Chapter 11 - Electricity Class 10 Notes (Science)
Ultimate NCERT Solutions for Chapter 11 Electricity Class 10 Notes
Updated Solution 2024-2025 Updated Solution 2024-2025
NCERT Solutions for Science, Chapter 11 Electricity Class 10 Notes, Question/Answers, Activity, Experiment’s & Science Projects
Chapter 11 Electricity
Questions
Q 1. What does an electric circuit mean?
Ans 1: An electric circuit is a closed path that allows electricity to flow. It usually includes a power source (like a battery), wires, and devices (like bulbs or motors) that use the electricity. If the path is complete, the current flows; if it’s broken, the flow stops.
Q 2. Define the unit of current.
Ans 2: The unit of current is the ampere (A). It measures the flow of electric charge through a conductor. One ampere equals the flow of 1 coulomb of charge per second.
Q 3. Calculate the number of electrons constituting one coulomb of charge.
Ans 3: To find the number of electrons in one coulomb of charge, we use the charge of a single electron:
The charge of one electron is = 1.6 × 10⁻¹⁹ coulombs
To calculate the number of electrons (n) in 1 coulomb of charge:
n = Total charge / Charge of one electron
n = 1 / (1.6 × 10⁻¹⁹)
n = 6.25 × 10¹⁸ electrons
Conclusion: One coulomb of charge is made up of 6.25 × 10¹⁸ electrons.
Questions
Q 1. Name a device that helps to maintain a potential difference across a conductor.
Ans 1: A battery or a cell is a device that maintains a potential difference across a conductor, enabling the flow of electric current.
Q 2. What is meant by saying that the potential difference between two points is 1 V?
Ans 2: When we say the potential difference between two points is 1 volt (V), it means that 1 joule of energy is needed to move 1 coulomb of electric charge between those points. In simple terms, it’s a measure of how much energy is required to push a charge from one point to another in an electric field. A higher potential difference means more energy is needed.
Q 3. How much energy is given to each coulomb of charge passing through a 6 V battery?
Ans 3: To find the energy given to each coulomb of charge passing through a 6 V battery, we use the formula:
Energy (in joules) = Voltage (in volts) × Charge (in coulombs)
Here, the voltage is 6 V, and the charge is 1 coulomb (since we’re looking at the energy per coulomb).
So:
Energy = 6 V × 1 C = 6 joules
This means that for every coulomb of charge passing through the 6 V battery, 6 joules of energy are given to it.
Activity 11.1
- Set up a circuit as shown in Fig. 11.2, consisting of a nichrome wire XY of length, say 0.5 m, an ammeter, a voltmeter and four cells of 1.5 V each. (Nichrome is an alloy of nickel, chromium, manganese, and iron metals.)
- First use only one cell as the source in the circuit. Note the reading in the ammeter I, for the current and reading of the voltmeter V for the potential difference across the nichrome wire XY in the circuit. Tabulate them in the Table given.
- Next connect two cells in the circuit and note the respective readings of the ammeter and voltmeter for the values of current through the nichrome wire and potential difference across the nichrome wire.
- Repeat the above steps using three cells and then four cells in the circuit separately.
- Calculate the ratio of V to I for each pair of potential difference V and current I.
- Plot a graph between V and I, and observe the nature of the graph.
S. No. | Number of cells used in the circuit | Current through the nichrome wire, I (ampere) | Potential difference across the nichrome wire, V (volt) | V/I (volt/ampere) |
1 | 1 |
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2 | 2 |
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3 | 3 |
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4 | 4 |
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Questions And Answers related to Activity 11.1:
Q 1: What is the purpose of Activity 11.1?
Ans 1: The purpose of Activity 11.1 is to investigate the relationship between the current through a nichrome wire and the potential difference across it. This helps to verify Ohm’s Law, which states that V ∝ I, provided the physical conditions like temperature remain constant.
Q 2: Why is nichrome used in this experiment?
Ans 2: Nichrome is used because it is an alloy with high resistivity and stability at high temperatures, making it suitable for studying the relationship between current and potential difference.
Q 3: How is the ratio interpreted in this experiment?
Ans 3: The ratio represents the resistance of the nichrome wire. According to Ohm’s Law, this ratio should remain constant for a given wire if temperature and material properties are constant.
Q 4: What is the significance of plotting the graph between V and I?
Ans 4: The graph helps to visualize the relationship between V and I. If the graph is a straight line passing through the origin, it confirms that the nichrome wire obeys Ohm’s Law.
Q 5: Why do we use multiple cells in the circuit?
Ans 5: Using multiple cells allows us to vary the potential difference across the nichrome wire and observe the corresponding changes in current. This helps in studying the proportional relationship between V and I.
Q 6: What type of graph is expected between V and I?
Ans 6: The graph is expected to be a straight line passing through the origin, indicating a linear relationship between V and I, as predicted by Ohm’s Law.
Q 7: What precautions should be taken during the experiment?
Ans 7:
- Ensure all connections are tight and secure.
- Avoid heating the nichrome wire excessively, as temperature changes can affect resistance.
- Use the correct range for the ammeter and voltmeter to get accurate readings.
- Ensure the circuit is switched off while changing the number of cells.
Q 8: If the ratio is not constant, what could be the possible reasons?
Ans 8:
- The nichrome wire may have heated up, altering its resistance.
- Faulty readings from the voltmeter or ammeter.
- Poor connections or fluctuating power supply.
Q 9: What conclusions can be drawn from the observations?
Ans 9: If ratio remains constant and the V – I graph is a straight line, it confirms that the nichrome wire obeys Ohm’s Law, and the wire’s resistance is constant under the given conditions.
Q 10: What is the formula used to calculate resistance in this activity?
Ans 10: Resistance is calculated using Ohm’s Law: R = V / I
where R is resistance, V is potential difference, and I is current.
Q 11: Calculate the ratio of V to I for each pair of potential difference V and current I.
