*Electricity*

Electricity: Electric current, electric circuit, voltage or electric potential, resistance and (Ohm’s law).

Electric Current: The flow of electric charge is known as Electric Current, Electric current is carried by moving electrons through a conductor.

By convention, electric current flows in the opposite direction to the movement of electrons.

By convention, electric current flows in the opposite direction to the movement of electrons.

Electric Circuit: Electric circuit is a continuous and closed path of electric current.

Expression of Electric Current: Electric current is denoted by the letter ‘I’. Electric current is expressed by the rate of flow of electric charges. Rate of flow means, the amount of charge flowing through a particular area in unit time.

If a net electric charge (Q) flows through a cross-section of a conductor in time t, then,

Where I is electric current, Q is a net charge and t is a time in second.

If a net electric charge (Q) flows through a cross-section of a conductor in time t, then,

Where I is electric current, Q is a net charge and t is a time in second.

S.I. Unit of Electric Charge and Current: S.I. unit of electric charge is coulomb (C).

One coulomb is nearly equal to 6 × 1018 electrons. S.I. unit of electric current is ampere (A). Ampere is the flow of electric charge through a surface at the rate of one coulomb per second. This means, if 1 coulomb of electric charge flows through a cross section for 1 second, it would be equal to 1 ampere.

Therefore, 1 A = 1 C/1 s

One coulomb is nearly equal to 6 × 1018 electrons. S.I. unit of electric current is ampere (A). Ampere is the flow of electric charge through a surface at the rate of one coulomb per second. This means, if 1 coulomb of electric charge flows through a cross section for 1 second, it would be equal to 1 ampere.

Therefore, 1 A = 1 C/1 s

Small Quantity of Electric Current: Small quantity of electric current is expressed in milliampere and microampere. Milliampere is written as mA and microampere as pA.

1 mA (milliampere) = 10-3 A

1 pA (microampere) = 10-6 A

1 mA (milliampere) = 10-3 A

1 pA (microampere) = 10-6 A

Ammeter: An apparatus to measure electric current in a circuit.,

Charge: Like mass, the charge is the fundamental property of matter. There are two types of charge

(i) Positive charge.

(ii) Negative charge.

(i) Positive charge.

(ii) Negative charge.

Positive and Negative Charge: The charge acquired by a glass rod when rubbed with silk is called a positive charge and the charge acquired by an ebonite rod when rubbed with wool is called negative charge.

Properties of Electric Charge:

(i) Unlike charges attract each other and like charges repel each other.

(ii) The.force between two charges varies directly as the product of two charges and inversely as the square of the distance (r) between both charges (q1 and q2).

S.I. unit of charge is coulomb (C).

1 coulomb = 1 ampere × 1 second.

1C = 1A × 1s

Thus, the quantity of charge which flows through a circuit when one ampere of current flows through it in one second is known as a 1-coulomb charge.

(i) Unlike charges attract each other and like charges repel each other.

(ii) The.force between two charges varies directly as the product of two charges and inversely as the square of the distance (r) between both charges (q1 and q2).

S.I. unit of charge is coulomb (C).

1 coulomb = 1 ampere × 1 second.

1C = 1A × 1s

Thus, the quantity of charge which flows through a circuit when one ampere of current flows through it in one second is known as a 1-coulomb charge.

Electric Potential and Potential Difference

Electric Potential: The amount of electric potential energy at a point is called electric potential.

Potential Difference: The difference in the amount of electric potential energy between two points in an electric circuit is called electric potential difference.

Electric potential difference is known as voltage, which is equal to the amount of work done to move the unit charge between two points against static electric field.

Therefore, Voltage =

Voltage or electric potential difference is denoted by V’. Therefore, V =

Where, W = Work done and Q = Charge

Electric Potential: The amount of electric potential energy at a point is called electric potential.

Potential Difference: The difference in the amount of electric potential energy between two points in an electric circuit is called electric potential difference.

Electric potential difference is known as voltage, which is equal to the amount of work done to move the unit charge between two points against static electric field.

Therefore, Voltage =

Voltage or electric potential difference is denoted by V’. Therefore, V =

Where, W = Work done and Q = Charge

S.I. Unit of Electric Potential Difference (Voltage)

S.I. unit of electric potential difference is volt and denoted by ‘V’ This is named in honour of Italian Physicist Alessandro Volta.

S.I. unit of electric potential difference is volt and denoted by ‘V’ This is named in honour of Italian Physicist Alessandro Volta.

Since joule is the unit of work and Coulomb is the unit of charge, 1 volt of electric potential difference is equal to the 1 joule of work to be done to move a charge of 1 coulomb from one point to another in an electric circuit. Therefore

1V = 1Joule/1Coulomb = 1J/1C

1V = 1JC-1

1V = 1Joule/1Coulomb = 1J/1C

1V = 1JC-1

Voltmeter: An apparatus to measure the potential difference or electric potential difference between two points in an electric circuit.

Galvanometer: It is a device to detect current in an electric circuit.

Ohm’s Law: Ohm’s Law states that the potential difference between two points is directly proportional to the electric current, at a constant temperature.

This means potential difference V varies as electric current.

V ∝ I

V = RI

I =

R =

Where, R is constant for the given conductor at a given temperature and is called resistance.

This means potential difference V varies as electric current.

V ∝ I

V = RI

I =

R =

Where, R is constant for the given conductor at a given temperature and is called resistance.

Resistance: Resistance is the property of conductor which resists the flow of electric current through it.

S.I. unit of resistance is ohm. Ohm is denoted by Greek letter ‘Q’

S.I. unit of resistance is ohm. Ohm is denoted by Greek letter ‘Q’

1 Ohm: 1 ohm (Q) of resistance (R) is equal to the flow 1A of current through a conductor between two points having a potential difference equal to 1V.

This means; 1Ω =

From the expression of Ohm’s Law, it is obvious that electric current through a resistor is inversely proportional to resistance. This means electric current will decrease with an increase in resistance and vice versa. The graph of V (potential difference) versus I (electric current) is always a straight line.

Graph of Potential Difference (V) Vs Electric Current (I)

Voltage, i.e. Potential diffrence (V) = ?

