Class 10 Chapter 1 Electricity Physics Solution S.Chand. 1 to 3

 Very Short Answer Type Questions

1. By what other name is the unit joule/coulomb called?

(a) The unit joule/coulomb is also known as a volt.

2. Which of the following statements correctly defines a volt?

(b) A volt is a joule per coulomb.

3. (a) What do the letters p.d. stand for?

(a) The letters p.d. stand for potential difference.

(b) A voltmeter is used to measure potential difference (p.d.).

4. What is meant by saying that the electric potential at a point is 1 volt?

By saying that the electric potential at a point is 1 volt, it means that 1 joule of work is done in moving a charge of 1 coulomb from infinity to that point against the electric field.

5. How much work is done when one coulomb charge moves against a potential difference of 1 volt?

When one coulomb charge moves against a potential difference of 1 volt, 1 joule of work is done.

6. What is the SI unit of potential difference?

The SI unit of potential difference is the volt (V).

7. How much work is done in moving a charge of 2 C across two points having a potential difference of 12 V?

When a charge of 2 coulombs is moved across two points with a potential difference of 12 volts, the work done is calculated using the formula: Work = Charge × Potential Difference = 2 C × 12 V = 24 joules.

8. What is the unit of electric charge?

The unit of electric charge is the coulomb (C).

9. Define one coulomb charge.

One coulomb charge is defined as the amount of charge passing a point in an electric circuit in one second when the current is one ampere.

10.Fill in the following blanks with suitable words:

(a) Potential difference is measured in volts by using a voltmeter placed in parallel across a component. (b) Copper is a good conductor. Plastic is an insulator.

Short Answer Type Questions

11. What is meant by conductors and insulators ? Give two examples of conductors and two of insulators.

Conductors are materials that allow the flow of electric charge, while insulators are materials that inhibit the flow of electric charge. Examples of conductors include copper and silver, while examples of insulators include rubber and glass. 12. Which of the following are conductors and which are insulators ? Sulphur, Silver, Copper, Cotton, Aluminium, Air, Nichrome, Graphite, Paper, Porcelain, Mercury, Mica, Bakelite, Polythene, Manganin.

Conductors: Silver, Copper, Aluminium, Nichrome Insulators: Sulphur, Cotton, Air, Graphite, Paper, Porcelain, Mercury, Mica, Bakelite, Polythene, Manganin 13. What do you understand by the term “electric potential” ? (or potential) at a point ? What is the unit of electric potential ?

Electric potential at a point refers to the amount of electric potential energy per unit charge at that point in an electric field. The unit of electric potential is the volt (V). 14. (a) State the relation between potential difference, work done and charge moved. (b) Calculate the work done in moving a charge of 4 coulombs from a point at 220 volts to another point at 230 volts.

(a) The relation between potential difference ($�$), work done ($�$), and charge moved ($�$) is given by $�=�\cdot �$. (b) Given ${�}_1=220$ volts and ${�}_2=230$ volts, and $�=4$ coulombs, the work done is $�=�\cdot ({�}_2-{�}_1)=4\cdot (230-220)=40$ joules. 15. (a) Name a device that helps to measure the potential difference across a conductor. (b) How much energy is transferred by a 12 V power supply to each coulomb of charge which it moves around a circuit ?

(a) A device that helps measure the potential difference across a conductor is called a voltmeter. (b) The energy transferred by a 12 V power supply to each coulomb of charge which it moves around a circuit is $12$ joules per coulomb.

Long Answer Type Question

16. (a) What do you understand by the term “potential difference” ? (b) What is meant by saying that the potential difference between two points is 1 volt ? (c) What is the potential difference between the terminals of a battery if 250 joules of work is required to transfer 20 coulombs of charge from one terminal of battery to the other ? (d) What is a voltmeter ? How is a voltmeter connected in the circuit to measure the potential difference between two points. Explain with the help of a diagram. (e) State whether a voltmeter has a high resistance or a low resistance. Give reason for your answer.

16. (a) Potential difference refers to the difference in electric potential between two points in an electric field. It represents the amount of work done per unit charge in moving a charge between those two points.

(b) Saying that the potential difference between two points is 1 volt means that one joule of work is required to move one coulomb of charge between those two points.

(c) Using the formula $�=�\cdot �$, where $�$ is the work done, $�$ is the charge, and $�$ is the potential difference, we can rearrange to find $�=\frac{�}{�}$. Substituting the given values, $�=\frac{250}{20}=12.5$ volts.

(d) A voltmeter is a device used to measure the potential difference between two points in an electric circuit. It is connected in parallel across the two points where the potential difference is to be measured. Here's a diagram illustrating the connection:

(e) A voltmeter typically has a high resistance. This is because it needs to minimize the flow of current through itself so as not to affect the potential difference being measured. If it had a low resistance, it would draw a significant amount of current and distort the measurement.

