Class 10 Chapter 1 Electricity Physics Solution S.Chand. Part 7

 Part vii Page 58

Very Short Answer Type Questions

1. State two factors on which the electrical energy consumed by an electrical appliance depends.

 The electrical energy consumed by an electrical appliance depends on its power rating (in watts) and the duration of its usage (in hours).

2. Which one has a higher electrical resistance : a 100 watt bulb or a 60 watt bulb ? 

A 100 watt bulb has higher electrical resistance compared to a 60 watt bulb. Generally, higher wattage bulbs have lower resistance because they allow more current to flow.

3. Name the commercial unit of electric energy. 

The commercial unit of electric energy is the kilowatt-hour (kWh).

4. An electric bulb is rated at 220 V, 100 W. What is its resistance ?

Using Ohm's Law (V = IR) and the power formula (P = VI), we can find the resistance (R) of the bulb: $�=�\times �$ $100\text{W}=220\text{V}\times �$ $�=\frac{100\text{W}}{220\text{V}}=0.4545\text{A}$

Then, using Ohm's Law: $�=��$ $220\text{V}=0.4545\text{A}\times �$ $�=\frac{220\text{V}}{0.4545\text{A}}=484.848\mathrm{\Omega }$

So, the resistance of the bulb is approximately $484.848\mathrm{\Omega }$

484.848Ω.

5. What is the SI unit of (i) electric energy, and (ii) electric power ? 

The SI unit of electric energy is the joule (J), and the SI unit of electric power is the watt (W).

6. Name the quantity whose unit is (i) kilowatt, and (ii) kilowatt-hour. 

(i) The quantity whose unit is kilowatt is electric power. (ii) The quantity whose unit is kilowatt-hour is electric energy.

7. Which quantity has the unit of watt ? 

The quantity that has the unit of watt is electric power.

8. What is the meaning of the symbol kWh ? Which quantity does it represent ? 

kWh stands for kilowatt-hour. It represents the amount of energy consumed or produced when a device with a power rating of one kilowatt (1 kW) operates for one hour.

9. If the potential difference between the end of a wire of fixed resistance is doubled, by how much does the electric power increase ? 

If the potential difference (voltage) between the ends of a wire of fixed resistance is doubled, the electric power increases by a factor of four. This is because electric power is directly proportional to the square of the voltage according to the formula $�={�}^2\mathrm{/}�$.

10. An electric lamp is labelled 12 V, 36 W. This indicates that it should be used with a 12 V supply. What other information does the label provide ? 

The label on the electric lamp provides information about its voltage rating (12 V) and its power consumption (36 W). This indicates that the lamp should be used with a 12 V power supply and consumes 36 watts of power when operating.

11. What current will be taken by a 920 W appliance if the supply voltage is 230 V ?

1. Using the formula $�=��$: $920\text{W}=230\text{V}\times �$ $�=\frac{920\text{W}}{230\text{V}}=4\text{A}$

So, the current taken by the appliance is 4 amperes.

Short Answer Type Questions

12. Define watt. Write down an equation linking watts, volts and amperes.

12. Watt: A watt is the SI unit of power, representing the rate of energy transfer or consumption. It is equivalent to one joule of energy transferred per second. The equation linking watts, volts, and amperes is:

$\text{Power (W)}=\text{Voltage (V)}\times \text{Current (A)}$

13. Define watt-hour. How many joules are equal to 1 watt-hour ?

13. Watt-hour: A watt-hour (Wh) is a unit of energy equivalent to one watt of power expended for one hour. It represents the amount of energy consumed or produced over time.

$1\text{Watt-hour (Wh)}=3600\text{joules (J)}$

14. How much energy is consumed when a current of 5 amperes flows through the filament (or element) of a heater having resistance of 100 ohms for two hours ? Express it in joules.