Ans 11: This experiment aims to demonstrate the relationship between potential difference 𝑉 and current 𝐼 in a nichrome wire, verifying Ohm’s Law. Below is a hypothetical table illustrating the process, assuming measured data based on typical results.
Hypothetical Data Table
S. No. | Number of Cells Used | Current Through Nichrome Wire, I (A) | Potential Difference Across Nichrome Wire, V (V) | Ratio V / I (Ω) |
1 | 1 | 0.5 | 1.5 | 1.5 / 0.5 = 3.0 |
2 | 2 | 1.0 | 3.0 | 3.0 / 1.0 = 3.0 3 |
3 | 3 | 1.5 | 4.5 | 4.5 / 1.5 = 3.0 |
4 | 4 | 2.0 | 6.0 | 6.0 / 2.0 = 3.0 |
Activity 11.2
- Take a nichrome wire, a torch bulb, a 10 W bulb and an ammeter (0 – 5 A range), a plug key and some connecting wires.
- Set up the circuit by connecting four dry cells of 1.5 V each in series with the ammeter leaving a gap XY in the circuit, as shown in Fig. 11.4.
- Complete the circuit by connecting the nichrome wire in the gap XY. Plug the key. Note down the ammeter reading. Take out the key from the plug. [Note: Always take out the key from the plug after measuring the current through the circuit.]
- Replace the nichrome wire with the torch bulb in the circuit and find the current through it by measuring the reading of the ammeter.
- Now repeat the above step with the 10 W bulb in the gap XY.
- Are the ammeter readings different for different components connected in the gap XY? What do the above observations indicate?
- you may repeat this Activity by keeping any material component in the gap. Observe the ammeter readings in each case. Analyze the observations.
Activity 11.2 Answers:
Q 1. What is the procedure for measuring current using a nichrome wire in the gap XY?
Ans 1:
- Complete the circuit by connecting the nichrome wire in the gap XY.
- Plug the key to allow current to flow through the circuit.
- Observe and record the ammeter reading.
- After noting the reading, take out the key from the plug.
Q 2. How do you measure the current when the torch bulb is connected in the gap XY?
Ans 2:
- Replace the nichrome wire in the gap XY with a torch bulb.
- Plug the key to complete the circuit.
- Observe and record the ammeter reading.
- After measuring, remove the key from the plug.
Q 3. How do you measure the current when the 10 W bulb is connected in the gap XY?
Ans 3:
- Replace the torch bulb with the 10 W bulb in the gap XY.
- Plug the key to complete the circuit.
- Observe and record the ammeter reading.
- Remove the key after noting the reading.
Q 4. Are the ammeter readings different for the nichrome wire, torch bulb, and 10 W bulb?
Ans 4: Yes, the ammeter readings are different for each component.
Q 5. What do the different readings of the ammeter indicate?
Ans 5: The different readings indicate that the current through the circuit depends on the resistance of the component connected in the gap XY.
- Nichrome wire has a high resistance, resulting in lower current.
- The torch bulb has a medium resistance, so the current is relatively higher.
- The 10 W bulb, being designed for higher power, typically has lower resistance, resulting in the highest current among the three.
Q 6. What happens if you repeat the activity with different material components in the gap XY?
Ans 6: When the activity is repeated with different materials in the gap XY, the ammeter readings vary for each material.
- Conducting materials (e.g., copper wire) show higher current due to low resistance.
- Insulating materials (e.g., rubber or plastic) do not allow current to flow, resulting in zero ammeter reading.
- Semi-conducting materials show moderate current depending on their resistance.
Q 7. What conclusion can be drawn from the observations in this activity?
Ans 7: The observations indicate that:
- Current in a circuit depends on the resistance of the material/component connected.
- Different materials/components have different resistances, which affect the amount of current flowing through the circuit.
- Materials can be classified as conductors, insulators, or semi-conductors based on their ability to conduct electric current.
Chapter 11– Electricity Class 10 Notes, Question/Answer, Activity, Experiment’s & Projects
Updated Solutions 2024-2025 Updated Solutions 2024-2025
Updated Solutions 2024-2025 Updated Solutions 2024-2025
Activity 11.3
- Complete an electric circuit consisting of a cell, an ammeter, a nichrome wire of length l and a plug key, as shown in Fig. 11.5.
- Now, plug the key. Note the current in the ammeter.
- Replace the nichrome wire by another nichrome wire of same thickness but twice the length, that is 2l [marked (2) in the Fig. 11.5].
- Note the ammeter reading.
- Now replace the wire by a thicker nichrome wire, of the same length l [marked (3)]. A thicker wire has a larger cross-sectional area. Again, note down the current through the circuit.
- Instead of taking a nichrome wire, connect a copper wire [marked (4) in Fig. 11.5] in the circuit. Let the wire be of the same length and same area of cross-section as that of the first nichrome wire [marked (1)]. Note the value of the current.
- Notice the difference in the current in all cases.
- Does the current depend on the length of the conductor?
- Does the current depend on the area of cross-section of the wire used?
Activity 11.3 Base Question Answers:
Q 1. What happens to the current when the length of the nichrome wire is doubled?
Ans 1: When the length of the nichrome wire is doubled, the resistance increases because resistance is directly proportional to the length of the wire. As a result, the current decreases.
Q 2. How does the thickness (or cross-sectional area) of the wire affect the current in the circuit?
Ans 2: A thicker wire (with a larger cross-sectional area) reduces the resistance, as resistance is inversely proportional to the area of cross-section. Therefore, the current increases when a thicker wire is used.
Q 3. How does the material of the wire (nichrome vs copper) influence the current in the circuit?
Ans 3: Copper has a lower resistivity compared to nichrome. This means copper allows more current to flow through the same length and thickness of wire, while nichrome offers more resistance. Hence, the current is higher in the copper wire compared to the nichrome wire.
Q 4. Does the current depend on the length of the conductor?