Multiple Choice Questions (MCQs)

17. The work done in moving a unit charge across two points in an electric circuit is a measure of :

(a) current (b) potential difference (c) resistance (d) power

18. The device used for measuring potential difference is known as :

(a) potentiometer (b) ammeter (c) galvanometer (d) voltmeter

19. Which of the following units could be used to measure electric charge ?

(a) ampere (b) joule (c) volt (d) coulomb

20. The unit for measuring potential difference is :

(a) watt (b) ohm (c) volt (d) kWh

21. One coulomb charge is equivalent to the charge contained in :

(a) 2.6 × 1019 electrons (b) 6.2 × 1019 electrons

(c) 2.65 × 1018 electrons (d) 6.25 × 1018 electrons

Part 2

 Very Short Answer Type Questions

1. By what other name is the unit joule/coulomb called ?

The unit joule/coulomb is also called the volt.

2. Which of the following statements correctly defines a volt ?

(a) a volt is a joule per ampere.

(b) a volt is a joule per coulomb.

(b) A volt is a joule per coulomb.

3. (a) What do the letters p.d. stand for ?

(b) Which device is used to measure p.d. ?

The letters "p.d." stand for potential difference.

A voltmeter is used to measure potential difference.

4. What is meant by saying that the electric potential at a point is 1 volt ?

It means that one joule of work is done in moving one coulomb of charge against the electric field to that point.

5. How much work is done when one coulomb charge moves against a potential difference of 1 volt ?

One joule of work is done.

6. What is the SI unit of potential difference ?

The SI unit of potential difference is the volt (V).

7. How much work is done in moving a charge of 2 C across two points having a potential difference of 12 V ?

The work done is 24 joules.

8. What is the unit of electric charge ?

The unit of electric charge is the coulomb (C).

9. Define one coulomb charge.

One coulomb charge is the amount of electric charge carried by approximately 6.24 × 10^18 elementary charges, such as electrons or protons.

Short Answer Type Questions

16. (a) Name a device which helps to maintain potential difference across a conductor (say, a bulb). (b) If a potential difference of 10 V causes a current of 2 A to flow for 1 minute, how much energy is transferred ?

16. (a) A device that helps maintain potential difference across a conductor is a voltage source, such as a battery or a power supply.

(b) The energy transferred can be calculated using the formula:

$\text{Energy}=\text{Voltage}\times \text{Current}\times \text{Time}$

Given: Voltage ($�$) = 10 V Current ($�$) = 2 A Time ($�$) = 1 minute = 60 seconds

Substituting the values: $\text{Energy}=10\text{V}\times 2\text{A}\times 60\text{s}=1200\text{J}$

So, 1200 Joules of energy is transferred.

17. (a) What is an electric current ? What makes an electric current flow in a wire ? (b) Define the unit of electric current (or Define ampere).

17. (a) Electric current is the flow of electric charge carriers (usually electrons) through a conductor. It flows when there is a potential difference (voltage) across the conductor.

(b) The unit of electric current is Ampere (A), defined as the flow of charge per unit time. 1 Ampere is equivalent to 1 Coulomb of charge passing through a point in a circuit per second.

18. What is an ammeter ? How is it connected in a circuit ? Draw a diagram to illustrate your answer.

An ammeter is a device used to measure electric current in a circuit. It is connected in series with the circuit element whose current is to be measured. Here's a diagram illustrating the connection:

19. (a) Write down the formula which relates electric charge, time and electric current. (b) A radio set draws a current of 0.36 A for 15 minutes. Calculate the amount of electric charge that flows through the circuit.

19. (a) The formula relating electric charge ($�$), time ($�$), and electric current ($�$) is:

$�=�\times �$

(b) Given: Current ($�$) = 0.36 A Time ($�$) = 15 minutes = 15 × 60 seconds = 900 seconds

Using the formula: $�=0.36\text{A}\times 900\text{s}=324\text{C}$

So, 324 Coulombs of electric charge flows through the circuit.

20. Why should the resistance of : (a) an ammeter be very small ? (b) a voltmeter be very large ?

20. (a) The resistance of an ammeter should be very small to minimize its impact on the circuit's current. If the resistance is too high, it could significantly alter the current being measured.

(b) The resistance of a voltmeter should be very large to ensure that it draws minimal current from the circuit being measured. If the resistance is too low, it could create a parallel path for current, affecting the accuracy of voltage measurement.

21. Draw circuit symbols for (a) fixed resistance (b) variable resistance (c) a cell (d) a battery of three cells (e) an open switch (f) a closed switch.

