Class 10 Chapter 1 Electricity Physics Solution S.Chand. Part 8

 Part 8 Page 66

Very Short Answer Type Questions

1. How does the heat H produced by a current passing through a fixed resistance wire depend on the magnitude of current I ?

1. The heat produced by a current passing through a fixed resistance wire depends on the magnitude of the current squared, as described by Joule's law: $�={�}^2\times �$, where $�$ is the heat produced, $�$ is the current, and $�$ is the resistance of the wire.

2. If the current passing through a conductor is doubled, what will be the change in heat produced ? 

1. If the current passing through a conductor is doubled, the heat produced will increase by a factor of four. This is because of the square relationship between current and heat in Joule's law.

3. Name two effects produced by electric current. 

wo effects produced by electric current are:

• Heating effect: When electric current flows through a conductor with resistance, it produces heat due to the collision of charge carriers with the atoms of the conductor.
• Magnetic effect: Electric current creates a magnetic field around the conductor, as described by Ampere's law.

4. Which effect of current is utilised in an electric light bulb ? 

1. The effect of electric current utilized in an electric light bulb is the heating effect. The electric current passing through the filament of the bulb heats it up to a high temperature, causing it to emit light.

5. Which effect of current is utilised in the working of an electric fuse ?

 

1. The effect of electric current utilized in the working of an electric fuse is the heating effect. When an excessive current passes through the fuse, it heats up due to the resistance of the fuse wire, which eventually melts and breaks the circuit, protecting the rest of the circuit from damage.

6. Name two devices which work on the heating effect of electric current. 

1. Two devices that work on the heating effect of electric current are:

   • Electric heater: It uses a resistive element to generate heat when current passes through it.
   • Electric toaster: It also utilizes a resistive element to produce heat for toasting bread.

7. Name two gases which are filled in filament type electric light bulbs. 

1. Two gases which are filled in filament type electric light bulbs are:

   • Argon
   • Nitrogen

8. Explain why, filament type electric bulbs are not power efficient. 

1. Filament-type electric bulbs are not power-efficient because a significant portion of the electrical energy is converted into heat rather than light. Only a small fraction of the energy is converted into visible light, making them inefficient compared to more modern lighting technologies such as LED bulbs.

9. Why does the connecting cord of an electric heater not glow hot while the heating element does ?

1. The connecting cord of an electric heater does not glow hot while the heating element does because the cord has a much lower resistance compared to the heating element. The majority of the heat is generated in the heating element due to its higher resistance, whereas the cord has relatively low resistance, so it does not heat up significantly.

2. Short Answer Type Questions

3. 10. (a) Write down the formula for the heat produced when a current I is passed through a resistor R for time t. (b) An electric iron of resistance 20 ohms draws a current of 5 amperes. Calculate the heat produced in 30 seconds.

4. (a) The formula for the heat produced when a current $�$ is passed through a resistor $�$ for time $�$ is given by the equation: $�={�}^2\times �\times �$

   (b) Given $�=20$ ohms, $�=5$ amperes, and $�=30$ seconds, plug these values into the formula: $�=(5{)}^2\times 20\times 30=7500\text{ joules}$

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   11. State three factors on which the heat produced by an electric current depends. How does it depend on these factors ?

5. 11. Three factors on which the heat produced by an electric current depends are:
   • Magnitude of current $(�)$: Heat produced is directly proportional to the square of the current (${�}^2$).
   • Resistance of the conductor $(�)$: Heat produced is directly proportional to the resistance ($�$).
   • Time duration $(�)$: Heat produced is directly proportional to the time for which the current flows ($�$).

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   12. (a) State and explain Joule’s law of heating. (b) A resistance of 40 ohms and one of 60 ohms are arranged in series across 220 volt supply. Find the heat in joules produced by this combination of resistances in half a minute.

7. (a) Joule’s law of heating states that the heat produced in a conductor is directly proportional to the square of the current passing through it, the resistance of the conductor, and the time for which the current flows. Mathematically, it is expressed as $�={�}^2\times �\times �$, where $�$ is the heat produced, $�$ is the current, $�$ is the resistance, and $�$ is the time.

   (b) Given ${�}_1=40$ ohms, ${�}_2=60$ ohms, and $�=220$ volts, the total resistance ${�}_{\text{total}}={�}_1+{�}_2=40+60=100$ ohms. Now, use the formula $�={�}^2\times �\times �$: $�=(220{)}^2\times 100\times 0.5=2.42\times 10^6\text{ joules}$

8. 13. Why is an electric light bulb not filled with air ? Explain why argon or nitrogen is filled in an electric bulb.

9. 13. An electric light bulb is not filled with air because oxygen in the air would cause the filament to oxidize and burn out quickly. Instead, inert gases like argon or nitrogen are filled in the electric bulb. These gases do not react with the filament material (usually tungsten), thereby prolonging the life of the bulb.