Ans 4: Yes, the current depends on the length of the conductor. As the length of the wire increases, the resistance increases, which causes the current to decrease.
Q 5. Does the current depend on the area of cross-section of the wire used?
Ans 5: Yes, the current depends on the area of cross-section. A larger cross-sectional area reduces the resistance, allowing more current to flow. Conversely, a smaller cross-sectional area increases the resistance and reduces the current.
Questions
Q 1. On what factors does the resistance of a conductor depend?
Ans 1: The resistance of a conductor depends on these factors:
- Length of the conductor: Longer wires have more resistance.
- Cross-sectional area: Thicker wires have less resistance.
- Material of the conductor: Materials like copper have lower resistance compared to steel.
- Temperature: Higher temperatures usually increase resistance.
These factors determine how easily electric current can flow through a conductor.
Q 2. Will current flow more easily through a thick wire or a thin wire of the same material, when connected to the same source? Why?
Ans 2: Current will flow more easily through a thick wire than a thin one, even if both are made of the same material and connected to the same source. This is because a thicker wire has a larger cross-sectional area, allowing more room for the electrons to move through. In contrast, a thin wire has less space, causing more resistance and making it harder for the current to flow. So, with less resistance, the current can travel more freely through the thicker wire.
Q 3. 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?
Ans 3: When the resistance of an electrical component stays constant, the current depends directly on the potential difference (voltage), as per Ohm’s Law: I = V / R
If the voltage decreases to half while the resistance remains the same, the current will also decrease to half of its original value.
In short: If the voltage is halved, the current is also halved, provided the resistance doesn’t change.
Q 4. Why are coils of electric toasters and electric irons made of an alloy rather than a pure metal?
Ans 4: The coils of electric toasters and irons are made of alloys instead of pure metals because alloys have higher resistance, which helps them heat up efficiently. They also have a higher melting point and don’t oxidize (rust) easily at high temperatures, making them more durable and safer for long-term use.
Q 5. Use the data in Table 11.2 to answer the following –
(a) Which among iron and mercury is a better conductor?
Ans (a): ron has a resistivity of 10.0 × 10⁻⁸ Ω·m, while mercury has a resistivity of 94.0 × 10⁻⁸ Ω·m.
Since mercury has a higher resistivity, iron is a better conductor than mercury.
(b) Which material is the best conductor?
Ans (b): The best conductor from the materials listed is silver, which has the lowest resistivity of 1.60 × 10⁻⁸ Ω·m.
Activity 11.4
- Join three resistors of different values in series. Connect them with a battery, an ammeter and a plug key, as shown in Fig. 11.6. You may use the resistors of values like 1 Ω, 2 Ω, 3 Ω etc., and a battery of 6 V for performing this Activity.
Figure 11.6 resistor in series
- Plug the key. Note the ammeter reading.
- Change the position of ammeter to anywhere in between the resistors. Note the ammeter reading each time.
- Do you find any change in the value of current through the ammeter?
Here are some questions and answers based on the activity 11.4:
Q 1: What happens to the current when the ammeter is placed at different positions in the series circuit?
Ans 1: The current remains the same at all positions of the ammeter in a series circuit. This is because, in a series circuit, the current is constant throughout, regardless of where the ammeter is placed.
Q 2: Why is the current the same at different positions in a series circuit?
Ans 2: In a series circuit, the same current flows through all components because the same path is shared by the current. The resistors only affect the potential difference (voltage), not the current.
Q 3: What is the relationship between the total resistance and the individual resistances when resistors are connected in series?
Ans 3: The total resistance in a series circuit is the sum of the individual resistances: Rtotal = R1+R2+R3
For example, if three resistors of values 1 Ω, 2 Ω, and 3 Ω are connected in series, the total resistance will be 6 Ω.
Q 4: What can you conclude about the current in a series circuit when different resistors are used?
Ans 4: The current depends on the total resistance and the voltage applied. The higher the total resistance, the lower the current. For a constant voltage, increasing the total resistance decreases the current according to Ohm’s law: I = V / Rtotal
Q 5: How does the value of the ammeter reading change when different resistors are used in the series?
Ans 5: The ammeter reading will decrease if the total resistance is increased (by using resistors with higher values), and increase if the total resistance is decreased. This is in accordance with Ohm’s law.
Q 6: Why is it important to check the ammeter at different positions in a series circuit?
Ans 6: Checking the ammeter at different positions confirms that the current remains the same at all points in a series circuit, validating the principle of current conservation in series connections.
Q 7: What will happen if the key is opened (disconnected)?
Ans 7: If the key is opened, the circuit will be broken, and the current will stop flowing, leading to a zero reading on the ammeter.
Activity 11.5
- In Activity 11.4, insert a voltmeter across the ends X and Y of the series combination of three resistors, as shown in Fig. 11.6.
Figure 11.6 resistor in series
- Plug the key in the circuit and note the voltmeter reading. It gives the potential difference across the series combination of resistors. Let it be V. Now measure the potential difference across the two terminals of the battery. Compare the two values.
- Take out the plug key and disconnect the voltmeter. Now insert the voltmeter across the ends X and P of the first resistor, as shown in Fig. 11.8.
Figure 11.8
- Plug the key and measure the potential difference across the first resistor. Let it be V 1.
- Similarly, measure the potential difference across the other two resistors, separately. Let these values be V2 and V3, respectively.
- Deduce a relationship between V, V1, V2 and V3
Here are some questions and answers based on the activity 11.4:
Q 1. Why is a voltmeter used in the circuit?
Ans 1: A voltmeter is used to measure the potential difference across different components in the circuit.
Q 2. What does the reading of the voltmeter across the combination of resistors (V) indicate?
Ans 2: The voltmeter reading V indicates the total potential difference across the series combination of the three resistors.
Q 3. Why is it necessary to disconnect the voltmeter when measuring the potential difference across individual resistors?
