# Electricity : Complete Set of Question

This set of questions contains all the possible questions
which could be asked in the examination

Electric Current and Circuit

Q.1 What is electric charge?

Electric charge is the physical property of matter that causes it to experience a force when placed in an electromagnetic field. There are two types of electric charges: positive and negative.

Q.2 Describe the properties of electric charges?

Properties : -

(a) Opposite charges (or unlike charges) attract each other.

(b) Similar charge (or like charges) repel each other.

Q.3 What do you mean by static electricity and current electricity?

Static electricity is an imbalance of electric charges within or on the surface of a material. The charge remains until it is able to move away by means of an electric current or electrical discharge.

Current electricity is the set of physical phenomena associated with the presence and flow of electric charge.

Q.4 What do you mean by electric current?

The electric current is a flow of electric charges (called electrons or protons) in a conductor such as a metal wire.

Q.5 What is the S.I. unit of electric charge and current?

S.I. unit of

Charge – Coulomb

Current – Ampere

Q.6 State  the basic differences between conductor and insulator. Q.7 Which particles constitute the electric current in a metallic conductor?

Electrons are the negative charge particles constitute the electric current in metallic conductor.

Q.8 Calculate the number of electrons constituting one coulomb of charge? (Charge on one electron = 1.6 × 10-19C)

We know that charge over 1 electron = 1.6 × 10–19 coulomb

Thus, 1.6 × 10–19 C of charge = 1 electron

Therefore, 1 C of charge = 1/ 1.6 ×10−19 Electrons

= 1019 / 1.6 electrons = 10 × 1018 / 1.6 electrons = 6.25 × 1018 electrons

Q.9  Name the instrument / device that is used to measure electric current in circuit?

Ammeter is used to measure electric current in circuit.

Q.10 How is an ammeter connected in a circuit to measure current flowing through it?

An ammeter is connected in series in a circuit to measure current flowing through it.

Q.11 In an electric circuit, state the relationship between the direction of conventional current and the direction of flow of electrons?

The direction of conventional current is opposite to the direction of flow of electrons.

Q.12 What does an electric circuit mean? Distinguish between an open and a closed circuit?

A continuous and closed path of electric current is called electric circuit.

In an open circuit, the switch key is open. While in a closed circuit, the switch key is closed.

Q.13 Define electric current with its S.I. unit. Write its formula:

Electric Current is the amount of charge flowing through a particular area in unit time.

SI unit of current I = Ampere

Formula = I = Q / t = Coulomb/time

Q.14 The amount of charge passing through a cell in four seconds in 12 C. Find the current supplied by cell.

Given,

Charge Q = 12 C

Time t = 4 s

Current = I = Q / t = 12/ 4 = 3 A

Q.15 Calculate the number of electrons that could flow per second through the cross section of a wire when 4 A current flows in it.

Given,

I = 4 A, t = 1 s

I = Q / t

Then charge, Q = I × t = 4 A × 1 s = 4 C

1 C of charge = 1/ 1.6×10−19 Electrons

Therefore, Number of electrons in 4 C of charge = 4 C × 1/ 1.6 × 1019 = 2.500 × 1019

Q.16 A current of 10 A flows through a conductor for two minutes :

(a) Calculate the amount of charge passed through unit area of cross section of the conductor.

(b) If the charge of an electron is 1.6 × 10-19 C then calculate the total number of electrons flowing.

Given,

Current I = 10 A

Time = 2 minutes = 2 × 60 = 120 sec

(a) Then charge Q = I × t = 10 × 120 = 1200 C

(b) Number of electrons = 1200 C x 1/ 1.6×1019  = 7.5 x 1017

Q.17    What is the SI unit of charge? Is it a scalar or vector quantity?

SI unit of charge is coulomb (C) = It is scalar quantity

Q.18 Define 1 C or one coulomb?

1 C charge is the charge which when placed at a distance of 1 m from an equal like charge in vacuum, experiences a repulsive force on 1 N.

Q.19 Define 1 A or one Ampere?

When 1 coulomb of charge flows through any cross-section of a conductor in 1 sec., the electric current flowing through it is said to be 1 ampere i.e.