(a) Fixed resistance:

(b) Variable resistance:

(c) Cell:

(d) Battery of three cells:

(e) Open switch:

(f) Closed switch:

22. What is a circuit diagram ? Draw the labelled diagram of an electric circuit comprising of a cell, a resistor, an ammeter, a voltmeter and a closed switch (or closed plug key). Which of the two has a large resistance : an ammeter or a voltmeter ?

A circuit diagram is a graphical representation of an electrical circuit, showing how components are connected by lines. Here's a labeled diagram as described:

The ammeter (A) and voltmeter (V) are connected in series and parallel respectively, and the switch is closed. The voltmeter has a large resistance compared to the ammeter.
23. If the charge on an electron is 1.6 × 10–19 coulombs, how many electrons should pass through a conductor in 1 second to constitute 1 ampere current ?

23. To constitute 1 ampere current, a charge of $1\text{C}$ should pass through a point in 1 second. Given the charge on an electron (${�}_{�}$) is $1.6\times 10^{-19}\text{C}$, the number of electrons passing through in 1 second ($�$) can be calculated:

$1\text{C}=(1.6\times 10^{-19})\times �$

Solving for $�$:

$�=\frac{1}{1.6\times 10^{-19}}=6.25\times 10^{18}$

So, approximately $6.25\times 10^{18}$ electrons should pass through a conductor in 1 second to constitute 1 ampere current.

24. The p.d. across a lamp is 12 V. How many joules of electrical energy are changed into heat and light when : (a) a charge of 1 C passes through it ?

(b) a charge of 5 C passes through it ?

(c) a current of 2 A flows through it for 10 s ?

(a) To calculate the energy transformed when a charge of 1 Coulomb passes through the lamp, we use the formula:

$�=�\times �$

Where: $�$ = Energy transformed (in Joules) $�$ = Charge (in Coulombs) $�$ = Potential difference (in Volts)

Given $�=1$ C and $�=12$ V,

$�=1\text{C}\times 12\text{V}=12\text{J}$

So, when a charge of 1 Coulomb passes through the lamp, 12 Joules of electrical energy are transformed into heat and light.

(b) Similarly, when $�=5$ C passes through the lamp:

$�=5\text{C}\times 12\text{V}=60\text{J}$

So, when a charge of 5 Coulombs passes through the lamp, 60 Joules of electrical energy are transformed into heat and light.

(c) When a current of 2 A flows through the lamp for 10 seconds, we first find the total charge $�$ passed through:

$�=�\times �$

Where: $�$ = Current (in Amperes) $�$ = Time (in seconds)

Given $�=2$ A and $�=10$ s,

$�=2\text{A}\times 10\text{s}=20\text{C}$

Then, using the formula $�=�\times �$,

$�=20\text{C}\times 12\text{V}=240\text{J}$

So, when a current of 2 Amperes flows through the lamp for 10 seconds, 240 Joules of electrical energy are transformed into heat and light.

25. In 10 s, a charge of 25 C leaves a battery, and 200 J of energy are delivered to an outside circuit as a result. (a) What is the p.d. across the battery ? (b) What current flows from the battery ?

(a) Given that 200 J of energy are delivered to an outside circuit, and a charge of 25 C leaves the battery in 10 seconds, we use the formula for energy:

$�=�\times �$

$200\text{J}=25\text{C}\times �$

To find $�$, the potential difference (p.d.) across the battery:

$�=\frac{200\text{J}}{25\text{C}}=8\text{V}$

So, the potential difference across the battery is 8 Volts.

(b) To find the current flowing from the battery, we use Ohm's law:

$�=\frac{�}{�}$

However, we don't have resistance ($�$) given in the problem. Instead, we know the time ($�$) and charge ($�$).

$�=\frac{�}{�}$

Given $�=25$ C and $�=10$ s,

$�=\frac{25\text{C}}{10\text{s}}=2.5\text{A}$

So, the current flowing from the battery is 2.5 Amperes.

Long Answer Type Question

26. (a) Define electric current. What is the SI unit of electric current.

(b) One coulomb of charge flows through any cross-section of a conductor in 1 second. What is the current

flowing through the conductor ?

(c) Which instrument is used to measure electric current ? How should it be connected in a circuit ?

(d) What is the conventional direction of the flow of electric current ? How does it differ from the direction

of flow of electrons ?

(e) A flash of lightning carries 10 C of charge which flows for 0.01 s. What is the current ? If the voltage is

10 MV, what is the energy ?

(a) Electric current is defined as the flow of electric charge through a conductor. The SI unit of electric current is the Ampere, symbolized by "A".

(b) If one coulomb of charge flows through any cross-section of a conductor in 1 second, then the current flowing through the conductor is 1 Ampere, as 1 Ampere is defined as 1 Coulomb of charge flowing per second.

(c) An ammeter is used to measure electric current. It should be connected in series in a circuit, meaning it should be placed in the path of the current flow.