Ans 3: The voltmeter should be disconnected to avoid interference and ensure accurate readings for individual resistors.
Q 4. What relationship can be deduced between V, V1, V2, and V3?
Ans 4: In a series circuit, the total potential difference V is equal to the sum of the potential differences across each resistor.
V = V1 + V2 + V3
Q 5. How does the potential difference across the battery compare with the potential difference across the series combination of resistors?
Ans 5: The potential difference across the battery is equal to the total potential difference across the series combination of resistors, provided there is no significant internal resistance in the battery.
Q 6. Why are resistors connected in series in this experiment?
Ans 6: Resistors are connected in series to demonstrate how the total potential difference in the circuit is distributed across individual resistors based on their resistances.
Q 7. What is the significance of measuring the potential difference across each resistor separately?
Ans 7: Measuring the potential difference across each resistor helps verify the relationship that the total potential difference is the sum of the potential differences across all individual resistors in series.
Questions
Q 1. Draw a schematic diagram of a circuit consisting of a battery of three cells of 2 V each, a 5Ωresistor, an 8Ωresistor, and a 12Ωresistor, and a plug key, all connected in series.
Ans 1:
Q 2. Redraw the circuit of Question 1, putting in an ammeter to measure the current through the resistors and a voltmeter to measure the potential difference across the 12Ωresistor. What would be the readings in the ammeter and the voltmeter?
Ans 2:
Given:
Resistors in series: 5 Ω, 8 Ω, and 12 Ω
Total voltage (V) = 6 volts
Equivalent resistance = 5 + 8 + 12
R = 25Ω
We know, I = V / R
I = 6 /25
I = 0.24 ampere
P.D across 12 Ω resistor = IR
= 0.24 × 12
= 2.88 volts
Ammeter reads 0.24 A and voltmeter reads 2.88 V.
Activity 11.6
- Make a parallel combination, XY, of three resistors having resistances , , and , respectively. Connect it with a battery, a plug key and an ammeter, as shown in Fig. 11.10. Also connect a voltmeter in parallel with the combination of resistors.
- Plug the key and note the ammeter reading. Let the current be I. Also take the voltmeter reading. It gives the potential difference V, across the combination. The potential difference across each resistor is also V. This can be checked by connecting the voltmeter across each individual resistor (see Fig. 11.11).
- Take out the plug from the key. Remove the ammeter and voltmeter from the circuit. Insert the ammeter in series with the resistor , as shown in Fig. 11.11. Note the ammeter reading, . Figure 11.11
- Similarly, measure the currents through R₂ and R₃. Let these be I₂ and I₃, respectively. What is the relationship between I = I₁ = I₂ = I₃ ?
Questions and Answers related to Activity 11.6 on the parallel combination of resistors:
Q 1. What is the total current (I) in the circuit?
Ans 1: The total current I in the circuit is the sum of the individual currents I₁, I₂ and I₃ through the three resistors. Since the resistors are connected in parallel, the total current III is given by:
I = I₁ + I₂ + I₃
This relationship follows from the principle that in a parallel circuit, the total current is the sum of the currents through each branch.
Q 2. What is the potential difference across the parallel combination of resistors?
Ans 2: The potential difference V across the parallel combination of resistors is the same for each resistor in the combination. This can be verified by connecting a voltmeter across each individual resistor, as the voltage drop across each resistor in parallel is equal.
Q 3. What is the current through each resistor in a parallel circuit?
Ans 3: In a parallel circuit, the current through each resistor depends on the resistance of the individual resistors and the total potential difference V. The current through each resistor is given by Ohm’s Law:
I₁ = V / R₁
I₂ = V / R₂
I₃ = V / R₃
Here,
I₁, I₂, I₃ are the currents through R₁, R₂, and R₃, respectively.
V is the total voltage applied across the resistors.
Since the same current flows through all resistors in a series circuit, the above formulas show how the voltage affects the current through each individual resistor.
Q 4. Why is the current the sum of individual currents in a parallel combination?
Ans 4: In a parallel circuit, the total current is the sum of the individual currents because the current splits at the junctions where the resistors are connected. The current follows each path according to the resistance of each branch. Since the potential difference across all resistors is the same, the total current is simply the sum of the currents through each resistor.
Q 5. How do you verify that the potential difference across each resistor in a parallel combination is the same?
Ans 5: The potential difference across each resistor can be verified by using a voltmeter. First, measure the potential difference across the entire parallel combination. Then, connect the voltmeter across each individual resistor. You will find that the potential difference across each resistor is the same, confirming that the potential difference is identical for each resistor in a parallel arrangement.
Q 6. What is the relationship between the total current (I ) and the individual currents (I₁, I₂ and I₃)?
Ans 6: The total current I in a parallel circuit is the sum of the individual currents through each resistor. Mathematically, this is expressed as:
I = I₁ + I₂ + I₃
This relationship holds because in a parallel circuit, each resistor allows a portion of the total current to pass through based on its resistance.
Q 7. What would happen if one of the resistors in the parallel combination is removed?
Ans 7: If one of the resistors in the parallel combination is removed, the total current in the circuit would decrease, as the total resistance of the circuit would increase. This is because the total resistance of a parallel combination of resistors is less than the resistance of any individual resistor. Removing one resistor increases the equivalent resistance, reducing the total current.
Q 8: What is the effect of increasing the resistance of one resistor on the current through that resistor?
Ans 8: Increasing the resistance of one resistor will decrease the current through that resistor, as the current is inversely proportional to the resistance (according to Ohm’s Law: I₁ = V / R₁ However, the total current I will still be the sum of the currents through all resistors, and increasing one resistance may cause the currents through the other resistors to change as well.
Questions
Q 1. Judge the equivalent resistance when the following are connected in parallel – (a) 1 Ω and Ω, (b) 1 Ω and Ω, and Ω.