1 ampere = 1 coulomb/ 1 second

Or 1 A = 1 C/1 s

Q.20 Write down the formula which relates electric charges,time and electric current ?

The equation below shows the relationship between charge, current and time:

charge (coulomb, C) = current (ampere, A) × time (second, s)

Q = I × t

Q.21 By what name is the physical quantity coulombs / second called ?

Electric Current = Charge/Time = Coulombs/Second

The physical quantity coulombs / second  is called Electric current.

Q.22 What is a circuit diagram? Draw the labeled diagram of an electric circuit comprising of a cell, a resistor, an ammeter, a voltmeter and a switch (or plug key).

A circuit diagram is a diagram which indicates how different components in circuit have been connected by using the electrical symbols for every components, is called a circuit diagram. Q.23    What are the fundamental laws of Electrostatics?

Fundamental laws of electrostatics –

(a) There are two types of charges – one is positive and other is negative.

(b) Like charge repel each other and unlike charges attract each other.

Q.24 Is Ampere scalar or vector?

Ampere is scalar quantity.

Q.25 Name a device that helps to maintain a potential difference across a conductor.

Cell, battery, power supply etc. are the devices that help to maintain a potential difference across a conductor.

Q.26 A current of 0.75 A is drawn by a filament of an electric bulb for 5 minutes. Find the amount of electric charge that flows through the circuit.

Given,

Current I = 0.75 A

Time t = 5 min = 5 × 60 s = 300 sec.

Since, Q = I × t = 0.75 × 300 = 225 C.

Q.27 Why does an ammeter burn out when connected in parallel in an electric circuit?

An ammeter is a low Resistance device. So, it should always be connected in series to a circuit for measuring current effectively.

However, if it is connected in parallel, the resistance of the circuit reduces considerably and a large amount of current flows through the circuit which may lead to the burning of the ammeter.

Q.28 Name the instrument used to change the resistance of the circuit?

Rheostat is the instrument used to change the resistance of the circuit.

Electric Potential and Potential Different

Q.29 Name the device / instrument used to measure potential difference. How is it connected in an electric circuit?

Voltmeter is used to measure potential difference; it is connected in parallel in an electric circuit.

Q.30 What is the resistance of an ideal voltmeter?

The resistance of an ideal voltmeter is very high resistance, its value is infinity.

Q.31 Define the S.I. unit of potential difference?

Volt is the S.I. unit of potential difference.

Q.32 What is meant by saying that the potential difference between two points is 1 V?

Potential difference between two points is said to be 1 V when 1 joule of work is done to move a charge of 1 coulomb from one point to other point.

Q.33 Define electric potential?

It is the work done in carrying a unit positive charge from infinity to that point against any electric field.

It is denoted by symbols V. Its SI unit is volt.

i.e. V = W / q = Joules/Coulomb

Q.34 Define electric potential difference?

When a unit charge (= 1 C) is carried from one point to another point, the work done is called potential difference.

Example : Potential difference between A and B is : -

VAB = W (Work done to carry charge q from A to B) /q (charge in coulombs)

1 Volt = 1 Joule/1 C

Q.35 A charge of 2 C moves between two plates maintained at a potential difference of 1 V. What is the work done?

Given,

Potential difference (V) = 1 V

Charge (q) = 2 C

W = q.V = 2 × 1 = 2 J

Q.36 How much work is done in moving a charge of 5 C across two points having a potential difference 20 V?

Given,

Charge q = 5 C, potential difference V = 20v

Then, the work done in moving W = qV = 5 × 20 = 100 J

Q.37 How does current flow in a wire?

Electric current is the flow of electrons in a metal wire when a cell or battery is applied across its ends. A metal has a plenty of free electrons in it.

(i) When the metal wire is not connected to a source of electricity like a cell or a battery, the electrons present in it move at random in all directions between the atoms of the metal wire.

(ii) When a source of electricity like cell is connected between the ends of the metal wire. The electric force acts on the electrons present in wire. Since electrons are negatively charged, they start moving from negative end to the positive end of the wire. This flow of electrons constitutes the electric current in the wire.

Q.38 Define ‘1 volt’. State the relation between work, charge and potential different for an electric circuit. Calculate the potential difference between two terminals of the battery, if 100 joules of work is required to transfer 20 coulombs of charge from one terminal of the battery to the other.