(d) The conventional direction of the flow of electric current is considered to be from the positive terminal to the negative terminal of a battery or power source. This is known as the conventional current flow. However, in reality, electric current consists of the movement of negatively charged electrons, which flow from the negative terminal to the positive terminal in the opposite direction. So, the direction of flow of electrons is opposite to the conventional direction of electric current.

(e) Given:

• Charge ($�$) = 10 C
• Time ($�$) = 0.01 s
• Voltage ($�$) = 10 MV = $10\times 10^6$ V

To find the current ($�$), we use the formula $�=\frac{�}{�}$:

$�=\frac{10\text{C}}{0.01\text{s}}=1000\text{A}$

So, the current is 1000 Amperes.

To find the energy ($�$), we use the formula $�=�\times �$:

$�=10\text{C}\times 10\text{MV}=100\text{MJ}$

So, the energy is 100 MegaJoules.

Multiple Choice Questions (MCQs)

27. The other name of potential difference is :

(a) ampereage (b) wattage (c) voltage (d) potential energy

28. Which statement/statements is/are correct ?

1. An ammeter is connected in series in a circuit and a voltmeter is connected in parallel.

2. An ammeter has a high resistance.

3. A voltmeter has a low resistance.

(a) 1, 2, 3 (b) 1, 2 (c) 2, 3 (d) 1

The correct statement is:

1. An ammeter is connected in series in a circuit, and a voltmeter is connected in parallel.

Explanation:

- An ammeter measures current and needs to be connected in series because it measures the flow of current through a circuit.

- A voltmeter measures voltage and needs to be connected in parallel because it measures the potential difference between two points in a circuit.

- Ammeters have low resistance to ensure minimal interference with the circuit's current flow.

- Voltmeters have high resistance to minimize the amount of current they draw from the circuit being measured. 

So, the correct statement is (1).

29. Which unit could be used to measure current ?

(a) Watt (b) Coulomb (c) Volt (d) Ampere

30. If the current through a floodlamp is 5 A, what charge passes in 10 seconds ?

(a) 0.5 C (b) 2 C (c) 5 C (d) 50 C

To find the charge passing through the floodlamp in 10 seconds, we can use the formula:

$�=�\times �$

Where:

• $�$ is the charge (in Coulombs)
• $�$ is the current (in Amperes)
• $�$ is the time (in seconds)

Given $�=5$ A and $�=10$ seconds,

$�=5\text{A}\times 10\text{s}=50\text{C}$

So, the charge passing through the floodlamp in 10 seconds is 50 Coulombs.

31. If the amount of electric charge passing through a conductor in 10 minutes is 300 C, the current flowing is :

(a) 30 A (b) 0.3 A (c) 0.5 A (d) 5 A

To find the current flowing through the conductor, we'll use the formula:

$�=\frac{�}{�}$

Where:

• $�$ is the current (in Amperes)
• $�$ is the charge (in Coulombs)
• $�$ is the time (in seconds)

Given $�=300$ C and $�=10$ minutes, we need to convert the time to seconds:

$�=10\times 60=600$ seconds

Now, we can calculate the current:

$�=\frac{300\text{C}}{600\text{s}}=0.5\text{A}$

So, the current flowing through the conductor is $0.5$ Amperes.

Questions Based on High Order Thinking Skills (HOTS)

32. A student made an electric circuit shown here to measure the current

through two lamps.

(a) Are the lamps in series or parallel ?

(b) The student has made a mistake in this circuit.

What is the mistake ?

(c) Draw a circuit diagram to show the correct way to connect the circuit.

Use the proper circuit symbols in your diagram.

(a) In series

(b) Ammeter is connected in parallel with the lamps.

It should be connected in series.

(c)

33. Draw a circuit diagram to show how 3 bulbs can be lit from a battery

so that 2 bulbs are controlled by the same switch while the third bulb

has its own switch.

34. An electric heater is connected to the 230 V mains supply. A current of

8 A flows through the heater.

(a) How much charge flows around the circuit each second ?

(b) How much energy is transferred to the heater each second ?

To solve this problem, we can use the following formulas:

(a) The amount of charge flowing through the circuit each second can be calculated using the formula:

$�=�\times �$

Where:

• $�$ is the charge (in Coulombs)
• $�$ is the current (in Amperes)
• $�$ is the time (in seconds)

Given $�=8$ A, and we want to find the charge per second, we can set $�=1$ second:

$�=8\text{A}\times 1\text{s}=8\text{C}$

So, 8 Coulombs of charge flow through the circuit each second.