14. To find the energy consumed ($�$) when a current of 5 amperes flows through a heater with a resistance of 100 ohms for two hours, we'll use the formula:

    $�=�\times �$

    First, let's find the power ($�$) dissipated by the heater. We can use the formula $�={�}^2�$, where $�$ is the current and $�$ is the resistance:

    $�=(5\text{A}{)}^2\times 100\mathrm{\Omega }=2500\text{W}$

    Now, we have the power ($�$) and the time ($�$), which is $2$ hours. We can calculate the energy consumed:

    $�=2500\text{W}\times 2\text{hours}$

    $�=5000\text{Wh}$

    To express this energy in joules, we'll use the fact that $1\text{Wh}=3600\text{J}$:

    $�=5000\text{Wh}\times 3600\text{J/Wh}$

    $�=1.8\times 10^7\text{J}$

    So, the energy consumed by the heater when a current of 5 amperes flows through it for two hours is $1.8\times 10^7\text{J}$.

15. An electric bulb is connected to a 220 V power supply line. If the bulb draws a current of 0.5 A, calculate the power of the bulb.

15. Power can be calculated using the formula:

$\text{Power (W)}=\text{Voltage (V)}\times \text{Current (A)}$

Given: Voltage (V) = 220 V, Current (I) = 0.5 A.

$\text{Power (W)}=220\text{V}\times 0.5\text{A}=110\text{W}$

16. In which of the following cases more electrical energy is consumed per hour ? (i) A current of 1 ampere passed through a resistance of 300 ohms. (ii) A current of 2 amperes passed through a resistance of 100 ohms.

To determine which case consumes more electrical energy per hour, we can calculate the power ($�$) for each case and compare.

The power ($�$) can be calculated using the formula:

$�=�\times �$

where $�$ is the voltage and $�$ is the current. Since we're given the resistance ($�$) and the current ($�$), we can use Ohm's Law to find the voltage ($�$):

$�=�\times �$

Let's calculate the power ($�$) for each case:

For case (i): $�=1\text{A}\times 300\mathrm{\Omega }=300\text{V}$ $�=�\times �=300\text{V}\times 1\text{A}=300\text{W}$

For case (ii): $�=2\text{A}\times 100\mathrm{\Omega }=200\text{V}$ $�=�\times �=200\text{V}\times 2\text{A}=400\text{W}$

Comparing the powers, we can see that the power in case (ii) is higher ($400\text{W}$) compared to the power in case (i) ($300\text{W}$). Therefore, more electrical energy is consumed per hour in case (ii), where a current of 2 amperes passes through a resistance of 100 ohms.

17. An electric kettle rated at 220 V, 2.2 kW, works for 3 hours. Find the energy consumed and the current drawn

16. Energy consumed can be calculated using the formula:

$\text{Energy (kWh)}=\text{Power (kW)}\times \text{Time (h)}$

Given: Voltage (V) = 220 V, Power (P) = 2.2 kW, Time (t) = 3 hours.

$\text{Energy (kWh)}=2.2\text{kW}\times 3\text{h}=6.6\text{kWh}$

Current can be calculated using the formula $�=\frac{�}{�}$:

$�=\frac{2200\text{W}}{220\text{V}}=10\text{A}$

18. In a house two 60 W electric bulbs are lighted for 4 hours, and three 100 W bulbs for 5 hours everyday. Calculate the electric energy consumed in 30 days.

18. Energy consumed by each bulb can be calculated using the formula $\text{Energy (kWh)}=\frac{\text{Power (W)}\times \text{Time (h)}}{1000}$:

$\text{Energy consumed by each 60 W bulb}=\frac{60\times 4}{1000}\text{kWh}=0.24\text{kWh}$

$\text{Energy consumed by each 100 W bulb}=\frac{100\times 5}{1000}\text{kWh}=0.5\text{kWh}$

Total energy consumed in 30 days will be $(2\times 0.24+3\times 0.5)\times 30\text{kWh}$.

$=(0.48+1.5)\times 30\text{kWh}=1.98\times 30\text{kWh}=59.4\text{kWh}$

19. A bulb is rated as 250 V; 0.4 A. Find its : (i) power, and (ii) resistance.

(i) To find the power ($�$) of the bulb, we'll use the formula $�=��$, where $�$ is the voltage and $�$ is the current: $�=250\text{V}\times 0.4\text{A}=100\text{W}$

So, the power of the bulb is $100\text{W}$.

(ii) To find the resistance ($�$) of the bulb, we'll use Ohm's Law $�=\frac{�}{�}$: $�=\frac{250\text{V}}{0.4\text{A}}=625\mathrm{\Omega }$

So, the resistance of the bulb is $625\mathrm{\Omega }$.