Ans 1: To find the equivalent resistance when resistors are connected in parallel, we use the formula: = 1/𝑅ₑₓ = 1/𝑅₁ + 1/𝑅₂ + ……
Now, let’s solve the two cases:
(a) 1 Ω and Ω in parallel: 1/Rₑq = 1 / 1 + 1 / 10⁶ = 1
Since 10⁶ Ω is very large, it has little effect on the total resistance. So, the equivalent resistance is approximately 1 Ω.
(b) Parallel Resistance Calculation
For resistors 1 Ω, 10³ Ω, and 10⁶ Ω in parallel:
1/Rₑq = 1/1 + 1/10³ + 1/10⁶
= 1 + 0.001 + 0.000001
= 1.001001 Ω⁻¹
Equivalent resistance: Rₑq = 1 / 1.001001 ≈ 0.999 Ω
In both cases, the equivalent resistance is close to the smallest resistor in the parallel connection.
Q 2. An electric lamp of 100 Ω, a toaster of resistance 50 Ω, and a water filter of resistance 500 Ω 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?
Ans 2: To solve this, we first calculate the total current drawn by the three appliances when connected in parallel, then determine the resistance of the electric iron that would draw the same total current.
Step 1: Find the total current for the three appliances
Using Ohm’s law, I = V / R , we can find the current for each appliance.
1. Electric Lamp: I=
=2.2 A
2. Toaster: I =
= 4.4 A
3. Water Filter: I =
= 0.44 A
Now, add up the currents to get the total current:
Iₜₒₜₐₗ = 2.2 A + 4.4 A + 0.44 A = 7.04 A
Step 2: Find the resistance of the electric iron
The electric iron needs to draw the same total current of 7.04 A. Using Ohm’s law again, we can calculate the required resistance for the electric iron:
R = V/I = 220 V / 7.04 A ≈ 31.25 Ω
Step 3: The current through the electric iron
Since the electric iron is designed to draw 7.04 A, the current through it is 7.04 A.
So, the electric iron’s resistance is 31.25 Ω, and the current through it is 7.04 A.
Q 3: What are the advantages of connecting electrical devices in parallel with the battery instead of connecting them in series?
Ans 3: Connecting electrical devices in parallel with a battery has several advantages over connecting them in series:
- Consistent Voltage: In parallel, each device gets the full voltage of the battery, ensuring that they all work at their designed voltage. In a series connection, the voltage is divided among the devices, which may cause some devices to receive less voltage and not work properly.
- Independent Operation: Devices connected in parallel operate independently. If one device stops working, the others continue to function. In a series connection, if one device fails, the entire circuit is broken, and all devices stop working.
- More Control: Parallel connections allow you to add or remove devices without affecting the performance of others, making it more flexible and manageable.
In short, parallel connections ensure consistent performance and reliability for all devices.
Q 4. How can three resistors of resistances 2 Ω, 3 Ω, and 6 Ω be connected to give a total resistance of (a) 4 Ω, (b) 1 Ω?
Ans 4: To achieve a total resistance of 4 Ω and 1 Ω with resistors of 2 Ω, 3 Ω, and 6 Ω, you can connect them in different ways:
(a) Total resistance of 4 Ω:
- Connect the 2 Ω and 6 Ω resistors in series. The total resistance of two resistors in series is simply the sum:
2 Ω + 6 Ω = 8 Ω. - Now, connect this combination in parallel with the 3 Ω resistor. For resistors in parallel, the total resistance is given by:
1/Rₜₒₜₐₗ = 1/8 Ω + 1/3 Ω
(b) Total resistance of 1 Ω:
Connect the 2 Ω and 3 Ω resistors in parallel
1/Rₚ = 1/2 + 1/3 = 5/6,
So, Rₚ = 6/5 = 1.2 Ω
- Then, connect this 1.2 Ω combination in parallel with the 6 Ω resistor: 1/Rₜₒₜₐₗ = 1/1.2 + 1/6
Solving this gives: Rₜₒₜₐₗ = 1 Ω
Q 5. What is (a) the highest, (b) the lowest total resistance that can be secured by combinations of four coils of resistance 4 Ω, 8 Ω, 12 Ω, 24 Ω?
Ans 5: To find the highest and lowest total resistance using four coils with resistances of 4 Ω, 8 Ω, 12 Ω, and 24 Ω, we can combine them in series and parallel:
(a) Highest Total Resistance:
When resistors are connected in series, the total resistance is the sum of the individual resistances. So,
Highest Resistance = 4 Ω + 8 Ω + 12 Ω + 24 Ω = 48 Ω.
(b) Lowest Total Resistance:
When resistors are connected in parallel, the total resistance is always less than the smallest resistor. The formula for total resistance in parallel is:
1/Rₜₒₜₐₗ = 1/R₁ + 1/R₂ + 1/R₃ + 1/R₄
Using the given values:
1/Rₜₒₜₐₗ = 1/4 + 1/8 + 1/12 + 1/24
This gives the lowest total resistance as approximately 2.4 Ω.
So, the highest resistance is 48 Ω, and the lowest is 2.4 Ω.
Questions
Q 1. Why does the cord of an electric heater not glow while the heating element does?
Ans 1: The cord of an electric heater doesn’t glow because it is made of materials with low resistance, like copper, which allows electricity to flow through easily without much heat. In contrast, the heating element is made of high-resistance materials, like nichrome, which resist the flow of electricity, causing it to heat up and glow as it gets hot.
Q 2. Compute the heat generated while transferring 96000 coulombs of charge in one hour through a potential difference of 50 V.
Ans 2: To compute the heat generated, we can use the formula:
Heat (H) = V × I × t
Where:
- V is the potential difference (50 V),
- I is the current (in amperes),
- t is the time (in seconds).
First, we need to calculate the current using the formula: I = Q / t
Where:
- Q is the charge (96000 coulombs),
- t is the time (1 hour = 3600 seconds).