Potential difference between two points is 1 V when 1 Joule of work is done to move a charge of 1 coulomb from one point to the other.

Given,

Work W = 100 Joule,

Charge (Q) = 20 C

Therefore, potential difference (V) = W /Q  = 100/20 = 5V

Q.39 List in a tabular form two differences between a voltmeter and a ammeter ? Q.40 A large number of electrons are present in metals, yet no current flows in the absence of electric potential across it. Explain the statement with reason.

In a metal, large number of free electrons is found. These electrons are in constant random motion. Thus, current generated in one direction is cancelled out by the current generated in the opposite direction by these randomly moving electrons. Their resultant motion is zero therefore no current flows in the conductor.

But, under applied electric field, all electrons move in one direction resulting in net current. The electrons move only if difference of electric pressure-called the potential difference along the conductor. This potential difference sets the charges in motion in the conductor and produces an electric current.

Q.41 Find the number of electrons transferred between two points kept at a potential difference of 20 V. If 40 J of work is done.

Given, potential difference = 20 V, Work = 40 J.

We know,

Work (W) = Q × V

Q = 40J/20V = 2C

Then, number of electrons = 2C × 1/ 1.6 × 1019  = 1.25 × 1019 electrons

Ohm’s Law

Factors on which resistance depends

Q.42 State the law which relates the current in a conductor to the potential difference across its ends.

Ohm’s law state : - At constant temperature, the current flowing through a conductor is directly proportional to the potential difference across its ends.

Or, V α I

$\Rightarrow {V \over I} = R..........\left( i \right)$

$\Rightarrow {1 \over I} = {R \over V}$

$\Rightarrow I = {V \over R}..........\left( i \right)$

$\Rightarrow V = RI.........\left( {ii} \right)$

Where R is constant for the given conductor at a given temperature and called resistance. Resistance is the property of conductor which resists the flow of electric current through it.

Q.43 What is Ohm’s law? Explain how it is used to define the unit of resistance?

Ohm’s law establishes a relation between potential difference (V) and current (I).

It states that, at same physical condition, (i.e. constant temperature and pressure), the current flowing through a conductor is directly proportional to the potential difference across it.

Mathematically,

V α I

I = Constant = R

V = IR

Where, R is a constant called “resistance” of the conductor. The value of constant R depends on nature, length, area of cross-section and temperature of the conductor.

V / I  = R

V = Potential difference

I = Current

R = Resistance

SI unit of resistivity (ρ) is Ω m

Ohm = Volt/ampere

Q.44 Name the physical quantity whose unit is Ohm.

Resistance of substance.

Q.45 The potential difference across the terminal of a cell is 1.5 volt. It is connected with a resistance of 30 Ohm. Calculate the current flowing through the circuit.

According to Ohm’s law, V = IR

Given, V = 1.5 volt,    1.5 = I × 30

R = 30 Ohm

I = 1.5/30 = 0.05 A

Q.46 A bulb cannot be used in place of a resistor to verify ohm’s law. Justify this statement with reason?

A bulb cannot be used in place of a resistor to verify ohm’s law. It is because Ohm’s law holds at constant temperature only.

Q.47 What is meant by electric resistance of a conductor?

Resistance is the property of a conductor to resist of the flow of current through it.

Q.48 Two students perform experiments on two given resistors R1 and R2 and plot the given V – I Graph. If R1 > R2. Which of the two diagrams correctly represents the situation on the plotted curves? Justify your answer : - Diagram (1) represents the situation correctly because slope of the V.I. graph gives the value of resistance R and slope of R1 is greater than R2 in diagram 1, so diagram 1 is the correct representation of the given situation.

Q.49 State the physical quantity which is equal to the ratio of potential difference and current?

Resistance is the physical quantity which is equal to the ratio of potential difference and current.

Q.50 How much current will an electric bulb of resistance 1100 Ω draw from 220 V source ? If a heater resistance of 100 Ω is connected to the same source instead of the bulb. Calculate the current drawn by the heater resistance.