We know, from Ohm’s Law that,

R =

15 Ω =

V = 225V

This means; 1Ω =

From the expression of Ohm’s Law, it is obvious that electric current through a resistor is inversely proportional to resistance. This means electric current will decrease with an increase in resistance and vice versa. The graph of V (potential difference) versus I (electric current) is always a straight line.

Graph of Potential Difference (V) Vs Electric Current (I)

Voltage, i.e. Potential diffrence (V) = ?

We know, from Ohm’s Law that,

R =

15 Ω =

V = 225V

Resistance: Resistance is a property of conductor due to which it resists the flow of electric current through it. A component that is used to resist the flow of electric current in a circuit is called a resistor.

In practical application, resistors are used to increase or decrease the electric current.

In practical application, resistors are used to increase or decrease the electric current.

Variable Resistance: The component of an electric circuit which is used to regulate the current, without changing the voltage from the source, is called variable resistance.

Rheostat: This is a device which is used in a circuit to provide variable resistance.

Cause of Resistance in a Conductor: Flow of electrons in a conductor is electric current. The positive particles of conductor create hindrance to flow of electrons, because of attraction between them, this hindrance is the cause of resistance in the flow of electricity.

Factors on Which Resistance of a Conductor Depends: Resistance in a conductor depends on nature, length and area of cross section of the conductor.

(i) Nature of Material: Some materials create least hindrance and hence, are called good conductors. Silver is the best conductor of electricity. While some other materials create more hindrance in the flow of electric current, i.e. flow of electrons through them. Such materials are called bad conductors. Bad conductor are also known as insulators. Hard plastic is the one of the best insulators of electricity.

(i) Nature of Material: Some materials create least hindrance and hence, are called good conductors. Silver is the best conductor of electricity. While some other materials create more hindrance in the flow of electric current, i.e. flow of electrons through them. Such materials are called bad conductors. Bad conductor are also known as insulators. Hard plastic is the one of the best insulators of electricity.

(ii) Length of Conductor: Resistance (R) is directly proportional to the length of the conductor. This means, resistance increases with increase in length of the conductor. This is the cause that long electric wires create more resistance to the electric current. Thus, Resistance (R) ∝ length of conductor (l)

or, R ∝ l …(i)

or, R ∝ l …(i)

(iii) Area of Cross Section: Resistance R is inversely proportional to the area of cross section (A) of the conductor. This means R will decrease with an increase in the area of conductor and vice versa. More area of conductor facilitates the flow of electric current through more area and thus, decreases the resistance. This is the cause that thick copper wire creates less resistance to the electric current.

Thus, resistance (R) ∝ 1/Area of cross section of conductor (A)

or, R ∝ ….(ii)

From equations (i) and (ii)

R ∝

R = ρ

Where, ρ (rho) is the proportionality constant. It is called the electrical resistivity of the material of conductor.

From equation (iii) RA = ρl ⇒ ρ = ..(iv)

Thus, resistance (R) ∝ 1/Area of cross section of conductor (A)

or, R ∝ ….(ii)

From equations (i) and (ii)

R ∝

R = ρ

Where, ρ (rho) is the proportionality constant. It is called the electrical resistivity of the material of conductor.

From equation (iii) RA = ρl ⇒ ρ = ..(iv)

The S.I. of Resistivity: Since, the S.I. unit of R is Q, S.I. unit of area is m2 and S.I. unit of length is m. Hence, unit of resistivity (ρ) = = Ωm

Thus, S.I. unit of resistivity (ρ) is Ωm.

Thus, S.I. unit of resistivity (ρ) is Ωm.

Resistivity: It is defined as the resistance offered by a cube of a material of side 1m when current flows perpendicular to its opposite faces. It’s S.I. unit is ohm-meter (Ωm).

Resistivity, ρ =

Resistivity is also known as specific resistance.

Resistivity depends on the nature of the material of the conductor.

Materials having a resistivity in the range of 10-8 Ωm to 10-6 Ωm are considered as very good conductors. Silver has resistivity equal to 1.60 × 10-8 Ωm and copper has resistivity equal to 1.62 × 10-8 Ωm.

Rubber and glass are very good insulators. They have a resistivity in the order of 10-12 Ωm to 10-8 Ωm.

The resistivity of materials varies with temperature.

Resistivity, ρ =

Resistivity is also known as specific resistance.

Resistivity depends on the nature of the material of the conductor.

Materials having a resistivity in the range of 10-8 Ωm to 10-6 Ωm are considered as very good conductors. Silver has resistivity equal to 1.60 × 10-8 Ωm and copper has resistivity equal to 1.62 × 10-8 Ωm.

Rubber and glass are very good insulators. They have a resistivity in the order of 10-12 Ωm to 10-8 Ωm.

The resistivity of materials varies with temperature.

Combination of resistors (Series and Parallel combination), the heating effect of electric current and electric power.

Combination of Resistors

(i) Series combination

(ii) Parallel combination.

Combination of Resistors

(i) Series combination

(ii) Parallel combination.

1. Resistors in Series: When resistors are joined from end to end, it is called in series. In this case, the total resistance of the system is equal to the sum of the resistance of all the resistors in the system.

Let, three resistors R1, R2, and R3 get connected in series.

Potential difference across A and B = V

Potential difference across R1, R2 and R3 = V1, V2 and V3

Current flowing through the combination = I

We, know that

V= V1 + V2 + V3 …. (i)

According to Ohm’s Law :

V1 = IR1, V2 = IR2 and V3 = IR3 ….. (ii)

Let, total resistance = Rs

Then, V = IRs …(iii)

From equations (i) and (ii) and (iii)

IRs = IR1 + IR2 + IR3

Rs = R1 + R2 + R3

When the resistors are connected in series, the current flowing through each resistor is the same and is equal to the total current.

Let, three resistors R1, R2, and R3 get connected in series.

Potential difference across A and B = V

Potential difference across R1, R2 and R3 = V1, V2 and V3

Current flowing through the combination = I

We, know that

V= V1 + V2 + V3 …. (i)

According to Ohm’s Law :

V1 = IR1, V2 = IR2 and V3 = IR3 ….. (ii)

Let, total resistance = Rs

Then, V = IRs …(iii)

From equations (i) and (ii) and (iii)

IRs = IR1 + IR2 + IR3

Rs = R1 + R2 + R3

When the resistors are connected in series, the current flowing through each resistor is the same and is equal to the total current.

