CBSE Class 10 Physics Electricity Assignment Set B

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 : What is the SI unit of electric charge? How many electrons make one coulomb of charge?

Question : Define the SI unit of current.

Question : Which instrument is used to measure the current flowing in a circuit? How is it connected in the circuit and why?

Question : In which smaller units can we measure the small amount of current flowing in the circuit? How are they related to the SI unit of current?

Question : What makes the electric charge to flow?

Question : State the energy conversion taking place in

(a) Electric cell (b) Electric torch

Question : Define the electric potential difference between two points in an electric circuit carrying some current.

Question : Define the SI unit of Electric potential.

Question : Which instrument is used to measure the electric potential difference between two points in a circuit? How is it connected in the circuit and why?

Question : State the law relating the potential difference across a conductor and the current through it?