Given,

Voltage V = 220 V

Resistance R =1100 Ω

Current for electric bulb (I) = V / R = 220/1100

= 0.2A

Resistance of heater =100 Ω

Current for heater = 220/100= 2.2A

Q.51 Explain why a conductor offers resistance to the flow of current?

Motion of electrons in an electric circuit constitutes an electric current and these electrons are not completely free to move within the conductor. They are restrained by the attraction of the atoms among which they move. that why a conductor offers resistance to the flow of current.

Q.52 Calculate the resistance of a resistor if the current flowing through it is 200 mA. When the applied potential difference is 0.8 V.

Given,

Current (I) = 200 mA = 200/1000 = 0.2A

Potential difference (V) = 0.8 V

V = IR

0.8/0.2 = R

R = 4 Ω

Q.53 Describe an activity to find relationship between the potential difference (V) across two ends of a conductor and the current (I) flowing through it. Include in your answer:

(a) The diagram of the electric circuit

(b) A V- I graph

Activity : -

(1) Set up an electric circuit which is consisting of a nichrome wire XY of length 0.5m, an ammeter, a voltmeter and four cells of 1.5 V each as in fig. (2) First we connect on cell in circuit and note down the reading of ammeter A for current and voltmeter V for potential differences across the nichrome wire XY.

(3) Now, connect two, three and four cells in the circuit one by one and note down the reading of ammeter and voltmeter in each case.

(4) Plot a graph b/w I and V observation: V – I graph is a straight line passing through the origin. Result : - V is directly proportional to I.

Q.54 Differentiate between conductor, resistor and insulator?

Conductor : The substances that are capable of readily transmitting heat, electricity etc. is called a conductor.

Resistor : A conductor having some appreciable resistance is called a resistor.

Insulator : Those substances there are not capable of readily transmitting heat, electricity etc. are called an insulator.

Q.55 Potential difference between two points A and B of a conductor is 15 V. Current flowing through the conductor AB is 3A. Calculate the work done when the current flow for 2 sec. ?

Given,

I = 3A, t = 2 s, v = 15 V

then, Q = I × t = 3 A × 2 s = 6 C.

W = VQ = 15 v × 6 C = 90 Joule.

Q.56 Keeping the potential difference constant, the resistance of a circuit is doubled. By how much does the current change?

According to Ohm’s law: -

V = IR

So, I = V/R……………..(1)

Now, by the given condition

V = I' x 2 R

I'  =  V /2 R

I' = I/2………..(from(1))

So, the current becomes half.

Q.57 Define 1 Ohm ?

1 Ohm is the resistance of a conductor such that when a potential difference of 1 volt is applied to its ends, a current of 1 ampere flows through it.

Q.58 What is electrical resistivity? Also write S.I. unit of resistivity?

Electrical resistivity (also known as resistivity, specific electrical resistance) quantities how strongly a given material opposes the flow of electric current. It is the resistance of a unit length with unit area of cross section of the material of the conductor. A low resistivity indicates a material that readily allows the movement of electric charge. Resistivity is commonly represented by the Greek letter ρ (rho).The SI unit of electrical resistivity is ohm-metre(Ω⋅m)

Q.59 A wire of resistivity ρ is pulled to double its length. What will be its new resistivity?

Resistivity remains the same.

Q.60 (a) Resistivity of iron is 10 × 10-8 Ω m and of mercury is 94 × 10-8 Ω m. Which among iron and mercury is a better conductor? Give reason.

(b) Resistivity of a material A is in the range of 1010 – 1014 Ω m while that of B is in the range of 10-6 – 10-8 Ω m. Which of the two will behave as an insulator and why?

(a) Iron is a better conductor because it has lower resistivity.

(b) Material A, because insulator have higher resistivity.

Q.61 A wire of length L and resistance R is stretched so that the length is doubled and area of cross-section is halved. How will it : -

(a) Resistance change

(b) Resistivity change

Here, length is doubled and area of cross- section is halved. Thus, a wire of length l and area of cross –section A becomes of 2 l and area of cross section A/2

Since R = ρ l / A

New resistance, R' = ρ 2l/A/2

R' = 4 ρl/A= 4 x R

i.e., resistance becomes four times.