10. 14. Explain why, tungsten is used for making the filaments of electric bulbs.

11. 13. Tungsten is used for making the filaments of electric bulbs because it has a very high melting point and does not evaporate quickly at high temperatures. This property allows the filament to operate at high temperatures without burning out quickly, thus ensuring a longer lifespan for the bulb.

    14. 15. Explain why, the current that makes the heater element very hot, only slightly warms the connecting wires

12. leading to the heater.

13. 13. The current that makes the heater element very hot only slightly warms the connecting wires leading to the heater because the connecting wires have much lower resistance compared to the heater element. According to Ohm's law ($�=�\times �$), the majority of the voltage drop and heat is across the higher resistance element (the heater) rather than the lower resistance wires.

14. 16. When a current of 4.0 A passes through a certain resistor for 10 minutes, 2.88 × 104 J of heat are produced. Calculate : (a) the power of the resistor. (b) the voltage across the resistor.

15. (a) Given $�=10$ minutes, $�=2.88\times 10^4$ joules, use $�={�}^2\times �\times �$ to find the power ($�$) of the resistor: $�=\frac{�}{�}=\frac{2.88\times 10^4}{10\times 60}=48\text{ watts}$

    (b) Use $�=�\times �$ to find the voltage ($�$) across the resistor: $�=�\times �=4.0\times 10=40\text{ volts}$

    17. A heating coil has a resistance of 200 ohm. At what rate will heat be produced in it when a current of 2.5 A flows through it ?

16. 17. To find the rate at which heat is produced ($�$) in the heating coil, you can use the formula $�={�}^2\times �$, where $�=2.5$ A and $�=200$ ohms: $�=(2.5{)}^2\times 200=1250\text{ watts}$

    So, the rate at which heat is produced in the heating coil is 1250 watts.

    18. An electric heater of resistance 8 ohm takes a current of 15 A from the mains supply line. Calculate the rate at which heat is developed in the heater.

17. 18. For the electric heater with a resistance $�=8$ ohms and current $�=15$ A, the rate at which heat is developed can be calculated using the same formula: $�=(15{)}^2\times 8=1800\text{ watts}$

    So, the rate at which heat is developed in the heater is 1800 watts.

    19. A resistance of 25 ohm is connected to a 12 V battery. Calculate the heat energy in joules generated per minute.

18. 19. To calculate the heat energy ($�$) generated per minute, you can use the formula $�={�}^2\times �\times �$, where $�=\frac{�}{�}$, $�=12$ volts, $�=25$ ohms, and $�=1$ minute: $�=\frac{�}{�}=\frac{12}{25}=0.48\text{ amperes}$ $�=(0.48{)}^2\times 25\times 60=6.912\text{ joules}$

    So, the heat energy generated per minute is 6.912 joules.

    20. 100 joules of heat is produced per second in a 4 ohm resistor. What is the potential difference across the resistor ?

19. 20. Given that 100 joules of heat is produced per second in a 4 ohm resistor, you can use the formula $�={�}^2\times �$ to find $�$: $�=\sqrt{\frac{�}{�}}=\sqrt{\frac{100}{4}}=5\text{ amperes}$

    Then, you can use Ohm's law $�=�\times �$ to find the potential difference across the resistor: $�=�\times �=5\times 4=20\text{ volts}$

    So, the potential difference across the resistor is 20 volts.

Long Answer Type Question

21. (a) Derive the expression for the heat produced due to a current ‘I’ flowing for a time interval ‘t’ through a resistor ‘R’ having a potential difference ‘V’ across its ends. With which name is this relation known ? (b) How much heat will an instrument of 12 W produce in one minute if it is connected to a battery of 12 V ? (c) The current passing through a room heater has been halved. What will happen to the heat produced by it ? (d) What is meant by the heating effect of current ? Give two applications of the heating effect of current. (e) Name the material which is used for making the filaments of an electric bulb.

(a) To derive the expression for the heat produced ($�$) due to a current $�$ flowing for a time interval $�$ through a resistor $�$ with a potential difference $�$ across its ends, we can use the formula for power ($�$):

$�=��$

The heat produced ($�$) over time $�$ is given by integrating power over time:

$�={\int }_0^{�}���$

Substituting the expression for power ($�=��$):

$�={\int }_0^{�}����$

$�=�{\int }_0^{�}���$

Since $�$ and $�$ are constant, we can factor $�$ out of the integral:

$�=�\left({\int }_0^{�}���\right)$

$�=�{[�\cdot �]}_0^{�}$

$�=�(�\cdot �-0)$

$�=���$

So, the expression for the heat produced due to a current $�$ flowing for a time interval $�$ through a resistor $�$ with a potential difference $�$ across its ends is $�=���$. This relation is known as Joule's law.

