Class 10 Physics Chapter 5 Notes Kerala Syllabus Electric Energy: Consumption and Conservation Questions and Answers

The comprehensive approach in SCERT Kerala Syllabus 10th Standard Physics Textbook Solutions and Class 10 Physics Chapter 5 Electric Energy: Consumption and Conservation Notes Questions and Answers English Medium ensure conceptual clarity.

SSLC Physics Chapter 5 Notes Questions and Answers Pdf Electric Energy: Consumption and Conservation

SCERT Class 10 Physics Chapter 5 Electric Energy: Consumption and Conservation Notes Pdf

SSLC Physics Chapter 5 Questions and Answers – Let’s Assess

Question 1.
An electric heater operating at 230 V draws 2 A current.
a) What is the resistance of the heater?
b) Calculate the heat produced when this heater operates for 10 minutes.
c) What is the power of this appliance?
Answer:
V = 230 V, I = 2A, t = 10 × 60 s
a) R = \(\frac{V}{I}\) = \(\frac{230}{2}\) = 115 Ω

b) H = VIt
H = 230 × 2 × 600
= 276000 J

c) H = Pt
p = \(\frac{276000}{600}\) = 460 W

Question 2.
a) A heating appliance operating at 230 V supply consumes 2 A current. What is the quantity of heat produced in five minutes?
b) What is the energy consumed by this appliance in five minutes while operating at 115 V?
Answer:
a) V = 230 V, I = 2 A, t = 5 × 60 = 300 s
H = 230 × 2 × 300
= 138000J

b) V = 1115 V
R = \(\frac{230}{2}\) = 115 Ω
I = \(\frac{V}{R}\) = \(\frac{115}{115}\) = 1A
H = 115 × 1 × 300
= 34500 J

Question 3.
In a house, a 500 W electric iron operates for one hour, two 40 W fans for 8 hours, and five 12 W LED bulbs for 10 hours. Calculate in unit the energy consumption per day in that house.
The tariff of electricity for domestic consumers in Kerala is given below (subject to change).

Answer:
Electric energy (in kilo watt hour)
= \(\frac{\text { Power in watt } x \text { time in hour }}{1000}\)
Energy consumed by electric iron (in kWh)
= \(\frac{500 \times 1}{1000}\) = 0.5 kWh = 0.5 unit
Energy consumed by fan (in kWh)
= \(\frac{2 \times 40 \times 8}{1000}\) = 0.64 kWh = 0.64 unit
Energy consumed by LED bulb (in kWh)
= \(\frac{5 \times 12 \times 10}{1000}\) = 0.6 kWh = 0.6 unit
Total energy consumption in a day = 0.5 + 0.64 + 0.6 = 1.74 kWh
Total energy consumption in a day in units =1.74 unit

Question 4.
In a house, a 600 W grinder operates for one hour, three 60 W fans for six hours, and ten 10 W LED bulbs for 10 hours daily.
a) What will be the electric energy consumed in units per month?
b) If the appliances operate in the same way, calculate the electricity bill for two months in that house.
(Use the tariff given in table 5.8)
The tariff of electricity for domestic consumers in Kerala is given below (subject to change).

Answer:
a) Energy consumed by grinder
= \(\frac{600 \times 1}{1000}\) = 0.6 kWh = 0.6 unit
Energy consumed by fans
= \(\frac{3 \times 60 \times 6}{1000}\) = 1.08 kWh = 1.08 unit
Energy consumed by bulbs
= \(\frac{10 \times 10 \times 10}{1000}\) = 1 kWh = 1 unit
Total energy consumption in a day
= 0.6 + 1.08 + 1 = 2.68 kWh
= 2.68 unit
Total energy consumption in a month (30 days) = 80.4 units

b) Total energy consumption in for two months (60 days) = 80.4 × 2 = 160.8 units
For up to 250 units of energy consumption it is known as Telescopic and above that it is non telescopic.
For 2 months the energy consumed is 160.8 units. Here the energy consumption comes under the telescopic section.

Check for the energy consumption charge for each range of unit.
For first 50 units of the total consumption ,the charge = unit × energy charge per unit = 50 × 3.35 = 167.5
For the second 50 units of the total consumption,
the charge = 50 × 4.25 = 212.5
For the third 50 units of the total consumption,
the charge = 50 × 5.35 = 267.5
For the remaining 10.8 .units the total
consumption charge = 10.8 × 7.2 = 77.76

So the total consumption charge = Rs. 725.26 The charge for the single phase (usually for household connections) for a month lies in the electricity consumption range of 50-100, is also added up with this to give the total electricity bill.

Total electricity bill = 725.26 + (85 × 2) = 725.26 + 170 = Rs. 895.26 (other charges may also add up to the bill you receive. In the given bill only energy charge per unit and single phase fixed charge is added up based on the tariff | given in table 5.8)

Question 5.
Two heaters A and B operate at 230 V. Heater A draws 2 A current and heater B draws 2.5 A current.

a) Calculate the power of the heaters in figure 5.12 (a) and (b).
b) Which heater has higher resistance?
c) If both heaters operate for 5 minutes each, which heater will produce more heat?
Answer:
a) Power, P = VI
For heater A
V = 230 V, I = 2A
P = 230 × 2 = 460 W
For heater B
V = 230 V, I = 2.5 A
P = 230 × 2.5 = 575 W

b) Resistance of heater A, R = \(\frac{V}{I}\) = \(\frac{230}{2}\) = 115 Ω
Resistance of heater B, R = \(\frac{V}{I}\) = \(\frac{230}{2.5}\) = 92 Ω
Heater A has more resistance

c) H =VIt
t = 5 × 60 = 300 s
For heater A, H = 230 × 2 × 300 = 138000 J
For heater B, H = 230 × 2.5 × 300 = 172500 J
Heater B produces more heat.
When the time of operation is same, the heater with more power produces more heat.

Question 6.
Which of the given statements is correct with regard to the heating element?