(ii) Resistivity of a substance does not depend on its length or area of cross- section. It depends on the nature of the substance and temperature. Hence, there is no change.

Q.62 Calculate the resistance of a 1 km long aluminium wire of diameter 5 mm. Find resistance for the wire made of nichrome of same length and diameter also.

Resistivity of Al = 2.8 x10-8 Ω m , Diameter of the wire= 5 mm=5x 10-3

Cross section area of wire (A) = πD2/4 =3.14 x 25 x10-6 /4 = 19.63 x 10-6 m2 .

Resistivity of nichrome = 100 x 10-6 Ω m

Resistance = resistivity x length/cross sectional area of the wire

R = ρ· l / A

Resistance of Al = ρ· l / A

= 2.8 × 10–8x 1000m/19.63× 10 - 6m2

= 1.43 Ω

Resistance of nichrome wire = 100 x10-6 × 1000m/19.63 × 10- 6m2

= 5094.24 Ω

Q.63 (a) A fuse wire melts at 5 A. If it is desired that the fuse wire of same material melts at 10 A then whether the new fuse wire should be of smaller or larger radius than the earlier one? Give reason for your answer.

(b) If the radius of a current carrying conductor is halved, how does current through it change?

(a) The new fuse wire should have larger radius than the previous one. This is because since area of the conducting wire is directly proportional to current and inversely proportional to resistance , a larger radius also means a smaller resistance.

V = IR …..(i)

According to Ohm’s law,

R = V/I ……(ii)

Also from Eq. (i) we get

I = V/R

On applying Ohm’s law [Equation (i) – (iii)], we observe that the resistance of the conductor depends (a) on its length, (b) on its area of cross-section, and (c) on the nature of its material. Precise measurements have shown that resistance of a uniform metallic conductor is directly proportional to its length (l) and inversely proportional to the area of cross-section (A). That is,

R α l……..(iv)

and R α 1 / A……(v)

Combining Eqs. (iv) and (v) we get

R α l / A

Or R = ρ(l / A) …..(vi)

(b) R α 1 / A

If the radius of the conductor is halved, its resistance will increase. If resistance is increased, current will decrease .

Resistance of a system of resistors

Q.64 Find the ratio of equivalent resistance of series combination of n equal resistances to the equivalent resistance in parallel combination of n resistances.

Total resistance of n equal resistor each of resistance R in series is : - R = R1 + R2 + R3 + …………+ Rn = nR

Total resistance of n equal resistor each of resistance R in parallel is : - ${1 \over R} = {1 \over {{R_1}}} + {1 \over {{R_2}}} + {1 \over {{R_3}}} + ..... + {1 \over {{R_n}}}$

R ρ = Rs / Rp(Rs- Total resistance of n equal resistor each of resistance R in series and Rp- Total resistance of n equal resistor each of resistance R in parallel)

Now ratio = nR/ 1/Rn = n2R / R = n2

Q.65 What are the (a) highest (b) lowest resistances that can be secured by combining four coils of resistance 4 Ω, 8 Ω, 12 Ω and 24 Ω?

(a) Highest resistance can be obtained by connecting the four coils in series.

Then, R = 4 + 8 + 12 + 24 = 48 Ω

(b) Lowest resistance can be obtained by connecting the four coils in parallel

${1 \over R} = {1 \over 4} + {1 \over 8} + {1 \over {12}} + {1 \over {24}} = {{12} \over {24}} = {1 \over 2},R = 2\Omega$

Q.66 A number of bulbs are to be connected to a single source: When will they provide more illumination when connected in parallel or in series? Give reason your answer and also list two advantages of these types of arrangement.

Bulbs will provide more illumination when the resistance of the circuit is minimum. So, to get the minimum resistance, the resistors are connected in parallel combination.

Advantages of parallel combination : -

(1) Each devices gets different current to operate properly.

(2) Even if a device is broken then the other component will continue to work properly.

Q.67 Three resistor of resistances R1, R2 and and R3 are connected in parallel to a source of potential difference V. Draw the schematic circuit diagram. Find the equivalent resistance of circuit. In parallel combination, the total current I is equal to the sum of the currents through each branch :

I = I1 + I2+ I3

I = V /R1 + V /R2 + V /R3

I / V = 1 /R1 + 1 /R2 + 1 /R3

R ρ be the equivalent resistance of the parallel combination of resistors.