(b) Given $�=12$ W and $�=12$ V, and $�=1$ minute, we can use the formula $�=\frac{�}{�}$ to find $�$:

$�=�\times �=12\times 60=720\text{ joules}$

So, the instrument will produce 720 joules of heat in one minute.

(c) If the current passing through a room heater is halved, the heat produced by it will also be halved. This is because heat produced ($�$) is directly proportional to the current ($�$) flowing through the resistor according to Joule's law ($�={�}^2��$).

(d) The heating effect of current refers to the phenomenon where electrical energy is converted into heat when an electric current passes through a conductor with resistance. Two applications of the heating effect of current are:

1. Electric heaters: They use the heating effect of current to generate heat for warming rooms or spaces.
2. Electric stoves: They utilize the heating effect of current to produce heat for cooking food.

(e) The material used for making the filaments of an electric bulb is tungsten. Tungsten is chosen for its high melting point and resistance to evaporation at high temperatures, which allows the filament to glow brightly without burning out quickly.

Multiple Choice Questions (MCQs)

22. The heat produced by passing an electric current through a fixed resistor is proportional to the square of : (a) magnitude of resistance of the resistor (b) temperature of the resistor (c) magnitude of current (d) time for which current is passed

23. The current passing through an electric kettle has been doubled. The heat produced will become : (a) half (b) double (c) four times (d) one-fourth 24. An electric fuse works on the : (a) chemical effect of current (b) magnetic effect of current (c) lighting effect of current (d) heating effect of current 25. The elements of electrical heating devices are usually made of : (a) tungsten (b) bronze (c) nichrome (d) argon 26. The heat produced in a wire of resistance ‘x’ when a current ‘y’ flows through it in time ‘z’ is given by : (a) x2 × y × z (b) x × z × y2 (c) y × z2 × x (d) y × z × x 27. Which of the following characteristic is not suitable for a fuse wire ? (a) thin and short (b) thick and short (c) low melting point (d) higher resistance than rest of wiring 28. In a filament type light bulb, most of the electric power consumed appears as : (a) visible light (b) infra-red-rays (c) ultraviolet rays (d) fluorescent light 29. Which of the following is the most likely temperature of the filament of an electric light bulb when it is working on the normal 220 V supply line ? (a) 500°C (b) 1500°C (c) 2500°C (d) 4500°C 30. If the current flowing through a fixed resistor is halved, the heat produced in it will become : (a) double (b) one-half (c) one-fourth (d) four times

Questions Based on High Order Thinking Skills (HOTS)

31. The electrical resistivities of four materials P, Q, R and S are given below : P 6.84 × 10–8 Ωm Q 1.70 × 10–8 Ωm R 1.0 × 1015 Ωm S 11.0 × 10–7 Ωm Which material will you use for making : (a) heating element of electric iron (b) connecting wires of electric iron (c) covering of connecting wires ? Give reason for your choice in each case.

(a) For making the heating element of an electric iron, we need a material with high resistivity to ensure efficient conversion of electrical energy into heat. Therefore, we would choose Material R, which has the highest resistivity (1.0 × 10^15 Ωm) among the given options. This high resistivity will result in a higher resistance, allowing the heating element to generate sufficient heat when current passes through it.

(b) For making the connecting wires of an electric iron, we need a material with low resistivity to minimize the loss of electrical energy due to heating of the wires. Therefore, we would choose Material Q, which has the lowest resistivity (1.70 × 10^-8 Ωm) among the given options. This low resistivity will result in lower resistance in the wires, minimizing heat loss and ensuring efficient transmission of electrical energy.

(c) For the covering of connecting wires, we need a material that provides insulation and protection against electrical hazards. Material S, with resistivity of 11.0 × 10^-7 Ωm, would be a suitable choice. It has a moderate resistivity, which allows for efficient insulation while not causing excessive heat generation. Additionally, it offers adequate mechanical strength and durability for covering the wires.

32. (a) How does the wire in the filament of a light bulb behave differently to the other wires in the circuit when the current flows ? (b) What property of the filament wire accounts for this difference ?

(a) The wire in the filament of a light bulb behaves differently from the other wires in the circuit when the current flows because it heats up and emits light. While the other wires in the circuit may conduct electricity without significantly heating up or emitting light, the filament wire experiences a substantial increase in temperature, causing it to glow and produce light.

(b) The property of the filament wire that accounts for this difference is its high resistance. The filament wire has a much higher resistance compared to the other wires in the circuit. This high resistance causes the filament wire to heat up significantly when current flows through it, resulting in the emission of light and the functioning of the light bulb.

33. Two exactly similar heating resistances are connected (i) in series, and (ii) in parallel, in two different circuits, one by one. If the same current is passed through both the combinations, is more heat obtained per minute when they are connected in series or when they are connected in parallel ? Give reason for your answer.