2. Resistors in Parallel: When resistors are joined in parallel, the reciprocal of the total resistance of the system is equal to the sum of reciprocal of the resistance of resistors.

Let three resistors R1, R2 and R3 connected in parallel.

Potential difference across point A and B = V

Total current flowing between point A and B = I

Currents flowing through resistors R1, R2 and R3 = I1, I2 and I3 respectively.

We, know that,

I = I1 + I2 + I3 …….(i)

Since, the potential difference across R1, R2, and R3 is the same = V

According to Ohm’s Law,

In parallel combination, the potential difference across each resistor is the same and is equal to the total potential difference.

The total current through the circuit can be calculated by adding the electric current through individual resistors.

Itotal = 6A + 48A + 30A + 12A + 24A = 120A

Let three resistors R1, R2 and R3 connected in parallel.

Potential difference across point A and B = V

Total current flowing between point A and B = I

Currents flowing through resistors R1, R2 and R3 = I1, I2 and I3 respectively.

We, know that,

I = I1 + I2 + I3 …….(i)

Since, the potential difference across R1, R2, and R3 is the same = V

According to Ohm’s Law,

In parallel combination, the potential difference across each resistor is the same and is equal to the total potential difference.

The total current through the circuit can be calculated by adding the electric current through individual resistors.

Itotal = 6A + 48A + 30A + 12A + 24A = 120A

Heating Effect of Electric Current: When electric current is supplied to a purely resistive conductor, the energy of electric current is dissipated entirely in the form of heat and as a result, resistor gets heated. The heating of resistor because of dissipation of electrical energy is commonly known as Heating Effect of Electric Current. Some examples are as follows : When electric energy is supplied to an electric bulb, the filament gets heated because of which, it gives light. The heating of electric bulb happens because of heating effect of electric current.

Cause of Heating Effect of Electric Current: Electric current generates heat to overcome the resistance offered by the conductor through which it passes. Higher the resistance, the electric current will generate higher amount of heat. Thus, generation of heat by electric current while passing through a conductor is an inevitable consequence. This heating effect is used in many appliances, such as electric iron, electric heater, electric geyser, etc.

Joule’s Law Of Heating: Let, an electric current, I is flowing through a resistor having resistance = R.

The potential difference through the resistor is = V.

The charge, Q flows through the circuit for the time, t

Thus, work done in moving of charge (Q) of potential difference (V),

W = V × Q

Since this charge, Q flows through the circuit for time t

Therefore, power input (P) to the circuit can be given by the following equation :

P =

P = V × …..(i)

We know, electric current, I =

Substituting = I in equation (i), we get,

P = VI …(ii)

i.e., P = VI

Since, the electric energy is supplied for time ?, thus, after multiplying both sides of equation (ii) by time t, we get,

P × t = VI × t = VIt ……(iii)

i.e., P = VIt

Thus, for steady current I, the heat produced (H) in time t is equal to VIt

H = VIt i.e., H = VIt

We know, according to Ohm’s Law,

V = IR

By substituting this value of V in equation (iii), we get,

H = IR × It

H = I2Rt ……(iv)

The expression (iv) is known as Joule’s Law of Heating, which states that heat produced in a resistor is directly proportional to the square of current given to the resistor, directly proportional to the resistance for a given current and directly proportional to the time for which the current is flowing through the resistor.

The potential difference through the resistor is = V.

The charge, Q flows through the circuit for the time, t

Thus, work done in moving of charge (Q) of potential difference (V),

W = V × Q

Since this charge, Q flows through the circuit for time t

Therefore, power input (P) to the circuit can be given by the following equation :

P =

P = V × …..(i)

We know, electric current, I =

Substituting = I in equation (i), we get,

P = VI …(ii)

i.e., P = VI

Since, the electric energy is supplied for time ?, thus, after multiplying both sides of equation (ii) by time t, we get,

P × t = VI × t = VIt ……(iii)

i.e., P = VIt

Thus, for steady current I, the heat produced (H) in time t is equal to VIt

H = VIt i.e., H = VIt

We know, according to Ohm’s Law,

V = IR

By substituting this value of V in equation (iii), we get,

H = IR × It

H = I2Rt ……(iv)

The expression (iv) is known as Joule’s Law of Heating, which states that heat produced in a resistor is directly proportional to the square of current given to the resistor, directly proportional to the resistance for a given current and directly proportional to the time for which the current is flowing through the resistor.

Electric Bulb: In an electric bulb, the filament of bulb gives light because of the heating effect of electricity. The filament of bulb is generally, made of tungsten metal, having melting point equal to 3380°C.

Electric Iron: The element of electric iron is made of alloys having high melting poir^ Electric heater and geyser work on the same mechanism.

Electric Fuse: Electric fuse is used to protect the electric appliances from high voltage if any. Electric fuse is made of metal or alloy of metals, such as aluminum, copper, iron, lead, etc. In the case of flow of higher voltage than specified, fuse wire melts and protect the electric appliances.

Fuse of 1A, 2A, 3A, 5A, 10A, etc., used for domestic purpose.

Suppose, if an electric heater consumes 1000W at 220 V.

Then electric current in circuit

I =

I = = 4.5 A

Thus, in this case of 5A should be used to protect the electric heater in the flow of higher voltage.

Fuse of 1A, 2A, 3A, 5A, 10A, etc., used for domestic purpose.

Suppose, if an electric heater consumes 1000W at 220 V.

Then electric current in circuit

I =

I = = 4.5 A

Thus, in this case of 5A should be used to protect the electric heater in the flow of higher voltage.

Electric Power

S.I. unit of electric power is watt (W).

1W = 1 volt × 1 ampere = 1V × 1A

I kilowatt or 1kW = 1000 W

Consumption of electricity (electric energy) is generally measured in kilowatt.

Unit of electric energy is kilowatt-hour (kWh).

1 kWh = 1000 watt × 1 hour = 1 unit = 1000 W × 3600 s

1 kWh = 3.6 x 106 watt second = 3.6 × 106 J

S.I. unit of electric power is watt (W).