20. For a heater rated at 4 kW and 220 V, calculate : (a) the current, (b) the resistance of the heater, (c) the energy consumed in 2 hours, and (d) the cost if 1 kWh is priced at ` 4.60.

19. (a) To find the current ($�$) drawn by the heater, we'll use Ohm's Law $�=\frac{�}{�}$, where $�$ is the power and $�$ is the voltage: $�=\frac{4000\text{W}}{220\text{V}}=18.18\text{A}$

    (b) To find the resistance ($�$) of the heater, we'll use Ohm's Law $�=\frac{�}{�}$: $�=\frac{220\text{V}}{18.18\text{A}}=12.1\mathrm{\Omega }$

    (c) To find the energy consumed ($�$) in 2 hours, we'll use the formula $�=�\times �$: $�=4000\text{W}\times 2\text{hours}=8000\text{Wh}=8\text{kWh}$

    (d) To find the cost of energy consumed, we'll multiply the energy consumed by the cost per kWh: $\text{Cost}=8\text{kWh}\times \text{cost per kWh}$ $\text{Cost}=8\text{kWh}\times \text{₹}4.60\mathrm{/}\text{kWh}=\text{₹}36.80$

    So, for part (a), the current drawn by the heater is $18.18\text{A}$. For part (b), the resistance of the heater is $12.1\mathrm{\Omega }$. For part (c), the energy consumed in 2 hours is $8\text{kWh}$. For part (d), the cost if 1 kWh is priced at ₹4.60 is ₹36.80.

21. An electric motor takes 5 amperes current from a 220 volt supply line. Calculate the power of the motor and electrical energy consumed by it in 2 hours.

To calculate the power ($�$) of the motor, we use the formula $�=��$, where $�$ is the voltage and $�$ is the current: $�=220\text{V}\times 5\text{A}=1100\text{W}$

So, the power of the motor is $1100\text{W}$.

To calculate the electrical energy consumed ($�$) by the motor in 2 hours, we use the formula $�=�\times �$: $�=1100\text{W}\times 2\text{hours}=2200\text{Wh}=2.2\text{kWh}$

So, the electrical energy consumed by the motor in 2 hours is $2.2\text{kWh}$.

22. Which uses more energy : a 250 W TV set in 1 hour or a 1200 W toaster in 10 minutes ?

For the TV set: $\text{Energy consumed}=\text{Power}\times \text{Time}=250\text{W}\times 1\text{hour}=250\text{Wh}$

For the toaster: $\text{Energy consumed}=\text{Power}\times \text{Time}=1200\text{W}\times \frac{10}{60}\text{hour}=200\text{Wh}$

Since $250\text{Wh}>200\text{Wh}$, the TV set uses more energy.

23. Calculate the power used in the 2  resistor in each of the following circuits : (i) a 6 V battery in series with 1  and 2  resistors. (ii) a 4 V battery in parallel with 12  and 2  resistors.

(i) For the series circuit: Total voltage (${�}_{\text{total}}$) = 6 V Total resistance (${�}_{\text{total}}$) = $1\mathrm{\Omega }+2\mathrm{\Omega }=3\mathrm{\Omega }$

Using Ohm's Law, $�=\frac{{�}^2}{�}$: $�=\frac{(6\text{V}{)}^2}{2\mathrm{\Omega }}=\frac{36\text{W}}{2\mathrm{\Omega }}=18\text{W}$

(ii) For the parallel circuit: Voltage across the 2 Ω resistor is $4\text{V}$ (since it's in parallel with the battery). $�=\frac{(4\text{V}{)}^2}{2\mathrm{\Omega }}=\frac{16\text{W}}{2\mathrm{\Omega }}=8\text{W}$

24. Two lamps, one rated 40 W at 220 V and the other 60 W at 220 V, are connected in parallel to the electric supply at 220 V. (a) Draw a circuit diagram to show the connections. (b) Calculate the current drawn from the electric supply. (c) Calculate the total energy consumed by the two lamps together when they operate for one hour.

(a) The circuit diagram for the lamps connected in parallel is as follows:

________ Lamp 1 (40 W) | ----| 220 V | ________ Lamp 2 (60 W)

(b) Total power consumed by both lamps: $40\text{W}+60\text{W}=100\text{W}$.