So, the current ‘I’ is:
I = 96000C / 3600s = 26.67A
Now, substitute the values into the heat formula:
H = 50 V × 26.67A × 3600 s = 4,800,000 J
So, the heat generated is 4,800,000 joules (J).
Q 3. An electric iron of resistance 20Ωtakes a current of 5 A. Calculate the heat developed in 30 s.
Ans 3: To calculate the heat developed in the electric iron, we use Joule’s law of heating:
H = I ²Rt
Where:
- H is the heat developed (in joules),
- I is the current (in amperes) = 5 A,
- R is the resistance (in ohms) = 20 Ω,
- t is the time (in seconds) = 30 s.
Substituting the values:
H = (5)² × 20 × 30
H = 25 × 20 × 30 = 15,000 J
Final Answer:
The heat developed in 30 seconds is 15,000 joules (15 kJ).
Questions
Q 1. What determines the rate at which energy is delivered by a current?
Ans 1: The rate at which energy is delivered by a current is determined by electric power (P). It depends on:
- Current (I): The flow of electric charge.
- Voltage (V): The potential difference across the circuit.
- Resistance (R): The opposition to the flow of current.
The power is calculated using formulas like:
- P = V I (Power = Voltage × Current)
P = I²R (if resistance is known)
P = V²/R (if resistance is dominant).
This shows how energy delivery varies based on these factors.
Q 2. An electric motor takes 5 A from a 220 V line. Determine the power of the motor and the energy consumed in 2 h.
Ans 2: To find the power of the motor and energy consumed:
1. Power of the motor
The formula for power is: P = V × I
Where:
- V = 220 V (voltage)
- I = 5 A (current)
Substitute values: P = 220 × 5 = 1100 W (or 1.1 kW)
2. Energy consumed in 2 hours
The formula for energy is: E = P × t
Where:
- P = 1.1 kW (power)
- t = 2 hours
Substitute values: E = 1.1 × 2 = 2.2 kWh
Final Answer:
- Power of the motor: 1100 W or 1.1 kW
- Energy consumed in 2 hours: 2.2 kWh
Exercise
Q 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 R/R′ is–
(a) 1/25
(b) 1/5
(c) 5
(d) 25
Ans 1: (d) 25
Q 2. Which of the following terms does not represent electrical power in a circuit?
(a) I ²R
(b) I R²
(c) VI
(d) V²/R
Ans 2: (b) IR² – The term that does not represent electrical power in a circuit
Q 3. An electric bulb is rated 220 V and 100 W. When it is operated on 110 V, the power consumed will be –
(a)100 W
(b) 75 W
(c) 50 W
(d) 25 W
Ans 3: (d) 25 W
When the bulb is operated at 110 V, the power consumed is calculated using the formula:
P = V²/R
The resistance R of the bulb can be calculated using its rated values:
R= V²/P = 220²/100 = 484Ω
At 110 V : P = 110²/484 = 25 w
Q 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 combinations would be –
(a)1:2
(b) 2:1
(c) 1:4
(d) 4:1
Ans 4: The ratio of heat produced in series and parallel combinations is (c) 1:4.
Explanation of Heat (H) and Resistance (R) Relationship:
Basic Proportionality
When potential difference (V) is constant:
H ∝ 1/R
(Heat is inversely proportional to resistance)Series vs. Parallel Resistance
Series connection:
Total resistance = R₁ + R₂ = 2R
So, Hₛₑᵣᵢₑₛ = 1/(2R)
Parallel connection:
Total resistance = (1/R₁ + 1/R₂)⁻¹ = R/2
So, Hₚₐᵣₐₗₗₑₗ = 1/(R/2) = 2/R
Ratio Calculation
Hₛₑᵣᵢₑₛ : Hₚₐᵣₐₗₗₑₗ = 1/ 2R : 2/R = 1 : 4
Q 5. How is a voltmeter connected in the circuit to measure the potential difference between two points?
Ans 5: A voltmeter is connected in parallel with the two points where you want to measure the potential difference. This ensures it accurately measures the voltage without significantly affecting the circuit’s current flow.
Q 6. A copper wire has diameter 0.5 mm and resistivity of 1.6 × 10⁻⁸ 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?
Ans 6: To calculate the length of the copper wire with a given resistance, we use the formula: R = ρ⋅L/A
Where:
- R is the resistance (10 Ω),
- ρ is the resistivity (1.6 × 10⁻⁸ Ωm),
- L is the length (to be found),
- A is the cross-sectional area of the wire, A = πr²
Given the diameter d = 0.5 mm, the radius r = d / 2 = 0.25 mm = 0.25 × 10⁻³ m.
Step 1: Find the area
A = πr² = π(0.25×10⁻³ m)² ≈ 1.96 × 10⁻⁷ m²
Step 2: Calculate the length
Rearranging the formula to solve for L = RA/ρ = (10 Ω × 1.96×10⁻⁷ m²) / (1.6×10⁻⁸ Ω·m) ≈122.5m
So, the length of the wire is approximately 122.5 meters.
Step 3: Effect of doubling the diameter
If the diameter is doubled, the radius becomes twice as large. Since area A ∝ r² doubling the diameter will increase the area by a factor of 4. This means the resistance R will decrease by a factor of 4, because resistance is inversely proportional to the cross-sectional area.
Thus, if the diameter is doubled, the new resistance will be:
Rₙₑ𝓌 = R/4 = 10 Ω / 4 = 2.5 Ω
Q 7. The values 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 that resistor.
Ans 7: According to Ohm’s Law: V = IR
Where:
- V is the voltage (in volts),
- I is the current (in amperes),
- R is the resistance (in ohms).