1W = 1 volt × 1 ampere = 1V × 1A

I kilowatt or 1kW = 1000 W

Consumption of electricity (electric energy) is generally measured in kilowatt.

Unit of electric energy is kilowatt-hour (kWh).

1 kWh = 1000 watt × 1 hour = 1 unit = 1000 W × 3600 s

1 kWh = 3.6 x 106 watt second = 3.6 × 106 J

Conductor: The material which can allow the flow of electrons through itself is called the conductor. It has a large number of free electrons. It offers low opposition in the flow of current.

Insulator: The material which does not allow the flow of electrons through itself is called insulator. It has less or no free electrons. It offers high opposition in the flow of current.

Electric Current: The amount of flow charge through any cross-sectional area of a conductor in unity time is called Electric Current.

It is represented by ‘I’

I =

It is represented by ‘I’

I =

Unit of Electric Current: It is CS-1 (coulomb per second) or Ampere (A). Electric Current is a scalar quantity. It is measured by an ammeter.

Direction: The direction of conventional current (or practical current) is opposite to the flow of electrons.

Electric potential: Electric Potential at any point in the electric field is defined as the amount of work done to bring the unit positive charge from infinity (from outside the electric field) to that point.

V =, S.I. unit of Electric Potential is JC-1 or volt (V). It is a scalar quantity. The +ve charge flows from higher to lower potential. The -ve charge flows from lower to a higher potential. The difference of electric potential between any two points in the electric field is called Electric Potential difference. It is known as a voltage which is equal to the work done per unit charge between two points against the static electric field.

VAB = VA – VB =

Electric Potential difference is measured by a voltmeter.

V =, S.I. unit of Electric Potential is JC-1 or volt (V). It is a scalar quantity. The +ve charge flows from higher to lower potential. The -ve charge flows from lower to a higher potential. The difference of electric potential between any two points in the electric field is called Electric Potential difference. It is known as a voltage which is equal to the work done per unit charge between two points against the static electric field.

VAB = VA – VB =

Electric Potential difference is measured by a voltmeter.

Ohm’s Law: According to this law “Under the constant physical condition the potential difference across the conductor is directly proportional to the current flowing through the conductor.”

V ∝ I

V = IR …[Where R is proportionality constant called resistance of conductor]

⇒ I =

R depends upon nature, geometry and physical condition of the conductor.

V ∝ I

V = IR …[Where R is proportionality constant called resistance of conductor]

⇒ I =

R depends upon nature, geometry and physical condition of the conductor.

The heat generated by electric current: The potential difference between two points in an electrical field is equal to the work done in moving a unit charge from one point to another.

Then, work is done, W = VQ and Q = I × t

W = V × I × t

From Ohm’s Law, we know that

V = IR

W = IR × I × t = I2.Rt

Since heat produced by the electric current is equal to work done, W

H = W

⇒ H (heat) = I2Rt Joule.

Then, work is done, W = VQ and Q = I × t

W = V × I × t

From Ohm’s Law, we know that

V = IR

W = IR × I × t = I2.Rt

Since heat produced by the electric current is equal to work done, W

H = W

⇒ H (heat) = I2Rt Joule.

Resistance: Ratio of the applied voltage to the current flowing in the conductor is called resistance of the conductor.

⇒ R =

S.I. Unit of resistance is VA-1 or ohm (Ω).

Resistance is the opposition offered by the conductor in the flow of current.

Practically it is

R ∝ L (L is the length of a conductor)

R ∝ 1/A (A is the area of a conductor)

So, R ∝ L/A

R = ρL/A …[Where p is proportionality constant called specific resistance of conductor

It only depend upon nature (material) and temperature of conductor.

⇒ R =

S.I. Unit of resistance is VA-1 or ohm (Ω).

Resistance is the opposition offered by the conductor in the flow of current.

Practically it is

R ∝ L (L is the length of a conductor)

R ∝ 1/A (A is the area of a conductor)

So, R ∝ L/A

R = ρL/A …[Where p is proportionality constant called specific resistance of conductor

It only depend upon nature (material) and temperature of conductor.

Specific resistance or Resistivity = ρ = RA /L

It’s S.I. Unit is Qm

It’s S.I. Unit is Qm

Combination of resistance:

- In this combination the current across every component is same but potential across every component is different.
- If resistance R1, R2 and R3 are connected in series with a battery of Potential V, then equivalence resistance of the combination

R = R1 + R2 + R3

The parallel combination of resistance:

- In this combination the current across every component is different. But potential across every component is the same.
- If resistance R1, R2 and R3 are connected in parallel with a battery of Potential V, then equivalence resistance of combination

Electric Energy is amount of work done to maintain the continuous flow of electric current in the circuit.

Its S.I. unit is joule (J).

Its S.I. unit is joule (J).

Electric power (P): The electric work done per unit time is called electric power.

Electric Power =

or P =

Electric power is also defined as the electric energy consumed per unit time.

P =

S.I. unit of electric power is Watt. When one joule of energy is used for one second, electric power is equal to one watt.

Electric Power =

or P =

Electric power is also defined as the electric energy consumed per unit time.

P =

S.I. unit of electric power is Watt. When one joule of energy is used for one second, electric power is equal to one watt.

Derivation of formula for electric power:

We know that electric work done, W = V × I × t or P =

P = VI

Electric power in watts = Volts × ampere

Also V = IR …[According to Ohm’s Law]

So P = IR × I

P = I2R

We know that I =

P = ()2 × R = Watt

The maximum value of electric current that can pass through an electric appliance without damaging electric appliance is called current rating of electric appliance.

We know that electric work done, W = V × I × t or P =

P = VI

Electric power in watts = Volts × ampere

Also V = IR …[According to Ohm’s Law]

So P = IR × I

P = I2R

We know that I =

P = ()2 × R = Watt

The maximum value of electric current that can pass through an electric appliance without damaging electric appliance is called current rating of electric appliance.

Question 1.

What does an electric circuit mean? [2011, 2013, 2014]

Answer.

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

What does an electric circuit mean? [2011, 2013, 2014]

Answer.

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

Question 2.