(a) Low melting point
(b) High resistivity
(c) Ability to remain in red hot state
(d) Low oxidation resistance
(e) High melting point
(f) High oxidation resistance
i. abfc
ii. abfe
iii. bdfe
iv. beef
Answer:
iv. bcef

Question 7.
Two bulbs, 230 V, 40 W and 230 V, 60 W are arranged as shown in the figures.

If 230 V is applied, which bulb will be brighter in each circuit?
Answer:
For 40 W bulb, R = \(\frac{V^2}{P}\) = \(\frac{230^2}{40}\) = 1322.5 Ω
For 60 W bulb, R = \(\frac{V^2}{P}\) = \(\frac{230^2}{60}\) = 881.7 Ω
40 W has higher resistance than 60 W.

In the parallel circuit (5.13(a)), the voltage across both bulbs is the same (equal to the source voltage, 230 V). The power dissipated by each bulb is given by P = \(\frac{V^2}{R}\). Since the voltage is the same, the bulb with lower resistance will dissipate more power and thus be brighter. Since resistance of 60 W bulb is lesser than resistance of 40 W bulb, the 60 W bulb will be brighter in the parallel circuit.

In the series circuit (5.13(b)), the current flowing through both bulbs is the same. The power dissipated by each bulb is given by P = VI = (IR)I = I²R. Since the current I is the same, the bulb with higher resistance will dissipate more power and thus be brighter. Since resistance of 40 W bulb is higher than resistance of 60 W bulb, 40 W bulb will be brighter in the series circuit.

In general, in a parallel connection, device of higher power will have lower resistance ,it will draw more current, produce more heat and glow brighter and in a series connection, device of lower power will have higher resistance, it will share more voltage and glow brighter.

Question 8.
To reduce electricity consumption in our houses, we can maximise the use of sunlight dining the day. Do you agree with this statement? Explain how.
Answer:
Yes, I agree with the statement.
I do believe that utilising sunlight during the day lowers the amount of electricity used. One free and natural light source that we may use in our home is sunlight. By opening the windows and drapes ,rooms can be made brighter without the use of artificial lighting. Our electricity costs are reduced as a result. To maximize daylight, we should choose locations for work or study next to windows. Mirrors and light-colored walls can improve the room’s ability to reflect sunlight. When we rely upon natural light, it reduces the reliability on electricity from power plants, thus contributing to pollution reduction. Using sunlight to save electricity at home is easy and environmentally beneficial.

Question 9.
In the context of the energy crisis, write down any two suggestions that can be implemented to reduce energy consumption in newly constructed houses.
Answer:

• New houses should have large windows to use sunlight and fresh air instead of lights and fans.
• Solar panels can be installed to produce electricity and lower electricity bills.
• Energy-saving appliances like LED lights and 5-star rated appliances should be used.
• Good insulation in walls and roofs can keep the house cool or warm without using much electricity.

Question 10.
What is the necessity for reducing carbon footprint?
Answer:
Both direct and indirect emissions of greenhouse gases come from people, families, businesses, events, services, and goods. The amount of greenhouse gases that are released, transformed, expressed as equivalent to the measure of carbon dioxide, is carbon footprint.

Reducing the carbon footprint is important because it helps slow down climate change by lowering greenhouse gas emissions. In turn, this reduces the occurrence of severe weather phenomena like droughts and floods. Preserving natural habitats for plants and animals and limiting pollution are two further ways in which it protects the environment. Decreased emissions lead to cleaner air and water, which enhances public health by lowering respiratory and other illnesses. By conserving limited natural resources like fossil fuels, more efficient energy use promotes sustainability. We contribute to a safer and better Earth for coming generations by reducing carbon emissions.

Physics Class 10 Chapter 5 Notes Kerala Syllabus Electric Energy: Consumption and Conservation

Question 1.
Which are the appliances in houses that operate using electricity?

Answer:

• Tube light
• Electric kettle
• Fan
• Electric iron
• Induction cooker
• Electric drill machine
• Refrigerator

Different electrical appliances are used for different purposes. These appliances convert electric energy into different forms of energy.

Question 2.
What forms of energy are produced when a mixie operates? Write them down.
Answer:

• Mechanical energy
• Sound energy
• Heat energy

The electric energy we supplied is transformed into various forms when the mixie is operated. Effect of electric current-The useful form of energy into which an appliance mainly converts electric energy is considered the effect of electric current in that appliance.

Question 3.
Observe the appliances in figure 5.1 and the appliances already listed. Based on their function, identify the effect of electric current utilised in each and complete table 5.1 given below.

Answer:

Name of the appliance	Function	Energy conversion	Effect of electric current
Electric kettle	Heats water	Electric energy → heat energy	Heating effect
Mixie	Grind or mix ingredients	Electrical energy → mechanical energy	Mechanical effect
Electric iron	To iron clothes	Electric energy → heat energy	Heating effect
Tube light	Provides light	Electric energy → Light energy	Lighting effect

Question 4.
From the table, you may have understood that electric current can produce various effects. Which of these appliances produce heat?
Answer:

• Electric iron
• Electric kettle

Question 5.
Connect a 5 cm long nichrome wire to a 9 V battery as shown in Figure 5.2.

Switch on and observe the nichrome wire.
What change do you notice?
Answer:
When current passes through the nichrome wire, heat energy is produced.

The process of production of heat, when electricity flows through a conductor is Joule heating or Ohmic heating. The part used to produce heat energy in heating appliances is the heating element.

Observe the pictures of the given electric heating appliances.

You may have noticed that the part common to all these appliances is the heating coil (heating element).
The heating coil is made of an alloy, nichrome.
(There are also heating appliances which does not make use of heated coils. Microwave ovens and
induction cookers are a few examples.)

Question 6.
What are the characteristics of nichrome that make it suitable for a heating element?
Answer:

• High oxidation resistance (oxidation resistance is the ability of a material to resist corrosion due to contact with oxygen or other oxidisers at high temperature).
• Ability to provide heat energy for a long time in a red hot state.
• High resistivity (due to this property even a nichrome wire of short length can provide sufficient resistance)
• High melting point(due to this property, nichrome is able to tolerate the high temperatures needed
  for efficient heating without melting or deforming.)