To calculate the resistance R, we can rearrange the formula: R = V / t
Step-by-step Calculation:
1. For I = 0.5 A, V = 1.6V:
R = 1.6 / 0.5 = 3.2Ω
2. For I = 1.0 A, V = 3.4 V:
R = 3.4 / 1.0 = 3.4Ω
3. For I = 2.0 A, V = 6.7V:
R = 6.7 / 2.0 = 3.35Ω
4. For I = 3.0A, V=10.2V:
R = 10.2 / 3.0 = 3.4Ω
5. For I = 4.0A, V = 13.2V:
R = 13.2 / 4.0 = 3.3Ω
Resistance:
The resistance values calculated are approximately consistent, ranging between 3.2 Ω and 3.4 Ω. The average resistance can be taken as:
Rₐᵥ₉ = (3.2 Ω + 3.4 Ω + 3.35 Ω + 3.4 Ω + 3.3 Ω) / 5 = 3.34Ω
Plotting the Graph:
Now, let’s plot the graph of V (voltage) on the y-axis and I (current) on the x-axis. In this case, the graph should show a linear relationship, confirming that the resistor follows Ohm’s Law.
Here is the graph of voltage (V) versus current (I), which shows a linear relationship, confirming that the resistor follows Ohm’s Law. The slope of this graph gives the resistance of the resistor.
Resistance Calculation:
From the graph, the slope of the line (which represents R = V / I ) is approximately 3.34 Ω, which matches the average value we calculated earlier.
Q 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.
Ans 8: To find the resistance of the unknown resistor, we can use Ohm’s Law, which states: V = I × R
Where:
- V is the voltage (12 V),
- I is the current (2.5 mA = 0.0025 A),
- R is the resistance (which we need to find).
Rearranging the formula to solve for R:
R = V / I
Now, substitute the known values:
R = 12 / 0.0025 = 4800Ω
So, the resistance of the resistor is 4800 ohms.
Q 9. A battery of 9 V is connected in series with resistors of 0.2Ω, 0.3Ω, 0.4Ω, 0.5Ωand 12Ω, respectively. How much current would flow through the 12Ωresistor?
Ans 9: To find the current flowing through the 12Ω resistor, first, we need to calculate the total resistance in the circuit.
The total resistance is the sum of all resistors since they are connected in series:
Rₜₒₜₐₗ= 0.2Ω + 0.3Ω + 0.4Ω + 0.5Ω + 12Ω = 13.4Ω
Next, use Ohm’s Law to calculate the current:
I = V / Rₜₒₜₐₗ
where:
- V = 9V (battery voltage)
- Rₜₒₜₐₗ = 13.4Ω (total resistance)
Substitute the values:
I = 9V/ 13.4Ω ≈ 0.67A
So, the current flowing through the 12Ω resistor is approximately 0.67 amps.
Q 10. How many 176Ω resistors (in parallel) are required to carry 5 A on a 220 V line?
Ans 10: Calculate the total resistance required using Ohm’s law:
Rₜₒₜₐₗ = V / I = 220 / 5 = 44Ω
Calculate the number of resistors in parallel needed to achieve 44Ω:
1/Rₜₒₜₐₗ = 1/R + 1/R ……… (n resistors in parallel)
1/44 = n/176
n = 144/76 = 4
So, 4 resistors of 176Ω are required in parallel.
Q 11. Show how you would connect three resistors, each of resistance 6Ω, so that the combination has a resistance of (i) 9Ω, (ii) 4Ω.
Ans 11: To connect three 6Ω resistors for the desired resistances:
(i) For 9Ω:
- Connect two resistors in series: 6Ω + 6Ω = 12Ω.
- Then connect this combination in parallel with the third resistor:
1/Rₜₒₜₐₗ = 1/12Ω + 1/6Ω = Rₜₒₜₐ ₗ = 9Ω
(ii) For 4Ω:
- Connect two resistors in parallel:
1/Rₜₒₜₐₗ = 1/6Ω + 1/6Ω = Rₜₒₜₐ ₗ = 3Ω
- Then connect the third resistor in series with this parallel combination
Rₜₒₜₐₗ = 3Ω + 6Ω = 4Ω
Q 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?
Ans 12: To solve this, first calculate the current drawn by each bulb:
Power P = 10 W, Voltage V=220 V.
Using the formula P = V × I, we get:
I = P / V = 10/220 = 0.045 A
Now, the maximum allowable current is 5 A. To find how many bulbs can be connected in parallel, divide the total current by the current drawn by one bulb:
Number of bulbs = 5 /0.045A ≈111 bulbs
So, approximately 111 bulbs can be connected in parallel.
Q 13. A hot plate of an electric oven connected to a 220 V line has two resistance coils A and B, each of 24Ωresistance, which may be used separately, in series, or in parallel. What are the currents in the three cases?
Ans 13: Given:
- Voltage (V) = 220 V
- Resistance of each coil (R₁ = R₂) = 24 Ω
- Coils used separately:
- Current in each coil: I = V/R = 220/24 =9.17A
- Coils in series:
- Total resistance: Rₜₒₜₐₗ = R1 + R2 = 24 + 24 = 48Ω
- Current: I = V/Rₜₒₜₐₗ = 220/48 =4.58A
- Coils in parallel:
- Total resistance: Rₜₒₜₐₗ = (R₁ × R₂)/(R₁ + R₂) = (24 Ω × 24 Ω)/(24 Ω + 24 Ω) = 12 Ω
- Current: I = V / Rₜₒₜₐₗ = 220/12 = 10.33A
So, the currents are:
- Separate: 9.17 A
- Series: 4.58 A
- Parallel: 18.33 A
Q 14. Compare the power used in the 2Ωresistor in each of the following circuits:
(i) A 6 V battery in series with 1Ωand 2Ωresistors.
(ii) A 4 V battery in parallel with 12Ωand 2Ωresistors.
Ans 14: To compare the power used in the 2Ω resistor in both circuits, let’s calculate it step by step.