Define the unit of current.

Answer.

One ampere is the current flowing through a circuit, when a charge of one coulomb flows through it in one second.

Define the unit of current.

Answer.

One ampere is the current flowing through a circuit, when a charge of one coulomb flows through it in one second.

Question 3.

Calculate the number of electrons constituting one coulomb of charge. [2015]

Answer.

Total charge, Q = n x e

Calculate the number of electrons constituting one coulomb of charge. [2015]

Answer.

Total charge, Q = n x e

Question 4.

Name a device that helps to maintain a potential difference across a conductor?

Answer.

With the help of a cell or a battery.

Name a device that helps to maintain a potential difference across a conductor?

Answer.

With the help of a cell or a battery.

Question 5.

What is meant by saying that potential difference between two points is 1 volt? [2010]

Answer.

It means that 1 joule work is done in moving 1 coulomb of positive charge from one point to the other in an elecfric field.

What is meant by saying that potential difference between two points is 1 volt? [2010]

Answer.

It means that 1 joule work is done in moving 1 coulomb of positive charge from one point to the other in an elecfric field.

Question 6.

How much energy is given to each coulomb of charge passing through a 6 V battery?

Answer.

Potential difference (V) = 6 V

Charge (Q) = 1 coulomb

Energy = Q x V x 6 = 6 J

How much energy is given to each coulomb of charge passing through a 6 V battery?

Answer.

Potential difference (V) = 6 V

Charge (Q) = 1 coulomb

Energy = Q x V x 6 = 6 J

Question 7.

On what factors does the resistance of a conductor depend? [2011-2014]

Answer.

Resistance of a conductor depends on its material, length and area of cross-section.

On what factors does the resistance of a conductor depend? [2011-2014]

Answer.

Resistance of a conductor depends on its material, length and area of cross-section.

Question 8.

Let the resistance of an electrical component remains constant while the potential difference across the two ends of the component decreases to half of its former value. What change will occur in the current through it?

Answer.

According to Ohm’s law, we have:

I ∝ V or V ∝ I

Since the potential difference (V) decreases to half, i.e.

∴

Hence, the current flowing through the component will become half.

Let the resistance of an electrical component remains constant while the potential difference across the two ends of the component decreases to half of its former value. What change will occur in the current through it?

Answer.

According to Ohm’s law, we have:

I ∝ V or V ∝ I

Since the potential difference (V) decreases to half, i.e.

∴

Hence, the current flowing through the component will become half.

Question 9.

Why are coils of electric toasters and electric irons made of an alloy rather than a pure metal? [2011,2012]

Answer.

This is because of the high resistivity and high melting point of the alloy. Also, they do not oxidise when red hot.

Why are coils of electric toasters and electric irons made of an alloy rather than a pure metal? [2011,2012]

Answer.

This is because of the high resistivity and high melting point of the alloy. Also, they do not oxidise when red hot.

Question 10.

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. [2011]

Answer.

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. [2011]

Answer.

Question 11.

Will current flow more easily through a thick wire or a thin wire of the same material when connected to the same source? Why?

Answer.

The current will flow more easily through a wire which have comparatively less resistance. According to the relation , the thin wire has less resistance. So, the current will flow more easily through thin wire.

Will current flow more easily through a thick wire or a thin wire of the same material when connected to the same source? Why?

Answer.

The current will flow more easily through a wire which have comparatively less resistance. According to the relation , the thin wire has less resistance. So, the current will flow more easily through thin wire.

Question 12.

Judge the equivalent resistance when the following are connected in parallel: [2013]

(a) 1 Ω and 106 Ω

(b) 1 Ω and 103 Ω, and 106 Ω

Answer.

(a) Approximately but less than 1 Ω, because in a parallel combination of resistors, the equivalent resistance is less than the least resistance.

(b) Same as in previous part of the question.

Judge the equivalent resistance when the following are connected in parallel: [2013]

(a) 1 Ω and 106 Ω

(b) 1 Ω and 103 Ω, and 106 Ω

Answer.

(a) Approximately but less than 1 Ω, because in a parallel combination of resistors, the equivalent resistance is less than the least resistance.

(b) Same as in previous part of the question.

Question 13.

Redraw the circuit of previous question 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.

Total resistance, Rs = 5 + 8 + 12 = 25 Ω

Redraw the circuit of previous question 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.

Total resistance, Rs = 5 + 8 + 12 = 25 Ω

Question 14.

An electric lamp of resistance 100 12, a toaster of resistance 50 11 and a water filter of resistance 500 12 are connected in parallel to a 220 V source. What is the resistance of an electric iron connected to the same source that takes as much current as all three appliances, and what is the current through it?

Answer.

Given: Resistance of electric lamp, R1 = 100Ω

Resistance of toaster, R2 = 50Ω

Resistance of water filter, R3 = 500Ω

Voltage, V = 220 V

Let Rp be the resistance of electric iron. Then:

An electric lamp of resistance 100 12, a toaster of resistance 50 11 and a water filter of resistance 500 12 are connected in parallel to a 220 V source. What is the resistance of an electric iron connected to the same source that takes as much current as all three appliances, and what is the current through it?

Answer.

Given: Resistance of electric lamp, R1 = 100Ω

Resistance of toaster, R2 = 50Ω

Resistance of water filter, R3 = 500Ω

Voltage, V = 220 V

Let Rp be the resistance of electric iron. Then:

Question 15.

What is (a) the highest, (b) the lowest total resistance that can be secured by combinations of four coils of resistances 4 Ω, 8 Ω, 12 Ω, 24 Ω?

Answer.

(a) The highest resistance can be secured by connecting all the four coils in series.

∴ Equivalent resistance of series combination, Rs= 4 + 8 + 12+ 24 = 48 Ω

Hence, the highest resistance that can be secured by given coils is 48 Ω.

What is (a) the highest, (b) the lowest total resistance that can be secured by combinations of four coils of resistances 4 Ω, 8 Ω, 12 Ω, 24 Ω?

Answer.

(a) The highest resistance can be secured by connecting all the four coils in series.

∴ Equivalent resistance of series combination, Rs= 4 + 8 + 12+ 24 = 48 Ω

Hence, the highest resistance that can be secured by given coils is 48 Ω.