Total current drawn from the electric supply: $�=\frac{�}{�}=\frac{100\text{W}}{220\text{V}}\approx 0.4545\text{A}$

(c) Total energy consumed by both lamps in one hour: $�=�\times �=100\text{W}\times 1\text{hour}=100\text{Wh}$

25. An electric kettle connected to the 230 V mains supply draws a current of 10 A. Calculate : (a) the power of the kettle. (b) the energy transferred in 1 minute.

(a) Power ($�$) of the kettle: $�=�\times �=230\text{V}\times 10\text{A}=2300\text{W}$

(b) Energy transferred in 1 minute: $\text{Energy}=�\times �=2300\text{W}\times \frac{1}{60}\text{hour}=38.33\text{Wh}$

26. A 2 kW heater, a 200 W TV and three 100 W lamps are all switched on from 6 p.m. to 10 p.m. What is the total cost at Rs. 5.50 per kWh ?

To find the total cost of running the appliances, we first need to calculate the total energy consumed by each appliance and then find the total energy consumed by all appliances combined. Then, we can calculate the total cost based on the given price per kWh.

Let's calculate the total energy consumed by each appliance:

1. Heater: $\text{Power}=2\text{kW}=2000\text{W}$ $\text{Time}=4\text{hours}$

$\text{Energy consumed by heater}=\text{Power}\times \text{Time}$ $=2000\text{W}\times 4\text{hours}$ $=8000\text{Wh}$ $=8\text{kWh}$

2. TV: $\text{Power}=200\text{W}$ $\text{Time}=4\text{hours}$

$\text{Energy consumed by TV}=\text{Power}\times \text{Time}$ $=200\text{W}\times 4\text{hours}$ $=800\text{Wh}$ $=0.8\text{kWh}$

3. Lamps (each): $\text{Power}=100\text{W}$ $\text{Time}=4\text{hours}$

$\text{Energy consumed by one lamp}=\text{Power}\times \text{Time}$ $=100\text{W}\times 4\text{hours}$ $=400\text{Wh}$ $=0.4\text{kWh}$

Since there are three lamps, the total energy consumed by all lamps is $0.4\text{kWh}\times 3=1.2\text{kWh}$.

Now, let's find the total energy consumed by all appliances combined: $\text{Total energy consumed}=8\text{kWh}+0.8\text{kWh}+1.2\text{kWh}$ $=10\text{kWh}$

Finally, let's calculate the total cost: $\text{Total cost}=\text{Total energy consumed}\times \text{Cost per kWh}$ $=10\text{kWh}\times ��\mathrm{.}5.50\mathrm{/}\text{kWh}$ $=��\mathrm{.}55$

So, the total cost for running the appliances from 6 p.m. to 10 p.m. is Rs. 55.

27. What is the maximum power in kilowatts of the appliance that can be connected safely to a 13 A ; 230 V mains socket ?

The maximum power ($�$) in kilowatts of the appliance that can be connected safely to a $13\text{A}$, $230\text{V}$ mains socket can be calculated using the formula: $�=��=(13\text{A})\times (230\text{V})\times 10^{-3}\text{kW}=2.99\text{kW}$

So, the maximum power of the appliance that can be safely connected is $2.99\text{kW}$.

28. An electric fan runs from the 230 V mains. The current flowing through it is 0.4 A. At what rate is electrical energy transferred by the fan ?

To find the rate of electrical energy transferred by the fan, we use the formula $�=��$, where $�$ is the voltage and $�$ is the current: $�=230\text{V}\times 0.4\text{A}=92\text{W}$

So, the rate of electrical energy transferred by the fan is $92\text{W}$.

Long Answer Type Question

29. (a) What is meant by “electric power” ? Write the formula for electric power in terms of potential difference and current. (b) The diagram below shows a circuit containing a lamp L, a voltmeter and an ammeter. The voltmeter reading is 3 V and the ammeter reading is 0.5 A.

(i) What is the resistance of the lamp ? (ii) What is the power of the lamp ? (c) Define kilowatt-hour. How many joules are there in one kilowatt-hour ? (d) Calculate the cost of operating a heater of 500 W for 20 hours at the rate of ` 3.90 per unit.

(a) Electric power refers to the rate at which electrical energy is transferred or consumed in an electrical circuit. It is a measure of how quickly electrical energy is converted into other forms, such as heat, light, or mechanical energy, within an electrical device.