(i) Series Circuit (6V battery, 1Ω, and 2Ω resistors):
- Total resistance: Rₜₒₜₐₗ = 1Ω + 2Ω = 3Ω
- Current: I = V/Rₜₒₜₐₗ = 6V/ 3Ω =2A
- Power in 2Ω resistor:
P = I² × R = (2 A)² × 2 Ω = 8 W
(ii) Parallel Circuit (4V battery, 12Ω, and 2Ω resistors):
- Total resistance: Use the parallel formula: 1/Rₜₒₜₐₗ = 1/12Ω + 1/2Ω , which gives Rₜₒₜₐₗ= 1.714Ω.
- Current: I = V/Rₜₒₜₐₗ = 4V/1.714Ω = 2.33A
- Power in 2Ω resistor: Since the voltage across the 2Ω resistor is 4V, P = V²/R = (4 V)²/2 Ω = 8 W
- Conclusion: In both circuits, the power used in the 2Ω resistor is 8W.
Q 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?
Ans 15: When two lamps are connected in parallel, they each receive the full supply voltage (220 V). To calculate the current drawn by each lamp, use the formula:
Power(P) = Voltage(V) × Current(I) I = P/V
1. Current drawn by the 100 W lamp:
I₁ = 100V / 220Ω = 0.4545 A
2. Current drawn by the 60 W lamp:
I₂ = 60V / 220Ω = 0.2727 A
Total current (since they are in parallel):
Iₜₒₜₐₗ = I₁ + I₂ = 0.4545 A + 0.2727 A = 0.7272 A
The total current drawn from the supply is approximately 0.73 A.
Q 16. Which uses more energy, a 250 W TV set in 1 hour, or a 1200 W toaster in 10 minutes?
Ans 16: we calculate energy used in kilowatt-hours (kWh):
Energy = Power × Time
- TV:
Power = 250 W = 0.25 kW
Time = 1 hour
Energy = 0.25 × 1 = 0.25 kWh - Toaster:
Power = 1200 W = 1.2 kW
Time = 10 minutes = 10/60 = 0.167 hours
Energy = 1.2 × 0.167 = 0.2 kWh
Result:
The TV (0.25 kWh) uses more energy than the toaster (0.2 kWh).
Q 17. An electric heater of resistance 44 Ω draws 5 A from the service mains for 2 hours. Calculate the rate at which heat is developed in the heater.
Ans 17: To calculate the rate at which heat is developed in the heater, use the formula for power:
P = I² × R
Where:
- P = Power (rate of heat developed)
- I = Current = 5 A
- R = Resistance = 44 Ω
Now substitute the values:
P = (5)² × 44 = 25 × 44 = 1100 watts
The rate at which heat is developed in the heater is 1100 watts (or 1.1 kW).
Q 18. Explain the following:
A. Why is the tungsten used almost exclusively for filament of electric lamps?
Ans A: Here’s why tungsten is the preferred choice for electric lamp filaments:
- High Melting Point: Tungsten has an extremely high melting point (about 3,422°C), allowing it to glow at high temperatures without melting.
- Durability: It resists wear and tear even under intense heat, making it long-lasting.
- High Resistance: Its resistance produces more heat and light when electricity flows through, enhancing efficiency.
- Low Vaporization: Tungsten evaporates slowly at high temperatures, reducing filament thinning and increasing lamp lifespan.
- Brightness: It emits bright, white light when heated, ideal for electric lamps.
These properties make tungsten the best material for electric lamp filaments.
B. Why are the conductors of electric heating devices, such as bread-toasters and electric irons, made of an alloy rather than a pure metal?
Ans B:
- Higher Melting Point: Alloys can withstand high temperatures without melting, making them ideal for heating devices.
- Resistance to Wear: They resist corrosion and oxidation better than pure metals, ensuring longer device life.
- Better Electrical Resistance: Alloys have higher resistance, which generates more heat efficiently.
- Durability: They are harder and stronger than pure metals, reducing wear and tear.
- Stable Performance: Alloys maintain consistent heating properties over time, ensuring reliability.
This combination of properties makes alloys perfect for bread-toasters, electric irons, and similar appliances.
C. Why is the series arrangement not used for domestic circuits?
Ans c: The series arrangement is not used for domestic circuits due to the following reasons:
- Uneven Voltage Distribution: Appliances in a series connection share the total voltage, so each gets less voltage, causing them to function inefficiently or not at all.
- Dependency on Each Appliance: If one appliance is switched off or fails, the entire circuit is disrupted, cutting power to all devices.
- Power Limitations: Series circuits cannot handle high loads effectively, leading to potential overheating or failure.
- Limited Control: Individual appliances cannot be controlled independently in a series connection.
- Safety Concerns: Overloading a series circuit increases the risk of fire or electrical hazards.
For these reasons, parallel arrangements are preferred in domestic circuits as they allow independent operation and ensure proper voltage for all appliances.
D. How does the resistance of a wire vary with its area of cross-section?
Ans D: The resistance of a wire is inversely proportional to its cross-sectional area. As the area increases, the resistance decreases. This is because a larger cross-sectional area allows more electrons to flow through the wire, reducing the opposition to the current.
E. Why are copper and aluminum wires usually employed for electricity transmission?
Ans E: Copper and aluminum wires are commonly used for electricity transmission due to the following reasons:
- Good Conductivity: Both copper and aluminum are excellent conductors of electricity, allowing efficient transmission of electrical energy.
- Low Resistance: These metals have relatively low resistance, minimizing energy loss as heat during transmission.
- Durability: Copper and aluminum are resistant to corrosion, ensuring long-lasting performance, especially in outdoor environments.
- Lightweight: Aluminum, in particular, is lightweight, making it easier and more cost-effective to handle and install.
- Cost-Effective: While copper is more expensive, aluminum is a cheaper alternative, offering a balance of performance and cost for large-scale transmission.
- Flexibility: Both materials can be easily shaped into wires, allowing them to be used in a variety of transmission systems.
Chapter 11 Electricity Class 10 Notes, Question/Answer, Activity & Science Projects
Updated Solution 2024-2025
This complete solution is prepared as per the latest syllabus of 2024-25. If you have any further queries, feel free to ask! 