(b) The lowest resistance can be secured by connecting all the four coils in parallel.

∴ Equivalent resistance of parallel combination,

or Rp = 2 Ω

Hence, the lowest resistance that can be secured by given coils is 2 Ω.

∴ Equivalent resistance of parallel combination,

or Rp = 2 Ω

Hence, the lowest resistance that can be secured by given coils is 2 Ω.

Question 16.

Why does the cord of an electric heater not glow while the heating element does?

[2010,2011,2013]

Answer.

The resistance of metal used in cord has much lower than that its element. So the heat produced by the element is much more than that produced by the cord while passing the current.

Why does the cord of an electric heater not glow while the heating element does?

[2010,2011,2013]

Answer.

The resistance of metal used in cord has much lower than that its element. So the heat produced by the element is much more than that produced by the cord while passing the current.

Question 17.

Compute the heat generated while transferring 96000 coulomb of charge in one hour through a potential difference of 50 V. [2012, 2013]

Answer.

Given, Q = 96000 C, V= 50 volt,

t = 1 hour = 60 x 60 = 3600 s

∴ Heat generated, H = VIt

= VQ = 50 x 96000

= 4800000 = 4.8 x 106 J

Compute the heat generated while transferring 96000 coulomb of charge in one hour through a potential difference of 50 V. [2012, 2013]

Answer.

Given, Q = 96000 C, V= 50 volt,

t = 1 hour = 60 x 60 = 3600 s

∴ Heat generated, H = VIt

= VQ = 50 x 96000

= 4800000 = 4.8 x 106 J

Question 18.

An electric iron of resistance 20 12 takes a current of 5 A. Calculate the heat developed in 30 s. [2010]

Answer.

Heat developed is equal to the work done (W).

Given: R = 20 Ω, I = 5 A and t = 30 s

∴ W = I2Rt = (5)2 x 20 x 30

= 25 x 20 x 30 = 15,000 J

Hence, the heat developed by electric iron is 15,000 J.

An electric iron of resistance 20 12 takes a current of 5 A. Calculate the heat developed in 30 s. [2010]

Answer.

Heat developed is equal to the work done (W).

Given: R = 20 Ω, I = 5 A and t = 30 s

∴ W = I2Rt = (5)2 x 20 x 30

= 25 x 20 x 30 = 15,000 J

Hence, the heat developed by electric iron is 15,000 J.

Question 19.

What determines the rate at which energy is delivered by a current?

Answer.

Electric power determines the rate at which energy is delivered by a current.

What determines the rate at which energy is delivered by a current?

Answer.

Electric power determines the rate at which energy is delivered by a current.

Question 20.

An electric motor takes 5 A from a 220 V line. Determine the power of the motor and the energy consumed in 2 hr.

Answer.

Given: Current, I = 5 A, Voltage, V = 220 V and t= 2 h

Electric power, P=VI = 220 x 5 = 1100 W

∴ Energy consumed = Pt = 1100 x 7200

= 7920000 = 792 x 104 J

An electric motor takes 5 A from a 220 V line. Determine the power of the motor and the energy consumed in 2 hr.

Answer.

Given: Current, I = 5 A, Voltage, V = 220 V and t= 2 h

Electric power, P=VI = 220 x 5 = 1100 W

∴ Energy consumed = Pt = 1100 x 7200

= 7920000 = 792 x 104 J

Question 21.

What are the advantages of connecting devices in parallel with the battery instead of connecting them in series?

Answer.

(a) When electrical devices are connected in parallel, each is operated at the same voltage and if one of them goes off, the others are not affected. But when electrical devices are connected in series, this does not happen because the circuit is broken when any one of the devices fails.

(b) In parallel connection, the current divides itself through different devices. This is helpful when each device has different resistance and requires different current to operate. But in series connection this does not happen because the same current flows through all the devices.

What are the advantages of connecting devices in parallel with the battery instead of connecting them in series?

Answer.

(a) When electrical devices are connected in parallel, each is operated at the same voltage and if one of them goes off, the others are not affected. But when electrical devices are connected in series, this does not happen because the circuit is broken when any one of the devices fails.

(b) In parallel connection, the current divides itself through different devices. This is helpful when each device has different resistance and requires different current to operate. But in series connection this does not happen because the same current flows through all the devices.

Question 22.

How can three resistors of resistances 2 Ω, 3 Ω and 6 Ω be connected to give a total resistance of (a) 4 Ω and

(b) 1 Ω ?

Answer.

(a) In order to get a resistance of 4 Ω, we can connect resistors of 3 Ω and 6 Ω in parallel and this parallel combination in series with 2 Ω resistor.

How can three resistors of resistances 2 Ω, 3 Ω and 6 Ω be connected to give a total resistance of (a) 4 Ω and

(b) 1 Ω ?

Answer.

(a) In order to get a resistance of 4 Ω, we can connect resistors of 3 Ω and 6 Ω in parallel and this parallel combination in series with 2 Ω resistor.

(b) In order to get 1 Ω resistance, we can connect all the three resistors in parallel.

Chapter End Questions

Question 1.

A piece of wire of resistance R is cut into five equal parts. These parts are then connected in parallel. If the equivalent resistance of this combination is R’, then the ratio is: [2012]

(a)

(b)

(c) 5

(d) 25

Answer.

(d) 25

A piece of wire of resistance R is cut into five equal parts. These parts are then connected in parallel. If the equivalent resistance of this combination is R’, then the ratio is: [2012]

(a)

(b)

(c) 5

(d) 25

Answer.

(d) 25

Question 2.

Which of the following terms does not represent the electrical power in a circuit?

(a) I2R

(b) IR2

(c) VI

(d)

Answer.

(b) IR2

Which of the following terms does not represent the electrical power in a circuit?

(a) I2R

(b) IR2

(c) VI

(d)

Answer.

(b) IR2

Question 3.

Two conducting wires of the same material and of equal lengths and equal diameters are first ‘ connected in series and then parallel in a circuit across the same potential difference. The ratio of heat produced in series and parallel combination would be:

(a) 1 : 2

(b) 2 : 1

(c) 1 : 4

(d) 4 : 1

Answer.