1/ R ρ = 1 /R1 + 1 /R2 + 1 /R3

R ρ = V / I

Q.68 Apply the Ohm’s law to obtain the relation for the combined resistance when three resistors R1, R2 and R3 are connected in series. List two disadvantages of connecting household appliances in series : - Let the potential difference across the resistors R1, R2 and R3 are V1, V2 and V3 respectively. Let I be the current through the circuit. In this case, the current through each resistor is also I and the potential difference across a combination of resistors is equal to the sum of potential differences across the individual resistors i.e.

V = V1 + V2 + V3

Applying the Ohm’s law to the entire circuit and three resistors separately,

V = IR

Where R = equivalent single resistor

V = potential difference across equivalent resistor.

V1 = IR1

V2 = IR2

V3 = IR3

From equation (1)

IR = IR1 + IR2 + IR3

R = R1 + R2 + R3

(a) One disadvantage of a series circuit is that if a bulb were broken or the pathway broken in any way, the other bulbs would go out too.

(b) If you connect them all in series, then none of them gets enough voltage to run properly

Q.69 You are supplied with a number of 100 Ω resistors. How could you combine some of these resistors to make a 250 Ω Resistor?

If two resistors of resistance 100 Ω are connected in parallel then total resistance becomes 50 Ω.

Now if two more 100 Ω resistors are connected in series. We will get total resistance of 250 Ω.

Q.70 You are given one hundred 1 Ω resistors. What is the smallest and largest resistance you can make in a circuit using these?

When all the one hundred 1 Ω resistors are connected in series then we will get largest resistance circuit of 100 Ω and if we connect these resistances in parallel then we will get smallest resistance circuit of 0.01 Ω.

Q.71 If two resistors of resistance 10 Ω and 20 Ω are connected in series with 60 V battery. Calculate the amount of current in resistor 20 Ω?

Two resistors of resistance 10 Ω and 20 Ω are connected in series.

Then total resistance of the circuit = 30 Ω.

Potential difference V = 60 V

Then amount of current I= V/R

I= 60/30 = 2 A

In series combination, current is equally flow in all the resistors.

So amount of current in 20 Ω resistor is 2 A.

Q.72 Why is a series arrangement not used for connecting domestic electrical appliances in a circuit?

A series arrangement is not used for connecting domestic electrical appliances in a circuit because different appliances have different resistances and they require different amount of current, if we supply equal amount of current in different appliances connected in series combination, then one of the appliances get damaged.

Q.73 Give three reasons why different electrical appliances in a domestic circuit are connected in parallel?

Three reasons why different electrical appliances in a domestic circuit are connected in parallel:

• If one device is damaged. It does not effect other electrical devices.

• It can provide unequal amount of current to different devices according to their requirement.

• Parallel circuit has law resistance.

Heating Effect of Current and Electric Power

Q.74 Give reason for the following: -

(i) Tungsten used almost exclusively for filament of electric lamp.

(ii) Copper and aluminum wires are usually employed for electricity transmission.

(i) Tungsten is used in making the filament of an electric bulb because

(a) Tungsten has high melting point

(b) Tungsten has high resistivity to retain much heat.

(ii) Copper and aluminum have low resistivity and they are good conductors of electricity.

Q.75 State a difference between the wire used in the element of an electric heater and in a fuse wire?

The wire used in the element of an electric heater has very high resistance. While a fuse wire has a low resistance.

Q.76 What is meant by the statement that the rating of a fuse in a circuit is 5A?

It means that the fuse wire melts when more than 5A current flows through it.

Q.77 State the factors on which the heat produced in a current carrying conductor depends. Give one practical application of this effect.

Heat produced in a current carrying wire depends upon : -

(a) Square of the current (I2)

(b) Resistance of the given conductor (R)

(c) Time of which the current flows (t)

I = I2Rt

This effect is used in an electric iron.