(b) The energy transferred to the heater each second can be calculated using the formula:

$�=�\times �$

Where:

• $�$ is the energy (in Joules)
• $�$ is the power (in Watts)
• $�$ is the time (in seconds)

Given that $�=��$, where $�$ is the voltage, $�$ is the current, we can calculate $�$:

$�=�\times �=230\text{V}\times 8\text{A}=1840\text{W}$

Now, using $�=1$ second:

$�=1840\text{W}\times 1\text{s}=1840\text{J}$

So, 1840 Joules of energy are transferred to the heater each second.

35. How many electrons are flowing per second past a point in a circuit in which there is a current of 5 amp ?

To find the number of electrons flowing per second past a point in a circuit, we can use the following steps:

1. Find the charge passing through the point per second.
2. Use the elementary charge to find the number of electrons equivalent to this charge.

Given:

• Current ($�$) = 5 A

We know that current is the rate of flow of charge ($�$) per unit time ($�$). So, we can rearrange the formula for current to solve for charge per second ($�$):

$�=\frac{�}{�}$

$�=�\times �$

Given that $�=5$ A, and $�=1$ second:

$�=5\text{C}$

Now, we need to find out how many electrons are equivalent to this charge.

The elementary charge, denoted by $�$, is the charge of a single electron, which is approximately $1.6\times 10^{-19}$ Coulombs.

So, to find the number of electrons:

$\text{Number of electrons}=\frac{�}{�}$

$\text{Number of electrons}=\frac{5\text{C}}{1.6\times 10^{-19}\text{C/electron}}$

$\text{Number of electrons}\approx 3.125\times 10^{19}$

So, approximately $3.125\times 10^{19}$ electrons are flowing per second past a point in the circuit with a current of 5 amperes.

Part 3

 Very Short Answer Type Questions

1. Name the law which relates the current in a conductor to the potential difference across its ends.

Ohm's Law. 2. Name the unit of electrical resistance and give its symbol.

The unit of electrical resistance is the ohm ($\mathrm{\Omega }$). 3. Name the physical quantity whose unit is “ohm”.

Resistance. 4. What is the general name of the substances having infinitely high electrical resistance ?

Insulators. 5. Keeping the resistance constant, the potential difference applied across the ends of a component is halved. By how much does the current change ?

If the potential difference is halved, the current will also be halved. 6. State the factors on which the strength of electric current flowing in a given conductor depends.

The strength of electric current flowing in a given conductor depends on the potential difference applied across its ends and the resistance of the conductor. 7. Which has less electrical resistance : a thin wire or a thick wire (of the same length and same material) ?

A thin wire has higher electrical resistance compared to a thick wire of the same material and length. 8. Keeping the potential difference constant, the resistance of a circuit is halved. By how much does the current change ?

If the resistance is halved while the potential difference remains constant, the current will double. 9. A potential difference of 20 volts is applied across the ends of a resistance of 5 ohms. What current will flow in the resistance ?

Using Ohm's Law, $�=\frac{�}{�}=\frac{20\text{V}}{5\mathrm{\Omega }}=4\text{A}$. 10. A resistance of 20 ohms has a current of 2 amperes flowing in it. What potential difference is there between its ends ?

using Ohm's Law, $�=��=2\text{A}\times 20\mathrm{\Omega }=40\text{V}$. 11. A current of 5 amperes flows through a wire whose ends are at a potential difference of 3 volts. Calculate the resistance of the wire.

Once more using Ohm's Law, $�=\frac{�}{�}=\frac{3\text{V}}{5\text{A}}=0.6\mathrm{\Omega }$.

12. Fill in the following blank with a suitable word : Ohm’s law states a relation between potential difference and ..........current............

Short Answer Type Questions

13. Distinguish between good conductors, resistors and insulators. Name two good conductors, two resistors and two insulators.

Good Conductors: Good conductors are materials that allow electric current to flow through them easily. They typically have low resistance. Examples include metals like copper and silver.

Resistors: Resistors are components that are designed to have a specific resistance to limit the flow of current in a circuit. They are used to control the amount of current in a circuit. Examples include carbon composition resistors and wirewound resistors.

Insulators: Insulators are materials that do not conduct electricity well. They have high resistance and do not allow electric current to flow through them easily. Examples include rubber and glass.

Examples:

• Good conductors: Copper, Silver
• Resistors: Carbon composition resistor, Wirewound resistor
• Insulators: Rubber, Glass

14. Classify the following into good conductors, resistors and insulators : Rubber, Mercury, Nichrome, Polythene, Aluminium, Wood, Manganin, Bakelite, Iron, Paper, Thermocol, Metal coin

Good Conductors: Mercury, Aluminium, Iron (although it is not as good as Mercury or Aluminium)

Resistors: Nichrome, Manganin

Insulators: Rubber, Polythene, Wood, Bakelite, Paper, Thermocol

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

15. Ohm's Law: Ohm's law states that the current ($�$) flowing through a conductor is directly proportional to the potential difference ($�$) applied across its ends, provided the temperature remains constant. Mathematically, Ohm's law is expressed as:

$�=��$

where:

• $�$ is the potential difference (voltage) across the conductor,
• $�$ is the current flowing through the conductor, and
• $�$ is the resistance of the conductor.