We know that all materials do not conduct electricity in the same way.

Resistivity is one of the intrinsic properties of a material that mainly determines the resistance which is the characteristic of a conductor. Resistivity is different for each material.

Resistivity and conductivity.
The resistance of a conductor with an area of cross section of one square metre and a length of one metre is the resistivity of the material of the conductor. It is denoted by the Greek letter ρ(rho). Its unit is ohmmetre (Ωm). Conductivity is the ability of a material to conduct electricity. This is the reciprocal of resistivity.

The factors which influence the resistance of a conductor are length of the conductor
(l), area of cross section (A) and the nature of the material (ρ). The relation of resistance (R) with these factors is given by the equation R = \(\frac{\rho l}{A}\)

Activity to find out the factors that influence the heat produced in a conductor carrying current.
• Construct a circuit which consists of a 5 cm long nichrome wire, 6 V battery eliminator, etc., as shown in figure 5.4. Take 20 mL of water in a boiling tube. Immerse the nichrome wire and a thermometer into it. Adjust the rheostat to vary the current. Record the temperature in table 5.2 every two minutes. Repeat the experiment by replacing the water in the boiling tube each time.

Observation
Temperature of water before starting the experiment = 31°C

Current (A)	Temperature (°C)
0.5	33
1	35
1.5	37

Analysis
On analysing table 5.2 given above we can understand that the temperature of water increases with increase in current. In this experiment a fixed quantity of water (20 mL) is heated by passing current. To increase the temperature of a fixed quantity of water more heat is required. Hence as the quantity of heat received by the water increases, the temperature also increases. Therefore, it is understood that when the current increases heat produced also increases.

• Adjust the rheostat to make the current 1 A. In table 5.3 given below record the temperature of water, every 2 minutes.
Observation

Time (in minutes)	Temperature (°C)
0	31
2	33
4	35
6	32

Analysis
Temperature increases because the current flowed for a longer time. We can understand that the heat produced increases as the time of flow of current increases.

Activity
Take nichrome wire and aluminium wire of equal length (5 cm) and equal area of cross section. Using them make a circuit as shown in the figure.

Pass current through the circuit for about 2 minutes using a 9 V, 50 W eliminator. Place a thin paper on both aluminium and nichrome wire.

Question 7.
What do you observe?
Answer:
Because the nichrome wire generates a lot more heat than the aluminium wire, we can see that the thin paper on the nichrome wire is likely to bum or char while the paper on the aluminium wire would either stay the same or only warm slightly.

Question 8.
Which conductor becomes hotter?
Answer:
Nichrome wire becomes hotter.

Question 9.
The magnitude of current in the nichrome wire and aluminium wire is (the same / different).
Answer:
Same

Question 10.
The resistance of nichrome wire and aluminium wire is (the same/different).
Answer:
Different
The nichrome wire become hotter because its resistance is greater than that of the aluminium wire. Resistance is another factor that influences the quantity of heat produced in a conductor carrying current.

Question 11.
From the experiments conducted, note down the three factors that influence the quantity of heat produced in a current carrying conductor:
Answer:

• Current (I)
• Resistance of the conductor(R)
• Time for which current flows(t)

The scientist James Prescott Joule discovered the relation between the heat produced in an electric conductor and the factors influencing the quantity of heat.

Joule’s law
The quantity of heat produced in a current carrying conductor is directly proportional to the square of the current (I²), the resistance of the conductor (R) and the time (t) for which the current flows. This is Joule’s law. If a current I flows through a conductor having resistance R for a time t, according to Joule’s law, the quantity of heat produced will be H = I²Rt. Its SI unit is joule (J).

Question 12.
A current of 0.5 A flows through a conductor with 100 Ω resistance for 5 minutes.
a) What will be the quantity of heat produced?
b) If a current of 1A flows through this conductor for 5 minutes, what will be the quantity of heat produced?
Ans:
a) R = 100 Ω
I = 0.5 A
t = 5 × 60s
H = ?
H = I²Rt
= 0.5 × 0.5 × 100 × 5 × 60 J
= 7500 J

b) I = 1A, H = I²Rt
H = 1 × 1 × 100 × 5 × 60 J
= 30000J

Question 13.
When the current was doubled, the heat produced in the conductor became 4 times. If so, what happens to the heat produced if the current is halved?
(halved/ quartered)
Answer:
quartered
The electrical appliances commonly used in houses operating at 230 V.

Other forms of Joule’s Law
We can formulate other equations to calculate the quantity of heat.
According to ohm’s law
V = IR
Hence I = \(\frac{V}{R}\)
On substituting the value of I in the equation
H = I²Rt
H = (\(\frac{V}{R}\))² Rt
So, we get H = \(\frac{V^2 t}{R}\)
Similarly, if we substitute R = \(\frac{V}{I}\) in the equation
H = I²Rt
We get H = VIt
H = I²Rt = \(\frac{V^2 t}{R}\) = VIt

Question 14.
If the resistance of a heating appliance operating at 230 V is 920 Ω , what is the quantity of heat produced in 10 minutes?
Answer:
V = 230 V, R = 920 Ω, t = 10 × 60 s
H = \(\frac{V^2 t}{R}\)
= \(\frac{230 \times 230}{920}\) × 10 × 60 = 34500 J

Question 15.
If 7500 J of heat is produced in 30 s in a conductor having 2500 resistance, what is the current through the conductor?
Answer:
H = 7500 J,
t = 30 s,
R = 250 Ω
H = I²Rt
\(\frac{H}{R t}\) = I²
I = \(\sqrt{\frac{H}{R t}}\) = \(\sqrt{\frac{7500}{250 \times 30}}\)
I = √1 = 1A