Q.78 Derive an expression for electric energy consumed in a device in terms of V, I and t, where V is potential difference applied to it. I is the current drawn by it and t is the time for which the current flows? Consider a current I flowing through a resistor of resistance R. Let the potential across the resistor is V. When charge Q moves against the potential difference V in time t, the amount of work is : -

W = Q × V……(1) where W= work done, Q= flow of charge and V=Potential difference

We also know,

Current(I)= Q/t where Q=flow of charge and t = time.

Therefore,Q = It……(2)

Putting value of Q from equation(2) to (1)

Work done or energy consumed = I x Vx t

Work done by battery for moving the charge in electric circuit (W) = VIt

Battery has chemical energy due to which it can do work. As battery does work, its chemical energy decreases. But total energy is always conserved. Chemical energy of battery is converted to heat energy in the resistor.

Therefore, heat produced in resistor (H) = VIt

By using Ohm’s law : -

H = (IR) It = I2Rt

This is known as Joule’s Law of Heating.

Q.79 State Joule’s law of heating? Mention two practical applications of heating effect of electric current?

Joule’s law of heating – Heat produced in a resistor is –

H = I2Rt

(i) Directly proportional to the square of current for given resistance.

(ii) Direct proportional to the time for which the current flow and through the conductor.

(iii) Directly proportional to the resistance for a given conductor.

Practical application: (i) To light an incandescent electric bulb.

(ii) Electric iron

(iii) Electric fused is used to protect the circuits.

Q.80 Calculate the amount of heat generated when 7200 coulombs of charge is transferred in an hour through a potential difference of 50 V.

Given charge Q = 7200 C, t = 1 h = 60 × 60 s, V = 50 V

I = Q / t = 7200 C/60 x 60 s = 2 A

R = V / I  = 50 V/2 A = 25 Ω

H = I2Rt

= 2 × 2 × 25 × 60 × 60

= 36 × 104 J or 360 K J

Q.81 (a) Write the expression for heat energy produced in time t in a wire of resistance R when a current I flows through it. Write the SI unit of this energy.

(b) Name the physical quantity whose SI unit is J/s. Write an expression for heat energy produced in a device of power P in time t.

(a) Expression for heat energy – H = I2Rt

S.I. unit of this energy – Joule (J)

(b) Power has S.I. unit J/s.

Expression for heat energy produced in a device of power P in time t.

H = Pt

Q.82 An electric iron is marked 1000 W – 220 V. Find the current drawn by iron and the resistance of the element of iron.

P = V2/ R or R = V2/ P

= 220 V x 220 V/1000 W = 48.4 Ω

Now from Ohm’s law : -

V = IR or I = V / R = 220/42.4 = 4.55 A

Q.83 Give three practical applications of heating effect of current.

For heating effect, the element of appliances must have high melting point to retain more heat.

Application : -

(a) Electric bulb : When electric energy is supplied to an electric bulb, the filament gets heated because of which it gives light.

(b) Electric iron : When an electric iron is connected to an electric circuit, the element of electric iron gets heated, which heats the electric iron.

(c) Electric fuse : Electric fuse is used to protect the electric appliances from high voltage. Electric fuse is made of metal or alloy of metals, such as aluminium etc. In the case of flow of higher voltage the specified, fuse wire melts and protect the electric devices.

Q.84 What is electric power ? Write the expression and S.I. unit for it ?

The rate of which electric work is consumed is called electric power.

We know, Heat produced = VIt

= V2/Rt = I2xRt

Power P = Heat produced per unit time = Heat produced/t

\$ = I 2 Rt/ t  =  I2 R = VI

SI unit of electric power is watt (W).

Q.85 Complete the heat generate while transferring 96000 coulombs of charge in one through a potential difference of 50 V between two ends of a conductor also calculate the power input by source.

Given Q = 96000 C, t = 1 hour = 3600 s

V = 50 V

I = Q / t = 96000 C/3600 = 26.67 A

R = V / I = 50 V/26.67 A = 1.87 Ω

Power P = VI = 50 V × 26.67 A = 1333.5 W

Q.86 Two lamps, one is rated 100 W at 200 V, and the other 60 W at 220 V, are connected in parallel to a 220 V supply. Find the current drawn from the supply line.

Total power of 2 bulbs in parallel (W)

= 100 + 60 = 160 W.