How Ohm's Law Defines the Unit of Resistance: Ohm's law can be rearranged to solve for resistance ($�$):

$�=\frac{�}{�}$

Thus, resistance ($�$) is defined as the ratio of potential difference ($�$) to current ($�$). Therefore, using Ohm's law, the unit of resistance ($�$) can be derived from the units of potential difference ($�$) and current ($�$). By definition, 1 ohm ($\mathrm{\Omega }$) is the resistance of a conductor in which a potential difference of 1 volt (V) produces a current of 1 ampere (A).

16. (a) What is meant by the “resistance of a conductor” ? Write the relation between resistance, potential difference and current. (b) When a 12 V battery is connected across an unknown resistor, there is a current of 2.5 mA in the circuit. Calculate the value of the resistance of the resistor.

(a) Resistance of a Conductor: The resistance of a conductor is a measure of its opposition to the flow of electric current. It depends on the material, length, and cross-sectional area of the conductor. The relation between resistance ($�$), potential difference ($�$), and current ($�$) is given by Ohm's law:

$�=\frac{�}{�}$

(b) Calculation of Resistance: Given $�=12$ V and $�=2.5$ mA ($2.5\times 10^{-3}$ A), we can use Ohm's law to find the resistance ($�$):

$�=\frac{�}{�}=\frac{12\text{V}}{2.5\times 10^{-3}\text{A}}=4800\mathrm{\Omega }$

So, the value of the resistance of the resistor is $4800\mathrm{\Omega }$.

17. (a) Define the unit of resistance (or Define the unit “ohm”). (b) What happens to the resistance as the conductor is made thinner ? (c) Keeping the potential difference constant, the resistance of a circuit is doubled. By how much does the current change ?

(a) Unit of Resistance (ohm): The unit of resistance is the ohm ($\mathrm{\Omega }$). One ohm is defined as the resistance between two points in a conductor where a potential difference of 1 volt (V) produces a current of 1 ampere (A) in the conductor.

(b) Effect of Thickness on Resistance: As the conductor is made thinner, its resistance increases. This is because a thinner conductor offers more opposition to the flow of electric current compared to a thicker conductor.

(c) Change in Current with Doubled Resistance: If the resistance of a circuit is doubled while keeping the potential difference constant, the current will be halved. This is because of the inverse relationship between resistance and current, as described by Ohm's law. Specifically, if resistance doubles, the current will decrease by half.

18. (a) Why do electricians wear rubber hand gloves while working with electricity ? (b) What p.d. is needed to send a current of 6 A through an electrical appliance having a resistance of 40 ohm

(a) Use of Rubber Hand Gloves: Electricians wear rubber hand gloves while working with electricity to insulate themselves from electric shocks. Rubber is an insulator and does not conduct electricity, so wearing rubber gloves prevents the flow of electric current through the body, thereby protecting the electrician from electric shock.

(b) Potential Difference Required: To send a current of 6 A through an electrical appliance with a resistance of 40 ohms ($�=40\mathrm{\Omega }$), we can use Ohm's law:

$�=��=6\text{A}\times 40\mathrm{\Omega }=240\text{V}$

Therefore, a potential difference of 240 volts ($�$) is needed to send a current of 6 amperes ($�$) through the electrical appliance.

19. An electric circuit consisting of a 0.5 m long nichrome wire XY, an ammeter, a voltmeter, four cells of 1.5 V each and a plug key was set up. (i) Draw a diagram of this electric circuit to study the relation between the potential difference maintained between the points ‘X’ and ‘Y’ and the electric current flowing through XY. (ii) Following graph was plotted between V and I values :

What would be the values of V/I ratios when the potential difference is 0.8 V, 1.2 V and 1.6 V respectively ?

What conclusion do you draw from these values ?

(iii) What is the resistance of the wire ?

(i) Diagram of the electric circuit:

(ii) The V-I graph indicates the relationship between potential difference (V) and current (I) for the nichrome wire XY. To find the V/I ratios when the potential difference is 0.8 V, 1.2 V, and 1.6 V respectively, we need to divide the potential difference by the corresponding current.

If the potential difference is 0.8 V: $\frac{�}{�}=\frac{0.8\text{V}}{{�}_1}$

If the potential difference is 1.2 V: $\frac{�}{�}=\frac{1.2\text{V}}{{�}_2}$

If the potential difference is 1.6 V: $\frac{�}{�}=\frac{1.6\text{V}}{{�}_3}$

(iii) To find the resistance of the wire, we can use Ohm's law, $�=\frac{�}{�}$, where $�$ is the potential difference and $�$ is the current.

