + R₂ + R₃ = 5 + 8 + 12 = 25Ω

 

Current through circuit I = V/R = 6/25 = 0.24A

Reading of ammeter = 0.24A

= I. R.= 0.24 x 12

= 2.88Ω

 

 

 

Question : Three resistors are connected as shown in Fig. Through a resistor of 5 ohms; a current of 1A is flowing. 

i. what is the current through the other two resistors?

ii. what is the potential difference (P.D.) across AB and AC?

iii. what is the total resistance?

 

Answer :  Let us first find the total resistance. Now resistance of 10 Ω and 15 Ω are in parallel.

If R_p is the effective resistance between B and C, then

 

Again AB and BC are in series.

Therefore, the total resistance = Resistance between A and B plus resistance between B and C.

i.e. Total resistance = 5 Ω + 6 Ω = 11 Ω

Potential difference between A and B = IR = 1 x 5 = 5 V

Potential difference between B and C = 1 x 6 = 6 V

Potential difference between A and C = 1 x 11 = 11 V

Current through A and B = 1 A

This current divides into two parts, one part I1 passing through 10 Ω and other part I₂

passing through 15 Ω each producing a P.D. of 6V (between B and C)

I₁ (10) = 6 or I₁ = 6/10 = 0.6 A and I₂(15) = 6 or I₂ = 6/15 = 0.4 A