Supply voltage = 220 V

∴ Current drawn from line (I) =P/V

= 160 W/220 V

= 0.727 A

Q.87 (a) State the commercial unit of electric energy and find its relation with its S.I. unit.

(b) The current through a resistor is made three times its initial value. Calculate how it will affect the heat produced in the resistor.

(c) Find the increase in the amount of heat generated in a conductor if another conductor of double resistance is connected in the circuit keeping all other factors unchanged.

(a) The commercial unit of electric energy is kilowatt hour kWh

SI unit of electrical energy is Joule (J)

1 kWh = 1000 watts × 60 min

= 1000 × 60 × 60 sec

= 3.6 × 106 watts seconds

= 3.6 × 106 Joules

(b) Initial production of Heat

H = I2Rt

With new resistance, heat produced

= I2 × 3 R × t

= 3I2Rt

So, the heat produced will be three times.

(c) Initial heat, H = I2Rt

New resistance = R + 2 R = 3 R

With new resistance, heat produced

= I2 × 3 R × t

= 3I2Rt

Therefore, the heat produced will be become thrice.

Q.88 Name the unit used in selling electric energy to consumers. An electric motor takes 5A from a 220 V line. Determine the power of the motor and energy consumed in 2 hrs. Calculate the cost of electric energy for the month of September at a rate of Rs. 4 / unit.

Kilowatt hour (kWh) unit is used in selling

Electric energy to consumers : -

Then power P = VI = 220 V × 5A = 1100 W

Energy consumed by motor = P × t

= 1100 W × 2 h = 2.2 kWh

Now, energy consumed = 2.2 kWh, Rate = Rs. 4/unit, No. of days = 30

Total Cost = 2.2 × 4 × 30 = Rs. 264.00

Q.89 Define Rating of devices with the example ?

Rating of device is the value of electric power and potential of that electric device. If we know the rating of any electrical device we can easily find the value of resistance.

For Example, an electric bulb is rated 220 V and 100 W, the resistance.

P = V2/ R

R = V2/ P = (220)2 /100  = 484

Q.90 What is heating effect of current ?

When current flows through a conductor, heat is developed and temperature of wire increases. This is known as heating effect of current.

Ex. (i) Electric iron / toaster

(ii) Electric Oven

Value Based Questions

Q.1 The electrical resistivity of five substances A, B, C, D and E are given below :

A : 5.20 × 10-8 Ω m

B : 110 × 10-8 Ω m

C : 2.60 × 10-8 Ω m

D : 10.00 × 10-8 Ω m

E : 1.70 × 10-8 Ω m

(a) Which substance is the best conductor of electricity? Why?

(b) Which one is a better conductor: A or C? Why?

(c) Which substance would you advise to be used for making heating elements of electric irons? Why?

(d) Which two substances should be used for making electric wires? Why?

(a) E substance is best conductor of electricity. It has least electrical resistivity.

(b) C is a better conductor than A. It has lesser electrical resistivity.

(c) B substance is used for making heating elements of electric irons. It has high electrical resistivity.

(d) C and E substances should be used for making electric wires. They have low electrical resistivity.

Q.2 In the circuit given below connect a nichrome wire of length ‘L’ between point X and Y and note the ammeter reading : - (i) When this experiment is repeated by inserting another nichrome wire of the same.

Thickness but twice the length (2 L), what changes are observed in the ammeter reading?

(ii) State the changes that are observed in the ammeter reading if we double the area of cross-section without changing the length in the above experiment.

(i) The ammeter reading will decrease (become half).

When increase in length, resistance of the circuit increases, hence current become half.

(ii) The ammeter reading will increase (become two times)

This is because, as area increases resistance decreases and current become two times.

Q.3 The electrical resistivity of three materials P, Q are given below : -

P => 2.3 × 103 Ω m

Q => 2.63 × 10-8 Ω m

R => 1.6 × 1015 Ω m

Which material will you use for making (a) electric wires  (b) handle for soldering iron and (c) solar cells. Give reason for your choice.

(a) Q material as will use for making electric wire because it has very low resistivity.

(b) R material we will use for making handle for soldering iron because it has very high resistivity.

(c) P material we will use for making solar cells – because it is semi conductor.

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