When two exactly similar heating resistances are connected, whether in series or parallel, the total resistance of the circuit and the current passing through the resistances will be different for each configuration.

(i) Series Connection: In a series connection, the total resistance (${�}_{\text{total}}$) is the sum of the individual resistances (${�}_1$ and ${�}_2$): ${�}_{\text{total}}={�}_1+{�}_2$

The current ($�$) passing through both resistances will be the same, as it is common to elements in series.

(ii) Parallel Connection: In a parallel connection, the reciprocal of the total resistance (${�}_{\text{total}}$) is the sum of the reciprocals of the individual resistances (${�}_1$ and ${�}_2$): $\frac{1}{{�}_{\text{total}}}=\frac{1}{{�}_1}+\frac{1}{{�}_2}$

The current ($�$) will split between the resistances according to their individual resistances; however, the total current through the circuit will still be the same as when they are connected in series.

Now, let's analyze the situation regarding the heat produced in each configuration:

The heat produced ($�$) in a resistor is given by $�={�}^2\times �$, where $�$ is the current and $�$ is the resistance.

(i) Series Connection: In a series connection, the total resistance is higher, so according to $�={�}^2\times �$, more heat is produced in each resistor compared to the parallel connection.

(ii) Parallel Connection: In a parallel connection, each resistor has the same voltage across it, but the current through each resistor is lower compared to the series connection. Therefore, according to $�={�}^2\times �$, less heat is produced in each resistor compared to the series connection.

Conclusion: More heat is obtained per minute when the resistances are connected in series because the higher total resistance results in more heat being produced in each resistor due to the same current passing through them.

34. An electric iron is connected to the mains power supply of 220 V. When the electric iron is adjusted at ‘minimum heating’ it consumes a power of 360 W but at ‘maximum heating’ it takes a power of 840 W. Calculate the current and resistance in each case.

To calculate the current and resistance in each case, we can use Ohm's law ($�=��$) and the formula for power ($�=��$).

Given:

• Mains power supply voltage ($�$) = 220 V
• Power at minimum heating (${�}_1$) = 360 W
• Power at maximum heating (${�}_2$) = 840 W

Let's start by finding the current ($�$) and resistance ($�$) at minimum heating:

1. For minimum heating:
   • Power (${�}_1$) = 360 W
   • Voltage ($�$) = 220 V

Using $�=��$, we can solve for the current (${�}_1$): ${�}_1=\frac{{�}_1}{�}=\frac{360}{220}\approx 1.636\text{ A}$

Now, using Ohm's law ($�=��$), we can solve for the resistance (${�}_1$): ${�}_1=\frac{�}{{�}_1}=\frac{220}{1.636}\approx 134.52\text{ ohms}$

So, at minimum heating:

• Current (${�}_1$) ≈ 1.636 A
• Resistance (${�}_1$) ≈ 134.52 ohms

Next, let's find the current ($�$) and resistance ($�$) at maximum heating:

2. For maximum heating:
   • Power (${�}_2$) = 840 W
   • Voltage ($�$) = 220 V

Using $�=��$, we can solve for the current (${�}_2$): ${�}_2=\frac{{�}_2}{�}=\frac{840}{220}=3.818\text{ A}$

Now, using Ohm's law ($�=��$), we can solve for the resistance (${�}_2$): ${�}_2=\frac{�}{{�}_2}=\frac{220}{3.818}\approx 57.64\text{ ohms}$

So, at maximum heating:

• Current (${�}_2$) ≈ 3.818 A
• Resistance (${�}_2$) ≈ 57.64 ohms

35. Which electric heating devices in your home do you think have resistors which control the flow of electricity ?

In many electric heating devices, resistors are used to control the flow of electricity and regulate the amount of heat generated. Some common electric heating devices in your home that likely use resistors for this purpose include:

1. Electric heaters: Portable electric heaters, baseboard heaters, and wall heaters often contain resistors to control the flow of electricity and adjust the heat output.

2. Electric stoves and ovens: The heating elements in electric stoves and ovens, whether they are coil-type burners or flat-top surfaces, usually contain resistors to regulate the heat produced for cooking.

3. Electric furnaces: Electric furnaces use resistors to generate heat, which is then circulated throughout the home via a forced air system.

4. Electric blankets and heating pads: These devices utilize resistors to control the flow of electricity and maintain a consistent temperature for comfort.

5. Electric water heaters: Electric water heaters contain resistors (often referred to as heating elements) that heat the water in the tank to the desired temperature.

6. Electric kettles: Electric kettles use resistors to heat the water quickly for boiling.

7. Toaster ovens and toaster: These appliances typically contain resistors in their heating elements to toast bread or cook food.

In each of these devices, the resistors play a crucial role in converting electrical energy into heat, allowing the device to function effectively for its intended purpose.