Question 16.
Calculate the quantity of heat produced if 2 A current flows for 10 minutes through a heating coil of an electric kettle having 100 Ω resistance?
Answer:
I = 2A,
t = 10 × 60 = 600 s,
R = 100 Ω
H = I²Rt
H = 2² × 100 × 600 = 240000 J

Question 17.
A potential difference of 230 V is applied across a circuit for 5 minutes. If the resistance in the circuit is as given below, calculate the current and heat in each case.
(a) 115 Ω
(b) 230 Ω
Answer:
V= 230 V, t = 5 × 60 = 300 s
(a) R = 115 Ω
Current, I = \(\frac{V}{R}\) = \(\frac{230}{115}\) = 2A
Heat, H = VIt = 230 × 2 × 300
H = 138000 J

(b) R = 230 Ω
Current, I = \(\frac{V}{R}\) = \(\frac{230}{230}\) = 1A
Heat, H = VIt = 230 × 1 × 300
H = 69000 J

Question 18.
Observe the circuits given below. Calculate the quantity of heat produced in each, in 5 minutes.

Answer:

Heat produced in circuit 5.6(a)	Heat produced in circuit 5.6(b)	Heat produced in circuit 5.6(c)
H = I2Rt = 0.25 × 0.25 × 48 × 300 = 900 J	H = I2Rt = 0.5 × 0.5 × 24 × 300 = 1800 J	H = I2Rt = 1 × 1 × 1 × 300 = 3600 J

Question 19.
Analysing the completed table, answer the following questions.
a) What is the difference in the quantity of heat in circuits 5.6 (a) and 5.6 (b)?
Answer:
Heat in circuit 5.6 (a) = 900 J
Heat in circuit 5.6 (b) = 1800 J
Difference in quantity of heat = 1800 – 900
= 900 J

b) Why is there a difference in the quantity of heat produced in these circuits?
Answer:
Even though the voltage and time are the same, different combinations of current and resistance cause a variation in the amount of heat generated. In particular, even with a lower resistance, the greater current in circuit 5.6 (b) -0.5 A when compared to 0.25 A in 5.6 (a) results in a considerably greater I² term, which generates more heat.

c) What is the resistance of the circuit in figure 5.6 (c)?
Answer:
12 Ω

d) What is the total heat produced in the circuit in figure 5.6 (c)?
Answer:
H = 3600 J

e) Less heat is produced in the resistor of higher resistance in the circuit in figure 5.6 (a). Why?
Answer:
When the current is constant, the heat generated is directly proportional to the resistance, as stated by Joule’s law (H = I²Rt). However, a larger resistance results in a smaller current (I = V/R) if the voltage is constant, for all these three circuits. The effect of square of current mainly contributes to heat generated because heat is dependent on I²R. Therefore, less heat is produced in the resistor of higher resistance in the circuit in figure 5.6 (a).

f) Among these circuits, in which resistor is the maximum quantity of heat produced?
Answer:
In the 12 resistor in circuit 5.6(c)

g) What are your inferences? Record them in the science diary.
Answer:
If voltage is constant, when the resistance in the circuit is decreased, the current increases. Hence the quantity of heat produced increases.

Question 20.
If the voltage remains constant, what change will occur in the quantity of heat produced when the resistance varies? Explain based on Joule’s law.
Answer:
Joule’s law is given by H = I²Rt. However, when the voltage (V) is constant, expressing the current using Ohm’s law and substituting in joule’s law, we get H = \(\frac{V^2 t}{R}\). This equation shows that the heat generated (H) is inversely proportional to the resistance (R) when V and t are constant. This indicates that as resistance increases, heat generated decreases and when resistance decreases heat generated increases.

Question 21.
Calculate the quantity of heat produced if 2 A current flows through an electric heater operating at 230 V for 15 minutes. What is the resistance of the heater?
Answer:
I = 2A, t = 15 × 60 = 900s, V = 230 V
Heat produced, H =VIt
H = 230 × 2 × 900 =414000 J
Resistance of the heater = R = \(\frac{V}{I}\)
R = \(\frac{230}{2}\) = 115

Question 22.
An electric iron operating at a potential difference of 230 V has a heating coil of resistance 100 £2. If this electric iron operates for half an hour, how much electric energy will be converted to heat energy? What is the current through the electric iron?
Answer:
V = 230 V, R = 100 Ω, t = 30 × 60 = 1800 s
H = \(\frac{V^2 t}{R}\)
H = \(\frac{230^2 \times 1800}{100}\)
Current through the electric iron, I = \(\frac{V}{I}\) = \(\frac{230}{100}\) = 2.3 A

Question 23.
A heating appliance having a resistance of 92 Ω operates at 230 V. Using various equations, calculate the heat produced by the appliance in 14 minutes and write them down in the table below. What resistance is needed to double the heat energy?

Answer:

H = I2Rt	H = \(\frac{V^2 t}{R}\)	H = VIt
V = 230 V R = 92 Ω t = 14 minute = 14 × 60 = 840s I = \(\frac{V}{R}\) = \(\frac{230 \mathrm{~V}}{92 \Omega}\) = 2.5 A H = I2Rt = (2.5)2 × 92 × 840 J = 483000 J	V = 230 V R = 92 Ω t = 14 minute = 14 × 60 = 840s H = \(\frac{V^2 t}{R}\) H = \(\frac{230^2 \times 840}{92}\) = 483000 J	V = 230 V I = \(\frac{V}{R}\) = \(\frac{230 \mathrm{~V}}{92 \Omega}\) = 2.5 A t = 14 minute = 14 × 60 = 840 s H = VIt H = 230 × 2.5 × 840 = 483000 J

As the voltage and time is remaining constant, we take H = \(\frac{V^2 t}{R}\)
For H to become 2H(to double the heat energy),
we need to find new resistance value-R_new
2H = \(\frac{\mathrm{V}^2 \mathrm{t}}{\mathrm{R}_{\text {new }}}\)
We have 2\(\frac{V^2 t}{R}\) = \(\frac{\mathrm{V}^2 \mathrm{t}}{\mathrm{R}_{\text {new }}}\), \(\frac{2}{R}\) = \(\frac{1}{\mathrm{R}_{\text {new }}}\)
R_new = \(\frac{R}{2}\) = \(\frac{92}{2}\) = 46 Ω

Question 24.
Observe the values given on the label of the appliance in figure 5.7. What does marking 750 W indicate?

