Kerala Syllabus 10th Standard Physics Solutions Chapter 1 Effects of Electric Current

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Kerala State Syllabus 10th Standard Physics Solutions Chapter 1 Effects of Electric Current

Effects of Electric Current Text Book Questions and Answers

Textbook Page No. 7

SSLC Physics Chapter 1 Question 1.
Some electrical devices are shown in the house of the child. What are they?
Answer:

  • Electric bulb
  • Electric fan
  • Mixi
  • Induction cooker
  • Microwave oven
  • Storage batten,
  • Inverter

Textbook Page No. 8

HssLive Physics Chapter 1 Question 2.
Write down the energy changes in them with respect to their use.
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Answer:
Kerala Syllabus 10th Standard Physics Solutions Chapter 1 Effects of Electric Current image 2

Heating Effect of Electric Current Class 10 Question 3.
which are the devices that give heating effect of electric current?
Answer:

  • Electric iron
  • Electric stove
  • Microwave oven
  • Heating coil
  • Induction cooker

Textbook Page No. 9

Heat Chapter Class 10 Questions and Answers Question 4.
Kerala Syllabus 10th Standard Physics Solutions Chapter 1 Effects of Electric Current image 3

→ How does the ni-chrome wire become red hot while passing electricity through the circuit?
Answer:
Due to the resistance of Ni-chrome wire.

→ In this case which form of energy was converted into heat energy?
Answer:
Electrical energy

→ How does this energy change occur?
Answer:
This based on the concept that energy can neither be created nor destroyed. It can only be converted from one form to another (Law of Conservation of Energy) As the resistance of Ni-chrome is higher it produced more heat.

Lighting Effect of Electric Current Question 5.
Kerala Syllabus 10th Standard Physics Solutions Chapter 1 Effects of Electric Current image 4
If the ammeter shows a current I ampere on applying a potential difference V across the resistor of resistance R Ω
Current \(I=\frac{Q}{t}\)
Then, the charge that flows through the conductor in t second,
Q = ……………coulomb.
Answer:
Q = I × t coulomb

Heat Chapter Class 10 Question 6.
factors influencing the heat developed when a current passes through a conductor.
Answer:

  • Electric current
  • Resistance of the conductor
  • Time of current flow

Textbook Page No. 11

Question 7.
Complete the following table on the basis of Joule’s Law. (Page no.11).
Kerala Syllabus 10th Standard Physics Solutions Chapter 1 Effects of Electric Current image 5
Answer:
Kerala Syllabus 10th Standard Physics Solutions Chapter 1 Effects of Electric Current image 6

Question 8.
Analyse the table and find out the factor that influences heat the most.
Answer:
Intensity of current

Question 9.
Experiment (Textbook Page no. 11)
Kerala Syllabus 10th Standard Physics Solutions Chapter 1 Effects of Electric Current image 7

→ Of the water in beakers A and B which one got heated more? Why?
Answer:
since nichrome has high resistance, more heat is produced. Hence temperature of water is increased.

→ What change is observed in the temperature of water in both the beakers when the current is increased using the rheostat?
Answer:
As the current increases, the heat produced is increases.

→ What was the change that happened to the temperature of water in both the beakers on increasing the time?
Answer:
Temperature is increased more

Textbook Page No. 12

Question 10.
Heat generated = 2400 J
If 4.2 J is one calorie then H = ……… calorie
Answer:
H = 2400 J = \(\frac{2400}{4.2}\) = 571.42 calorie

Question 11.
Let’s find out the heat developed in 3 minutes by a device of resistance 920 Ω working under 230 V.
Answer:
V = 230V,
R = 920W,
t = 3 × 60 sec
I = \(\frac { V }{ R }\) = \(\frac { 230 }{ 920 }\) = 0.25
H = I2Rt = 0.252 × 920 × 3 × 60 = 10350 J