(c) 1 : 4

Two conducting wires of the same material and of equal lengths and equal diameters are first ‘ connected in series and then parallel in a circuit across the same potential difference. The ratio of heat produced in series and parallel combination would be:

(a) 1 : 2

(b) 2 : 1

(c) 1 : 4

(d) 4 : 1

Answer.

(c) 1 : 4

Question 4.

An electric bulb is rated 220 V and 100 W. When it is operated on 110 V, the )aower consumed will be

(a) 100 W ,

(b) 75 W

(c) 50 W

(d) 25 W

Answer.

(d) 25 W

An electric bulb is rated 220 V and 100 W. When it is operated on 110 V, the )aower consumed will be

(a) 100 W ,

(b) 75 W

(c) 50 W

(d) 25 W

Answer.

(d) 25 W

Question 5.

How is a voltmeter connected in the circuit to measure the potential difference between two points? [2011,2012]

Answer.

A voltmeter is connected in parallel to the circuit.

How is a voltmeter connected in the circuit to measure the potential difference between two points? [2011,2012]

Answer.

A voltmeter is connected in parallel to the circuit.

Question 6.

A copper wire has diameter 0.5 mm and resistivity of 1.6 x 108 Ω m. What will be the length of this wire to make its resistance 10 Ω? How much does the resistance change, if the diameter is doubled?

Answer.

Given: Radius, r = 0.25 mm = 0.25 x 10-3 m

Length of the wire, l = 1

Resistance, R =10 Ω

A copper wire has diameter 0.5 mm and resistivity of 1.6 x 108 Ω m. What will be the length of this wire to make its resistance 10 Ω? How much does the resistance change, if the diameter is doubled?

Answer.

Given: Radius, r = 0.25 mm = 0.25 x 10-3 m

Length of the wire, l = 1

Resistance, R =10 Ω

Question 7.

The valued of current I flowing in a given resistor for the corresponding values of potential difference V across the resistor are given below:

The valued of current I flowing in a given resistor for the corresponding values of potential difference V across the resistor are given below:

I (amperes) | 0.5 | 1.0 | 2.0 | 3.0 | 4.0 |

V (volts) | 1.6 | 3.4 | 6.7 | 10.2 | 13.2 |

Plot a graph between V and I and calculate the resistance of the resistor.

Answer.

Answer.

Question 8.

When a 12 V battery is connected across an unknown resistor, there is a current of 2.5 mA in the circuit. Find the value of the resistance of the resistor.

Answer.

Given: V= 12 volts, I = 2.5 mA = 2.5 x 10-3 A

Resistance, R = ?

∴ Resistance = = = 4.8 x 103 Ω

When a 12 V battery is connected across an unknown resistor, there is a current of 2.5 mA in the circuit. Find the value of the resistance of the resistor.

Answer.

Given: V= 12 volts, I = 2.5 mA = 2.5 x 10-3 A

Resistance, R = ?

∴ Resistance = = = 4.8 x 103 Ω

Question 9.

A battery of 9 V is connected in series with resistors of 0.2 Ω, 0.3 Ω, 0.4 Ω, 0.5 Ω and 12 Ω, respectively. How much current would flow through the 12 Ω resistor? [2010]

Answer.

Given: V= 9 volts

Applying the formula R = R1 + R2+ R3+ R4 + Rs, we get

R = 0.2 + 0.3 + 0.4 + 0.5 + 12 = 13.4 Ω

By Ohm’s law, we have

V=RI or

or 9 = 13.4 x I

or

Hence, the current flows through the 12 Ω resistor is 0.67 A.

A battery of 9 V is connected in series with resistors of 0.2 Ω, 0.3 Ω, 0.4 Ω, 0.5 Ω and 12 Ω, respectively. How much current would flow through the 12 Ω resistor? [2010]

Answer.

Given: V= 9 volts

Applying the formula R = R1 + R2+ R3+ R4 + Rs, we get

R = 0.2 + 0.3 + 0.4 + 0.5 + 12 = 13.4 Ω

By Ohm’s law, we have

V=RI or

or 9 = 13.4 x I

or

Hence, the current flows through the 12 Ω resistor is 0.67 A.

Question 10.

How many 176 Ω resistors (in parallel) are required to carry 5 A on a 220 V line? [2012]

Answer.

Given: V= 220 volts, I= 5 A

Resistance of each resistor, R = 176 Ω

Number of resistors = n, We know that V = RpI

How many 176 Ω resistors (in parallel) are required to carry 5 A on a 220 V line? [2012]

Answer.

Given: V= 220 volts, I= 5 A

Resistance of each resistor, R = 176 Ω

Number of resistors = n, We know that V = RpI

Question 11.

Show how you would connect three resistors each of resistance 6 ft, so that the combination has a resistance of

(a) 9 Ω, and

(b) 4 Ω. ,

Answer.

(a) In order to get a resistance of 9 ft from the given three resistors, two resistors can be connected in parallel and this parallel combination in series with the third one.

(b) In order to get a resistance of 4 ft from the given three resistors, two resistors can be connected in series and this series in parallel with the third one.

Show how you would connect three resistors each of resistance 6 ft, so that the combination has a resistance of

(a) 9 Ω, and

(b) 4 Ω. ,

Answer.

(a) In order to get a resistance of 9 ft from the given three resistors, two resistors can be connected in parallel and this parallel combination in series with the third one.

(b) In order to get a resistance of 4 ft from the given three resistors, two resistors can be connected in series and this series in parallel with the third one.

Question 12.

Several electric bulbs designed to be used on a 220 V electric supply line, are rated 10 W. How many lamps can be connected in parallel with each other across the two wires of 220 V line, if the maximum allowable current is 5 A?

Answer.

Given: Potential, V = 220 volts, P = 10 W, I = 5 A

Let number of bulbs = n

Applying the formula P = VI, we get:

Hence, 110 bulbs can be connected is parallel.

Several electric bulbs designed to be used on a 220 V electric supply line, are rated 10 W. How many lamps can be connected in parallel with each other across the two wires of 220 V line, if the maximum allowable current is 5 A?

Answer.