Answer:
750 W indicates the power at which the appliance work.
Energy conversion occurs when an appliance operates, is the work done. The quantity of work done per unit time is power. Power is the rate of work based on time.

Question 25.
The power of each electric appliance is marked on it. Complete the table given below.

Answer:

Device	Power (P)W	Energy consume in 1 second (Pt = p × 1)
LED Bulb	9	= 9 × 1 = 9 J
Projector	100	= 100 × 1 = 100 J
Fan	60	= 60 × 1 = 60 J
Laptop	40	= 40 × 1 = 40 J

While labelling the power of an appliance, the voltage at which it is to be operated is also indicated. In our country, the power obtained while operating at 230 V is labelled on electrical appliances. If the appliance operates above or below 230 V, the power will vary accordingly. That is, if the voltage increases, the power increases and if the voltage decreases, the power also decreases. This may lead to the damage of appliances. (There are also some appliances that operate at 400 V).

Countries design electrical appliances and power according to their power supply voltage. In all electrical appliances, the voltage required for it and its power will be marked on it. The indicated power is available from an appliance only when it operates at the specified voltage.

If the voltage is constant, according to P = \(\frac{V^2}{R}\), appliances with higher resistance will have lower power, and those with lower resistance will have higher power. While solving numerical problems related to household appliances, it is more desirable to use the formula P = \(\frac{V^2}{R}\). In some foreign countries, the supply voltage is 110 V.

Question 26.
At what applied voltage will an electric heating appliance marked 800 W, 240 V produce a power of 200 W?
Answer:
P₁ = 800W, V₁ = 240 V
R = \(\frac{\mathrm{V}_1^2}{\mathrm{P}}\) = \(\frac{240^2}{800}\)
R = 72 Ω
P₂ = 200 W, V₂ = ?
V₂² = P₂ × R
V₂² = 200 × 72 = 14400
V₂ = \(\sqrt{14400}\) = 120 V
Whenever voltage becomes half, power becomes 1/4 times the original power.

Question 27.
Calculate the quantity of heat produced when a heating appliance of power 500 W operates for five minutes.
Answer:
P = 500 W
t = 5 min = 5 × 60 = 300 s
H = Pt = 500 × 300 J = 150000 J

Question 28.
How much heat energy is produced when an electric heater with 600 W power operates for 20 minutes?
Answer:
P = 600 W, t = 20 × 60 = 1200 s
H = Pt = 600 × 1200 = 720000 J

Question 29.
How much heat energy will be produced if a heating appliance with 250 W power operates for 10 minutes?
Ans:
P = 250 W, t = 10 × 6 = 600 s
H = Pt = 250 × 600 = 150000 J

Question 30.
An electric heater produced 450000 J of heat energy when operated for 5 minutes. What would be the power of this heater?
Answer:
H = 450000 J, t = 5 × 60 = 300 s
H = Pt, P = \(\frac{H}{t}\)
P = \(\frac{450000}{300}\) = 1500 W

Question 31.
On a heating appliance it is marked 529 W, 230 V.
a) What do each of these mean?
b) What would be the power of this appliance if 100 V is supplied?
Answer:
a) 529 W, 230 V indicates that the appliance operating at voltage of 230 V consumes 529 W power.

b) R = \(\frac{\mathrm{V}^2}{\mathrm{P}}[latex] = [latex]\frac{230^2}{529}[latex] = 100 Ω
P = [latex]\frac{\mathrm{V}^2}{\mathrm{R}}[latex]
P = [latex]\frac{100^2}{100}[latex] = 100 W

Question 32.
Calculate the energy consumed by a 1 kW power appliance in one hour.
Answer:
E = Pt = 1kW × 1h
= 1kWh

One kilowatt hour or one unit of electric energy is the energy consumed by an appliance with 1000 W (1 kW) power in one hour.

One unit of electric energy in joules

1000W = 1kW

Energy consumed,
E = H = Pt
P = 1000W, t = 1 h = 60 × 60s
El = 1000 × 60 × 60 J
= 3600000 J = 3.6 × 10⁶ J

1kWh = one unit of electric energy = 3.6 × 106 J

Question 33.
How much heat energy will be produced when a heating appliance of 60 W operates for one hour?
Answer:
H = Pt
= 60 × 1 × 60 × 60 J
= 216000 J

Question 34.
The power and operating time of some appliances are given in the table. Complete the table.

Answer:

Power of device (P)W	Time (t)h	Electric energy spent (Pt) J	Energy utilized in unit (kWh) = [latex]\frac{\text { Power in watt }}{1000}\) × hour
1000	1 hour	1000 × 60 × 60 = 3600000 J	1k Wh = 1 unit
2000	1 hour	2000 × 60 × 60 = 7200000 J	2k Wh = 2 unit
500	1 hour	500 × 60 × 60 = 1800000 J	0.5 kWh = 0.5 unit
500	2 hour	500 × 2 × 60 × 60 = 3600000 J	Ik Wh = 1 unit

Question 35.
Calculate the electric energy consumed if a 500 W grinder and a 600 W electric iron each operate for 2 hours.
Answer:
Total power in watt = 500 W + 600 W
Time = 2 h
Electric energy (in kilo watt hour)
= \(\frac{\text { Power in watt × time in hour }}{1000}\)
= \(\frac{1100 \mathrm{~W} \times 2 \mathrm{~h}}{1000}\)
= 2.2 kWh
= 2.2 unit
Using this formula, we can calculate the monthly consumption of electricity.