Given the potential difference $�$ and the length of the wire, we can find the current using the ammeter reading. Then we can calculate the resistance using Ohm's law.

Without specific values for potential difference, current, and the length of the wire, we cannot provide an exact value for the resistance. However, using the given setup and measured values from the experiment, one can calculate the resistance of the wire.

Long Answer Type Question

20. (a) What is the ratio of potential difference and current known as ?

(b) The values of potential difference V applied across a resistor and the correponding values of current I

flowing in the resistor are given below :

Potential difference, V (in volts) : 2.5 5.0 10.0 15.0 20.0 25.0

Current, I (in amperes) : 0.1 0.2 0.4 0.6 0.8 1.0

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

(c) Name the law which is illustrated by the above V–I graph.

(d) Write down the formula which states the relation between potential difference, current and resistance.

(e) The potential difference between the terminals of an electric iron is 240 V and the current is 5.0 A. What

is the resistance of the electric iron ?

(a) The ratio of potential difference and current is known as resistance.

(b) To plot the graph between potential difference ($�$) and current ($�$), and calculate the resistance, we can use the given values:

We can then use Ohm's law ($�=\frac{�}{�}$) to calculate the resistance for each pair of $�$ and $�$ values. The resistance is the slope of the graph.

(c) The law illustrated by the above $�-�$ graph is Ohm's Law.

(d) Ohm's Law states the relation between potential difference ($�$), current ($�$), and resistance ($�$) in an electrical circuit. The formula is:

$�=�\times �$

or equivalently:

$�=\frac{�}{�}$

(e) Using Ohm's law, we can find the resistance ($�$) of the electric iron:

Given:

• Potential difference, $�$ = 240 V
• Current, $�$ = 5.0 A

Using the formula $�=\frac{�}{�}$:

$�=\frac{240\text{V}}{5.0\text{A}}=48\mathrm{\Omega }$

So, the resistance of the electric iron is $48\mathrm{\Omega }$.

Multiple Choice Questions (MCQs)

21. The p.d. across a 3  resistor is 6 V. The current flowing in the resistor will be :

(a)

1 A

2 (b) 1 A (c) 2 A (d) 6 A

Given:

• $�=3\mathrm{\Omega }$ (resistance of the resistor)
• $�=6\text{V}$ (potential difference across the resistor)

We need to solve for $�$:

$�=\frac{�}{�}$

$�=\frac{6\text{V}}{3\mathrm{\Omega }}$

$�=2\text{A}$

So, the current flowing through the resistor is $2\text{A}$.

22. A car headlight bulb working on a 12 V car battery draws a current of 0.5 A. The resistance of the light bulb

is :

(a) 0.5Ω (b) 6Ω (c) 12Ω (d) 24Ω

Given:

• $�=12\text{V}$ (potential difference across the light bulb)
• $�=0.5\text{A}$ (current flowing through the light bulb)

We need to solve for $�$:

$�=\frac{�}{�}$

$�=\frac{12\text{V}}{0.5\text{A}}$

$�=24\mathrm{\Omega }$

So, the resistance of the light bulb is $24\mathrm{\Omega }$.

23. An electrical appliance has a resistance of 25 Ω. When this electrical appliance is connected to a 230 V

supply line, the current passing through it will be :

(a) 0.92 A (b) 2.9 A (c) 9.2 A (d) 92 A

Given:

• $�=25\mathrm{\Omega }$ (resistance of the electrical appliance)
• $�=230\text{V}$ (potential difference or voltage of the supply line)

We need to solve for $�$:

$�=\frac{�}{�}$

$�=\frac{230\text{V}}{25\mathrm{\Omega }}$

$�=9.2\text{A}$

So, the current passing through the electrical appliance will be $9.2\text{A}$.

24. When a 4 Ω resistor is connected across the terminals of a 12 V battery, the number of coulombs passing

through the resistor per second is :

(a) 0.3 (b) 3 (c) 4 (d) 12

Given:

• $�=4\mathrm{\Omega }$ (resistance of the resistor)
• $�=12\text{V}$ (potential difference or voltage of the battery)

We need to solve for $�$:

$�=\frac{�}{�}$

$�=\frac{12\text{V}}{4\mathrm{\Omega }}$

$�=3\text{A}$

Now, to find the number of coulombs passing through the resistor per second, we use the formula:

$\text{Charge}=�\times \text{Time}$

Given that the time interval is one second, we have:

$\text{Charge}=3\text{A}\times 1\text{s}=3\text{C}$

So, the number of coulombs passing through the resistor per second is $3\text{C}$.