Given: Potential, V = 220 volts, P = 10 W, I = 5 A

Let number of bulbs = n

Applying the formula P = VI, we get:

Hence, 110 bulbs can be connected is parallel.

Question 13.

A hot plate of an electric oven connected to a 220 volt line has two resistance coils A and B, each of 24 Ω resistance, which may be used separately, in series, or in parallel. What are the currents in the three cases?

Answer.

Given: Voltage, V = 220 V

A hot plate of an electric oven connected to a 220 volt line has two resistance coils A and B, each of 24 Ω resistance, which may be used separately, in series, or in parallel. What are the currents in the three cases?

Answer.

Given: Voltage, V = 220 V

Question 14.

Compare the power used in the 2 Ω resistor in each of the following circuits:

(a) A 6 V battery in series with 1 Ω and 2 Ω resistors, and

(b) a 4 V battery in parallel with 12 Ω and 2 Ω resistors.

Answer.

(a) Given: Voltage, V= 6 V

Equivalent resistance, R = 1 + 2 = 3 Ω

Compare the power used in the 2 Ω resistor in each of the following circuits:

(a) A 6 V battery in series with 1 Ω and 2 Ω resistors, and

(b) a 4 V battery in parallel with 12 Ω and 2 Ω resistors.

Answer.

(a) Given: Voltage, V= 6 V

Equivalent resistance, R = 1 + 2 = 3 Ω

(b) Since the battery is in parallel with 12 Ω and 2 Ω resistors

∴ Potential difference across 2 Ω resistor,

∴ Potential difference across 2 Ω resistor,

Question 15.

Two lamps, one rated 100 W at 220 V, and the other 60 W at 220 V, are connected in parallel to electric mains supply. What current is drawn from the line if the supply voltage is 220 V?

Answer.

Given: P1 = 100 W, P2 = 60 W and 7= 220 V

Resistance of first lamp,

Two lamps, one rated 100 W at 220 V, and the other 60 W at 220 V, are connected in parallel to electric mains supply. What current is drawn from the line if the supply voltage is 220 V?

Answer.

Given: P1 = 100 W, P2 = 60 W and 7= 220 V

Resistance of first lamp,

Question 16.

Which uses more energy, a 250 W TV set in 1 hr or a 1200 W toaster in 10 minutes?

Answer.

Energy used by TV set = Power x Time

Which uses more energy, a 250 W TV set in 1 hr or a 1200 W toaster in 10 minutes?

Answer.

Energy used by TV set = Power x Time

Question 17.

An electric heater of resistance 8 W draws a current of 15 A from the service mains for 2 hr. Calculate the rate at which heat is developed in the heater.

Answer.

Given: Resistance, R = 8 Ω, t = 2 hr, I = 15 A

Rate of heat developed,

Power = I2R = (15)2 x 8 = 1800 W

An electric heater of resistance 8 W draws a current of 15 A from the service mains for 2 hr. Calculate the rate at which heat is developed in the heater.

Answer.

Given: Resistance, R = 8 Ω, t = 2 hr, I = 15 A

Rate of heat developed,

Power = I2R = (15)2 x 8 = 1800 W

Question 18.

Explain the following:,

(a) Why is the tungsten used almost exclusively for filament of electric lamps?

(b) Why are the conductors of electric heating devices such as bread toaster and electric irons, made of an alloy rather than a pure metal?

(c) Why is the series arrangement not used for domestic circuits?

(d) How does the resistance of a wire vary with its area of cross-section?

(e) Why are copper and aluminium wires usually employed for electricity transmission?

Answer.

(a) Melting point of tungsten is very high. So it does not oxidise (burn) at high temperatures. Secondly, the resistivity of tungsten is high. So thin wire of tungsten which has high resistance which produces more heat enough to emit light.

Explain the following:,

(a) Why is the tungsten used almost exclusively for filament of electric lamps?

(b) Why are the conductors of electric heating devices such as bread toaster and electric irons, made of an alloy rather than a pure metal?

(c) Why is the series arrangement not used for domestic circuits?

(d) How does the resistance of a wire vary with its area of cross-section?

(e) Why are copper and aluminium wires usually employed for electricity transmission?

Answer.

(a) Melting point of tungsten is very high. So it does not oxidise (burn) at high temperatures. Secondly, the resistivity of tungsten is high. So thin wire of tungsten which has high resistance which produces more heat enough to emit light.

(b) This is because of the high resistivity and high melting point of the alloy. Also, they do not oxidise when red hot.

(c) Domestic circuits are not connected in series because they do not work at the same voltage and also if there is a short- circuiting in one distribution circuit, its fuse will not blow and other circuits will also be affected.

(d) Larger the area of cross-section means a large number of free electrons can move in the conductor, thereby decreasing the resistance.

(e) Copper and aluminium wires are good conductors of electricity. So they have low resistivity. They offer low opposition to the flow of current from one point to the other, hence employed for electricity.

### NCERT Notes for Class 10 Science

- Chapter 1 Chemical Reactions and Equations Class 10 Notes
- Chapter 2 Acids Bases and Salts Class 10 Notes
- Chapter 3 Metals and Non-metals Class 10 Notes
- Chapter 4 Carbon and its Compounds Class 10 Notes
- Chapter 5 Periodic Classification of Elements Class 10 Notes
- Chapter 6 Life Processes Class 10 Notes
- Chapter 7 Control and Coordination Class 10 Notes
- Chapter 8 How do Organisms Reproduce Class 10 Notes
- Chapter 9 Heredity and Evolution Class 10 Notes
- Chapter 10 Light Reflection and Refraction Class 10 Notes
- Chapter 11 Human Eye and Colourful World Class 10 Notes
- Chapter 12 Electricity Class 10 Notes
- Chapter 13 Magnetic Effects of Electric Current Class 10 Notes
- Chapter 14 Sources of Energy Class 10 Notes
- Chapter 15 Our Environment Class 10 Notes
- Chapter 16 Management of Natural Resources Class 10 Notes

**English Story Chapter Flamingo**

__English Vistas Class 12th CBSE__**Class 10 English First Flight**

__Class 10 English Footprints Without Feet__

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*NCERT Solutions for Class 10 English Extended Reading Text / Novels / Long Reading Text*

**Physical Education Class 12th**

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