The formula for electric power ($�$) in terms of potential difference ($�$) and current ($�$) is given by Ohm's Law: $�=��$ where:

• $�$ is the electric power measured in watts (W),
• $�$ is the potential difference (voltage) across the device measured in volts (V),
• $�$ is the current flowing through the device measured in amperes (A).

 (b) (i) 6  (ii) 1.5 W 

(c)A kilowatt-hour (kWh) is a unit of energy equivalent to the amount of energy consumed or produced by a one-kilowatt (1 kW) electrical device operating for one hour. It is commonly used as a billing unit by electric utilities for measuring electricity consumption.

To calculate the energy in kilowatt-hours, you multiply the power of the device in kilowatts by the time it operates in hours.

One kilowatt-hour is equal to 3.6 million joules (J). This conversion arises from the definition of a watt-hour (Wh), which is 1 watt of power expended for one hour. Since 1 kilowatt-hour is 1000 watt-hours, and 1 watt-hour is equal to 3600 joules, we have:

$1\text{kWh}=1000\text{Wh}=1000\times 3600\text{J}=3,600,000\text{J}$

So, there are 3.6 million joules in one kilowatt-hour.

(d) ` 39.00

Multiple Choice Questions (MCQs)

Questions Based on High Order Thinking Skills (HOTS)

41. State whether an electric heater will consume more electrical energy or less electrical energy per second when the length of its heating element is reduced. Give reasons for your answer.

When the length of the heating element in an electric heater is reduced, the heater will consume less electrical energy per second. This is because the length of the heating element directly affects its resistance.

According to Ohm's Law, $�=��$, where $�$ is the voltage, $�$ is the current, and $�$ is the resistance. If the resistance of the heating element decreases due to its reduced length, and assuming the voltage remains constant, the current flowing through the heating element will increase.

However, the power consumed by the heater is given by $�=��$. Since the voltage remains constant and the current increases, the overall power consumed by the heater will decrease.

Since power is directly proportional to the rate of energy consumption, reducing the length of the heating element will result in less electrical energy being consumed per second.

42. The table below shows the current in three different electrical appliances when connected to the 240 V mains supply : Appliance Current Kettle 8.5 A Lamp 0.4 A Toaster 4.8 A (a) Which appliance has the greatest electrical resistance ? How does the data show this ? (b) The lamp is connected to the mains supply by using a thin, twin-cored cable consisting of live and neutral wires. State two reasons why this cable should not be used for connecting the kettle to the mains supply. (c) Calculate the power rating of the kettle when it is operated from the 240 V mains supply. (d) A man takes the kettle abroad where the mains supply is 120 V. What is the current in the kettle when it is operated from the 120 V supply ?

(a) The appliance with the greatest electrical resistance can be determined by comparing the current values provided. According to Ohm's Law ($�=��$), if the voltage ($�$) remains constant, the appliance with the highest current will have the lowest resistance. Conversely, the appliance with the lowest current will have the highest resistance.

Given the current values:

• Kettle: 8.5 A
• Lamp: 0.4 A
• Toaster: 4.8 A

Since the lamp has the lowest current (0.4 A), it has the highest resistance among the three appliances.

(b) The reasons why the thin, twin-cored cable used for the lamp should not be used for connecting the kettle to the mains supply are:

1. The kettle draws a significantly higher current (8.5 A) compared to the lamp (0.4 A). The thin, twin-cored cable may not be designed to handle such high currents safely, leading to overheating and potential fire hazards.
2. The kettle likely requires a higher power rating, resulting in higher energy transfer through the cable. The cable's insulation and conductor size may not be adequate to handle the increased power without overheating or damage.

(c) To calculate the power rating of the kettle when operated from the 240 V mains supply, we use the formula $�=��$, where $�$ is the voltage and $�$ is the current: $�=240\text{V}\times 8.5\text{A}=2040\text{W}$

So, the power rating of the kettle when operated from the 240 V mains supply is $2040\text{W}$.

(d) When the kettle is operated from the 120 V supply, we can use the formula $�=��$ to find the power rating, assuming the resistance remains constant: $�=\frac{�}{�}=\frac{2040\text{W}}{120\text{V}}=17\text{A}$

So, the current in the kettle when operated from the 120 V supply is $17\text{A}$.