25. Ohm’s law gives a relationship between :

(a) current and resistance

(b) resistance and potential difference

(c) potential difference and electric charge

(d) current and potential difference

26. The unit of electrical resistance is :

(a) ampere (b) volt (c) coulomb (d) ohm

27. The substance having infinitely high electrical resistance is called :

(a) conductor (b) resistor (c) superconductor (d) insulator

28. Keeping the potential difference constant, the resistance of a circuit is doubled. The current will become :

(a) double (b) half (c) one-fourth (d) four times

29. Keeping the p.d. constant, the resistance of a circuit is halved. The current will become :

(a) one-fourth (b) four times (c) half (d) double

Questions Based on High Order Thinking Skills (HOTS)

30. An electric room heater draws a current of 2.4 A from the 120 V supply line. What current will this room

heater draw when connected to 240 V supply line ?

To find the current drawn by the room heater when connected to a 240 V supply line, we can use the concept of power being constant (assuming the resistance of the heater remains the same).

The formula for power ($�$) in an electrical circuit is given by:

$�=�\times �$

Where:

• $�$ is the power (constant)
• $�$ is the voltage
• $�$ is the current

Given that the power consumed by the heater remains constant when connected to different voltage supply lines, we can use the formula:

${�}_1={�}_2$

Where ${�}_1$ is the power when connected to the 120 V supply line, and ${�}_2$ is the power when connected to the 240 V supply line.

For the 120 V supply line: ${�}_1={�}_1\times {�}_1$ ${�}_1=120\times 2.4$ ${�}_1=288\text{W}$

Now, for the 240 V supply line: ${�}_2={�}_2\times {�}_2$ $288=240\times {�}_2$ ${�}_2=\frac{288}{240}$ ${�}_2=1.2\text{A}$

So, when connected to the 240 V supply line, the room heater will draw a current of $1.2\text{A}$.

31. Name the electrical property of a material whose symbol is “omega”.

The electrical property of a material whose symbol is "omega" is resistance.

The symbol for resistance in electrical circuits is represented by the Greek letter omega ($\mathrm{\Omega }$).

32. The graph between V and I for a conductor is a straight line passing through the origin.

(a) Which law is illustrated by such a graph ?

(b) What should remain constant in a statement of this law ?

(a) The law illustrated by a graph between voltage ($�$) and current ($�$) for a conductor that is a straight line passing through the origin is Ohm's Law.

(b) In a statement of Ohm's Law, the resistance ($�$) of the conductor should remain constant. Ohm's Law states that at a constant temperature, the current through a conductor between two points is directly proportional to the voltage across the two points. Mathematically, this can be expressed as:

$�=�\times �$

Where:

• $�$ is the potential difference (voltage),
• $�$ is the current, and
• $�$ is the resistance.

Therefore, to apply Ohm's Law, the resistance of the conductor must remain constant.

33. A p.d. of 10 V is needed to make a current of 0.02 A flow through a wire. What p.d. is needed to make a

current of 250 mA flow through the same wire ?

To find the potential difference (p.d.) needed to make a current of 250 mA flow through the same wire, we can use Ohm's Law, which states:

$�=�\times �$

Where:

• $�$ is the potential difference (voltage),
• $�$ is the current, and
• $�$ is the resistance.

Given that the resistance of the wire remains constant, we can use the ratio of potential difference to current to find the new potential difference:

${�}_1={�}_1\times �$ ${�}_2={�}_2\times �$

Since ${�}_1=10$ V and ${�}_1=0.02$ A, we can find the resistance using Ohm's Law:

$�=\frac{{�}_1}{{�}_1}=\frac{10\text{V}}{0.02\text{A}}=500\mathrm{\Omega }$

Now, using the same resistance ($�$) but a new current (${�}_2=250\text{mA}=0.25\text{A}$), we can find the new potential difference (${�}_2$):

${�}_2={�}_2\times �=0.25\text{A}\times 500\mathrm{\Omega }=125\text{V}$

So, a potential difference of $125\text{V}$ is needed to make a current of $250\text{mA}$ flow through the same wire.

34. A current of 200 mA flows through a 4 kΩ resistor. What is the p.d. across the resistor ?

To find the potential difference (p.d.) across the resistor, we can use Ohm's Law, which states:

$�=�\times �$

Where:

• $�$ is the potential difference (voltage),
• $�$ is the current, and
• $�$ is the resistance.

Given that the current $�$ flowing through the resistor is $200\text{mA}=0.2\text{A}$ and the resistance $�$ of the resistor is $4\text{k}\mathrm{\Omega }=4000\mathrm{\Omega }$, we can use Ohm's Law to find the potential difference $�$:

$�=�\times �=0.2\text{A}\times 4000\mathrm{\Omega }=800\text{V}$

So, the potential difference across the resistor is $800\text{V}$.

$$$\mathrm{\Omega }$$\mathrm{\Omega }$Ω