Question 36.
Considering the total power of the appliances used and their operating time, calculate the daily electricity consumption in your house. Similarly, calculate the monthly consumption of electricity and present in class the main findings and suggestions based on the project report.
Answer:
An example is given below
• Prepare a table listing the given information

Appliance	Power (Watt)	Hours used per day	Number of appliances
Fan	75 W	10 hours	3
Tube light	40 W	6 hours	4
Television	100 W	4 hours	1
Refrigerator	150 W	24 hours	1
Washing Machine	500 W	1 hour	1
Steamer	1000 W	0.5 hours	1

• Calculate energy consumption in a day
  Electric energy (in kilowatt hour)

= 8.21 kWh = 8.21 units

• Calculate energy consumption in a month (30 days) = 8.21 × 30 = 246.3 units
• Analyze the findings. Identify which were among the high consuming appliances.
• Compare with energy consumption pattern in the previous month.
• Take action if the energy consumption is increased very much.
• Some actions that can be taken are
  • Turn off appliances when not in use to save energy.
  • Use energy-efficient appliances (like LED lights).
  • Avoid using high-power devices unnecessarily (like steamer).
  • Use natural light during the day to reduce lighting load.
  • Consider installing solar panels for long-term savings.
• Stay alert of your energy consumption by tracking it,which helps you to take appropriate action whenever necessary.

All houses have the different electricity tariff. Electricity charges are levied at a higher rate from consumers who use more electricity and at a lower rate from consumers who use less electricity. (Table 5.8)

Now we can understand that one of the reasons for the increased consumption of electricity is the unwanted and careless consumption of electricity. The amount in the electricity bill can be reduced by lowering electricity consumption.

Question 37.
What can we do to reduce electricity consumption at home?
Answer:

• Use energy efficient electrical appliances.
• Turn off switches immediately after use.
• Use LED tubes and bulbs.
• Choose the size of the fan according to the size of the room.
• Use BLDC fans. [BLDC – Brush Less Direct Current]
• Electronic appliances should be unplugged while not in use.
• Switch to energy star rated appliances which helps in reducing electricity bills, have less environmental impact and government benefits are applicable to them. (Like offer rebates or incentives to purchase them)

Further activities to reduce energy consumption can be implemented under the leadership of the School Energy Club in collaboration with the Energy Management Centre (EMC).

Question 38.
Do large scale electricity generating power plants cause environmental impacts? Discuss.
Answer:
Large scale electricity generating power plants are important for providing power to cities, industries, and homes. They produce a large amount of electricity and offer a steady and reliable supply. However, many of these plants, especially those using coal or gas, cause serious environmental problems like air pollution, global warming, and water pollution. They often use large amounts of water and land, which can harm wildlife and displace communities. Large scale hydropower projects can cause environmental problems. They often lead to the destruction of forests, loss of wildlife habitats, and the shifting of people due to dam construction. These projects can also disrupt river ecosystems. But compared to thermal and nuclear power plants hydro power is a reliable alternative. There are some demerits in all, but ways are to be identified to tackle the disadvantages. Switching on to other renewable energy sources which causes less pollution can become a smart and sustainable solution to the energy and environmental challenges caused by other large scale power plants.

Observe figure 5.9.
There are ways to produce electricity without causing pollution.

You know that the sun is the basis of all energy on the earth.

Question 39.
List the devices that utilise solar energy.
Answer:
• Solar water heater
• Solar cooker
• Solar furnace

Solar cells, panels and power plants
Solar cells are devices that convert solar energy into electric energy. Only negligible voltage and current can be obtained from a single cell. Therefore, a solar panel is constructed by arranging many solar cells suitably to provide electricity as per the requirement. The electricity obtained from such panels can be stored in a storage battery and utilised when needed. Or it can be directly supplied to the electricity distribution agencies. Solar panels are used for energy requirements in artificial satellites. In addition to small scale requirements, large scale solar power plants are also in operation.

• Cochin International Airport Limited (CIAL) in Nedumbassery, Kerala, produces electricity required for its entire operations from such a solar power plant.
• Cochin International Airport is the world’s first airport to operate entirely on solar power.

Question 40.
List the benefits of installing solar panels.
Answer:

• Can produce electricity required for homes.
• Can reduce environmental pollution.
• Transmission and distribution losses are minimised as electricity is generated at the place of consumption.

It is understood that atmospheric pollution is reduced when solar panels are used. We know that the quantity of greenhouse gases in the atmosphere is increasing daily. An increase in such gases cause global warming, resulting in climate change. Each of our interventions in this matter is very important. Global warming can be reduced by reducing the production of major greenhouse gases like carbon dioxide, methane etc. We need activities that can reduce the quantity of carbon and carbon compounds emitted by individuals, organisations, and products.

Question 41.
What can be done to reduce carbon footprint?
Answer:

• Reduce domestic energy consumption.
• Avoid wasting food.
• Use public transport.
• Reduce waste by utilising reusable products.
• Educate society about reducing carbon footprint.
• Perform energy consuming activities in an energy saving manner.

Question 42.
Prepare a seminar paper on how to reduce individual carbon footprint and present it in class.
Answer:
An example is given below.
Introduction
One of the most significant issues our world is facing now is climate change. It is mostly brought on by gases like carbon dioxide that are emitted into the atmosphere as a result of things like using electricity, driving a car, and creating garbage. These gases raise Earth’s temperature by trapping heat in the atmosphere, which can result in hazardous weather and environmental changes. Every individual’s, everyday activities add to this issue, and their carbon footprint is the overall quantity of greenhouse gases they create. Lowering carbon footprint helps in slowing down climate change and thus helps in preserving the earth for future generations, so lowering it is cmcial. We may take action to lessen our carbon footprint by being aware of its origins.