43. A boy noted the readings on his home’s electricity meter on Sunday at 8 AM and again on Monday at 8 AM (see Figures below).

(a) What was the meter reading on Sunday ? (b) What was the meter reading on Monday ? (c) How many units of electricity have been used ? (d) In how much time these units have been used ? (e) If the rate is Rs. 5 per unit, what is the cost of electricity used during this time ?

(a) 42919 (b) 42935 (c) 16 units (d) 24 hours (e) Rs. 80

44. An electric bulb is rated as 10 W, 220 V. How many of these bulbs can be connected in parallel across the two wires of 220 V supply line if the maximum current which can be drawn is 5 A ?

To find out how many of these bulbs can be connected in parallel across the two wires of a 220 V supply line, we need to calculate the total current drawn by one bulb and then determine how many bulbs can be connected while staying within the maximum current limit of 5 A.

Given: Power rating of each bulb ($�$) = 10 W Voltage ($�$) = 220 V Maximum current limit (${�}_{\text{max}}$) = 5 A

Using the formula $�=��$, we can find the current drawn by one bulb: $�=\frac{�}{�}=\frac{10\text{W}}{220\text{V}}$

Now, we can calculate the current drawn by one bulb: $�=\frac{10}{220}\text{A}\approx 0.0455\text{A}$

Next, to find out how many bulbs can be connected in parallel while staying within the maximum current limit of 5 A, we divide the maximum current limit by the current drawn by one bulb: $\text{Number of bulbs}=\frac{5\text{A}}{0.0455\text{A/bulb}}$

$\text{Number of bulbs}\approx 109.89$

Since you cannot have a fraction of a bulb, you can connect a maximum of 109 bulbs in parallel across the two wires of the 220 V supply line without exceeding the maximum current limit.

45. Two exactly similar electric lamps are arranged (i) in parallel, and (ii) in series. If the parallel and series combination of lamps are connected to 220 V supply line one by one, what will be the ratio of electric power consumed by them ?

When two exactly similar electric lamps are arranged in parallel and in series, their power consumption will vary depending on the arrangement.

Let's denote the power consumed by one lamp as $�$. When lamps are arranged in parallel, each lamp receives the full voltage of the supply line (220 V). Thus, the power consumed by each lamp in parallel remains the same as $�$.

When lamps are arranged in series, the total voltage across both lamps is divided equally between them. In this case, each lamp receives half of the supply voltage (110 V). Therefore, the power consumed by each lamp in series can be calculated using $�=��$, where $�$ is the voltage across the lamp and $�$ is the current passing through it. Since the resistance of both lamps is the same, the current passing through each lamp in series will be the same. Thus, the power consumed by each lamp in series will be proportional to the square of the voltage across it.

Now, let's calculate the power consumed by each lamp when arranged in series: ${�}_{\text{series}}=(110\text{V}{)}^2\times \frac{1}{�}$ ${�}_{\text{series}}=\frac{(110\text{V}{)}^2}{�}$

Since the lamps are exactly similar, their resistances will be equal. So, we can simplify the expression: ${�}_{\text{series}}=\frac{(110\text{V}{)}^2}{�}=\frac{(110\text{V}{)}^2}{�}=\frac{(110\text{V}{)}^2}{�\mathrm{/}�}=\frac{110^2}{�}$

Now, let's find the ratio of power consumed by the lamps in parallel to the power consumed by the lamps in series: $\frac{{�}_{\text{parallel}}}{{�}_{\text{series}}}=\frac{�}{\frac{110^2}{�}}$ $\frac{{�}_{\text{parallel}}}{{�}_{\text{series}}}=\frac{{�}^2}{110^2}$

Since $�$ is the power consumed by one lamp, the ratio of power consumed by the lamps in parallel to the power consumed by the lamps in series will be the square of the ratio of the voltage across the lamps in series to the voltage across the lamps in parallel. $\frac{{�}_{\text{parallel}}}{{�}_{\text{series}}}={\left(\frac{220\text{V}}{110\text{V}}\right)}^2={\left(\frac{2}{1}\right)}^2=4$

So, the ratio of electric power consumed by the lamps when arranged in parallel to when arranged in series is $4:1$.