Methods for Lowering Carbon Footprint

• Conserve energy at home-When not in use, turn off fans, lights, and electrical equipment. Reduce the amount of electricity you use by using energy-efficient appliances and bulbs.
• Choose for greener transportation-Whenever feasible, take public transportation, walk, or cycling instead of driving a car. Carpooling with friends or travel mates reduces emissions as well. It is even better to use electric or fuel- efficient vehicles.
• Reducing, reusing, and recycling-Reduce the use of plastic bags and bottles. Use reusable bags and bottles instead of single-use plastics. Recycling glass, plastic, and paper contributes to a decrease in pollution and waste.
• Conserve water-As you brush your teeth, turn off the faucets and promptly address any leaks. Conserving water also reduces the energy required for water treatment.
• Encourage environmentally friendly products- Purchase goods that are manufactured sustainably and with less packaging. More green activities are encouraged when businesses that care about the environment are supported.
• Raise awareness-Discuss with loved ones the significance of lowering carbon footprints. To have a greater influence, join or form school-based environmental organizations.

Conclusion
It is not very easy to reduce carbon footprint, yet when many people make minor changes in their daily lives, the impact may be significant. Each of us may contribute to protecting the planet by minimizing trash, eating sensibly, adopting more environmentally friendly transportation, conserving energy, and raising awareness. Together we can make the planet a better and cleaner place to live.

‘Reducing atmospheric pollution and conserving energy is everyone’s duty. Energy conservation is equivalent to energy production. ’

‘Energy is precious, don’t waste it! ’

Std 10 Physics Chapter 5 Notes – Extended Activities

Question 1.
Observe the electricity bill of your house and find out the monthly electricity consumption. Present activities that can be implemented in your house to reduce electricity consumption in a seminar in the Energy Club. Also, prepare necessary posters for this.
Answer:
By observing the electricity bill one can find out the monthly electricity consumption. Total units consumed and amount to be paid can be identified and it can be compared with previous month to understand the pattern of energy consumption.

Activities that can be implemented to reduce electricity consumption

• Make a transition to energy-efficient appliances.
• Maximize the performance of your air conditioning and heating systems.
• Use LED bulbs-LEDs use less power and last longer than normal bulbs.
• Turn off electrical appliances when not in use-Always switch off fans, lights, and appliances when you leave a room.
• Unplug devices-Chargers, TVs, and microwaves still use power when plugged in. So electronics should be unplugged while not in use.
• Use natural light during the day-Open curtains and windows instead of turning on lights.
• Use washing machine for full loads only-This saves both electricity and water.
• Limit geyser use-Use it only for a short time and turn it off immediately after use.
• Check your meter weekly-Keep a track of electricity consumption which will make you aware of the
  energy usage.

Switch Off to Switch On a Better Future

Save Energy, Save Earth

Question 2.
List the activities that the School Science Club intends to undertake to reduce the school’s carbon footprint.
Answer:

• Plan tree-planting activities and create green spaces on the school’s property.
• To reduce waste, set up recycling facilities and encourage the use of reusable products.
• When not in use, remind employees and students to turn off lights and electronics.
• Encourage water-saving practices including repairing leaks and properly shutting off taps.
• Increase public knowledge of climate change and sustainability through seminars, discussions, and initiatives.
• Encourage environmentally sustainable modes of transportation, such as walking, bicycling, or ridesharing, to get to school.
• Initiate the process , of composting organic waste for school gardens.
• Examine the school’s energy usage on a regular basis and make recommendations for ways to reduce , electricity use.
• Join local or national green projects by working with environmental organizations.
• Reduce the amount of plastic used in the cafeteria and promote eco-friendly options for school supplies.

Electric Energy: Consumption and Conservation Class 10 Notes

Electric Energy: Consumption and Conservation Notes Pdf

• The conversion of electric energy mainly into heat energy is the heating effect of electric current. Appliances that utilise this effect are electric heating appliances. –
• The process of production of heat, when electricity flows through a conductor is Joule heating or Ohmic heating. The part used to produce heat energy in heating appliances is the heating element.
• The resistance of a conductor with an area of cross section of one square metre and a length of one metre is the resistivity of the material of the conductor. It is denoted by the Greek letter ρ(rho). Its unit is ohm metre (Ωm).
• Conductivity is the ability of a material to conduct electricity. This is the reciprocal of resistivity.
• The factors which influence the resistance of a conductor are length of the conductor (l), area of cross section (A) and the nature of the material (ρ). The relation of resistance (R) with these factors is given by the equation R = \(\frac{\rho l}{A}\)
• Factors that influence the quantity of heat produced in a current carrying conductor are
  • Current (I)
  • Resistance of the conductor (R)
  • Time for which current flows (t)
• The quantity of heat produced in a current carrying conductor is directly proportional to the square of the current (I²) the resistance of the conductor (R) and the time (t) for which the current flows. This is Joule’s law.
• If a current I flows through a conductor having resistance R for a time t, according to Joule’s law, the quantity of heat produced will be H = I²Rt. Its SI unit is joule (J).
• Power is the quantity of work done by an electrical appliance per unit time. The work done by an electrical appliance is the conversion of electric energy into another form of energy.
• The quantity of electric energy consumed in houses can be measured directly by a watt-hour meter connected to our household electric circuit. In this, electric energy is measured in kilowatt hour (kWh) units. This is the commercial unit of electric energy.
• One kilowatt hour or one unit of electric energy is the energy consumed by an appliance with 1000 W (1 kW) power in one hour.
• Energy crisis is the increase in demand for energy and the decrease in availability.
• Solar cells are devices that convert solar energy into electric energy.
• Greenhouse gases are directly and indirectly emitted by individuals, families, organizations, events, services, products, etc. The quantity of such emitted greenhouse gases, converted and expressed as equivalent to the measure of carbon dioxide, is carbon footprint.

INTRODUCTION

Electricity plays a crucial role in our everyday activities and is used in many forms. This chapter explores how electrical energy can be transformed into other types of energy, such as heat, light, and mechanical energy. A key topic is Joule heating, which explains how electric current produces heat in wires and devices. We also learn about electric power, which tells us the rate at which electrical energy is used, and how it is measured using a watt-hour meter. The electricity used in homes is measured in kilowatt-hours (kWh), which is the standard commercial unit. Excessive use of energy, especially from non-renewable sources, contributes to the energy crisis and increases our carbon footprint, damaging the environment. To solve these problems, switching to solar energy and other renewable sources is a cleaner and more sustainable option.

Heating effect of electric current

• The useful form of energy into which an appliance mainly converts electric energy is considered the effect of electric current in that appliance.
• The conversion of electric energy mainly into heat energy is the heating effect of electric current. Appliances that utilise this effect are electric heating appliances.
• The process of production of heat, when electricity flows through a conductor is Joule heating or Ohmic heating. The part used to produce heat energy in heating appliances is the heating element. The heating coil is made of an alloy, nichrome.

Resistivity and conductivity

• The resistance of a conductor with an area of cross section of one square metre and a length of one metre is the resistivity of the material of the conductor. It is denoted by the Greek letter ρ(rho). Its unit is ohmmetre (Ωm).
• Conductivity is the ability of a material to conduct electricity. This is the reciprocal of resistivity.
• The factors which influence the resistance of a conductor are length of the conductor (l), area of cross section (A) and the nature of the material (ρ). The relation of resistance (R) with these factors is given by the equation R = \(\frac{\rho l}{A}\)
• Factors that influence the quantity of heat produced in a current carrying conductor are
  • Current (I)
  • Resistance of the conductor(R)
  • Time for which current flows(t)

Joule’s law of heating

• The quantity of heat produced in a current carrying conductor is directly proportional to the square of the current (I²), the resistance of the conductor (R) and the time (t) for which the current flows. This is Joule’s law.
• If a current I flows through a conductor having resistance R for a time t, according to Joule’s law, the quantity of heat produced will be H = I²Rt. Its SI unit is joule (J).

Electric power

• Power is the quantity of work done by an electrical appliance per unit time. The work done by an electrical appliance is the conversion of electric energy into another form of energy.
• The quantity of electric energy consumed in houses can be measured directly by a watt-hour meter connected to our household electric circuit. In this, electric energy is measured in kilo watt hour (kWh) units. This is the commercial unit of electric energy.
• One kilowatt hour or one unit of electric energy is the energy consumed by an appliance with 1000 W (1 kW) power in one hour.

Energy crisis
• Though the demand for energy has increased many folds, production has not sufficiently increased. This situation is energy crisis. Simply, we say Energy crisis is the increase in demand for energy and the decrease in availability.

Solar cells, panels and power plants

• Solar cells convert sunlight into electrical energy, but a single cell produces only a small amount of power.
• To meet higher energy needs, many cells are combined to form solar panels, which can store electricity in batteries or supply it to the power grid.
• Solar panels are used in satellites, homes, and large-scale solar power plants for energy production with less pollution.

Carbon footprint

• Greenhouse gases are directly and indirectly emitted by individuals, families, organizations, events, services, products, etc. The quantity of such emitted greenhouse gases, converted and expressed as equivalent to the measure of carbon dioxide, is carbon footprint.
• Each individual should strive to reduce his/her personal carbon footprint. This can be achieved by paying attention to daily activities.

HEATING EFFECT OF ELECTRIC CURRENT
The conversion of electric energy mainly into heat energy is the heating effect of electric current. Appliances that utilise this effect are electric heating appliances.

ELECTRIC POWER
Power is the quantity of work done by an electrical appliance per unit time. The work done by an electrical appliance is the conversion of electric energy into another form of energy.

The function of electric heating appliances is to produce heat. You have learnt that the quantity of heat produced
H = I²Rt
Work = quantity of heat produced
W = H = I²Rt
Equation to find electric power
Power = \(\frac{Work}{time}\)
P = \(\frac{W}{t}\) = \(\frac{H}{t}\)
P = \(\frac{I^2 R t}{t}\)
P = I²R
According to Ohm’s law, I = \(\frac{V}{R}\)
Then P = \(\frac{V^2}{R}\)
We know that R = \(\frac{V}{I}\)
Hence P = VI
We will get the equations = P = \(\frac{V^2}{R}\), P = VI
P = I²R = \(\frac{V^2}{R}\) = VI
If we substitute the value of P in the equations used to calculate heat, H = I²Rt.We get
H = Pt
It is understood that the quantity of heat produced in a given time can be found if the power of an appliance is known.

Watt-hour meter

The quantity of electric energy consumed in houses can be measured directly by a watt hour meter connected to our household electric circuit. In this, electric energy is measured in kilowatt hour (kWh) units. This is the commercial unit of electric energy. Power is expressed in watt, time in second, and energy in joule.

The energy consumed if a 9 W bulb operates for one hour daily for 30 days is E = Pt = 9 × 30 × 3600J = 972000J (nine lakh seventy two thousand joules). So, if we consider the monthly energy consumption of all the appliances in a house, the total would be very large, making it inconvenient to record. Therefore, energy is calculated in kilowatt hour by measuring power in kilowatt and time in hour.

ENERGY CRISIS
Despite having many small scale electricity projects in addition to large power stations, the electricity they produce is not sufficient for our needs. Though the demand for energy has increased many folds, production has not sufficiently increased. This situation is energy crisis. This is why sometimes power cuts and load shedding are to be implemented.

‘Energy crisis is the increase in demand for energy and the decrease in availability.’

There are hydroelectric power plants, thermal power plants, nuclear power plants, etc., to produce electricity on a large scale. Yet, it is not easy to increase the electricity production.

Rooftop power project
Solar panels can be installed on rooftops of houses and other places receiving sunlight. The Rooftop Power Project is a scheme jointly initiated by the Central Government and KSEB for this purpose.

Carbon footprint
Greenhouse gases are directly and indirectly emitted by individuals, families, organizations, events, services, products, etc. The quantity of such emitted greenhouse gases, converted and expressed as equivalent to the measure of carbon dioxide, is carbon footprint.

Each individual should strive to reduce his/her personal carbon footprint. This can be achieved by paying attention to daily activities. How one travels, what food one eats, what clothes one uses, and how much waste is generated are all important.