→ Is there any difference in the amount of heat energy thus obtained?
Answer:
No

Textbook Page No. 13

→ How this problem can be solved using the relation, H = VIt.
Answer:
V = 230 V,
R=920 W,
t = 3 × 60 sec
I = \(\frac { V }{ R }\) = \(\frac { 230 }{ 920 }\) = 0.25
H = VIt = 230 × 0.25 × 3 × 60 = 10350 J

Question 12.
Let’s calculate the heat developed when 3 A current flows through an electric iron box designed to work under 230 V.
Answer:
V=230 V,
I = 3 A
t = 30 m = 30 × 60 = 1800 s
H = VIt =230 × 3 × 1800 = 1242000 J = 1242 kJ

Question 13.
Table 1.3
Kerala Syllabus 10th Standard Physics Solutions Chapter 1 Effects of Electric Current image 8

→ Why does the heater having low resistance get heated more?
Answer:
Intensity of electric current is high.

→ In which way does the change in resistance influence the heat developed?
Answer:
As resistance increases heat decreases.

→ Find out the current in the heaters A and B and compare the heat developed.
Answer:
I = \(\frac{V}{R}\)
Heater A: I = \(\frac { 230 }{ 1150 }\) = 0.2
Heater B: I = \(\frac { 230 }{ 460 }\) = 0.5
In heater A and B, I increase H increases.

→ How do the resistors bring about a change in the current in the circuit?
Answer:
As resistance increases, I decrease.

Textbook Page No. 14

Question 14.
Consider the fig 1.4(a) and 1.4(b).
Kerala Syllabus 10th Standard Physics Solutions Chapter 1 Effects of Electric Current image 9

Kerala Syllabus 10th Standard Physics Solutions Chapter 1 Effects of Electric Current image 10

→ In which circuit does the bulb glow with high intensity?
Answer:
Circuit A [figure 1.4(a)]

→ Remove one bulb from each circuit. What do you observe?
In figl.4 (a)
In figl.4 (b)
Answer:
In fig 1.4 (a) : bulb glows
In figl .4 (a) : bulb does not glow

→ Why do the bulbs in Fig 1.4 (a) glow with maximum brightness?
Ans: Low resistance

Textbook Page No. 15

Question 15.
Table 1.4.
Kerala Syllabus 10th Standard Physics Solutions Chapter 1 Effects of Electric Current image 11
Answer:
Kerala Syllabus 10th Standard Physics Solutions Chapter 1 Effects of Electric Current image 12

Question 16.
Analyse the table and tick (✓) the best suited
Kerala Syllabus 10th Standard Physics Solutions Chapter 1 Effects of Electric Current image 13
Answer:
Kerala Syllabus 10th Standard Physics Solutions Chapter 1 Effects of Electric Current image 14

Textbook Page No. 17

Question 17.
Complete Table 1.6 by analysing Tables 1.4 and 1.5.
Kerala Syllabus 10th Standard Physics Solutions Chapter 1 Effects of Electric Current image 15
Answer:

Resistors in series Resistors in parallel
1. Effective resistance increases 1. Effective resistance decreases
2. The current through each resistor is same as the main current. 2. The current through each resistor is different. It gets divided as per the value of resistors
3. The potential difference across each resistor is different. It gets divided as per the value of resistors. 3. The potential difference across each resistor will be the same
4. Each resistor cannot be controlled by separate switch, because the circuit is broken 4. Each resistor can be controlled by using separate switches.

Textbook Page No. 18

Question 18.
10 resistors of 2 Ω each are connected in parallel. Calculate the effective resistance.
Answer:
R = \(\frac { 2Ω }{ 10 }\) = 0.2 Ω

Textbook Page No. 19

Question 19.
A few heating appliances are shown in the figure(1.8). Examine any one of them and answer the following questions. Record the answers in the science diary.
Kerala Syllabus 10th Standard Physics Solutions Chapter 1 Effects of Electric Current image 16

→ Name the part in which electrical energy changes into heat energy.
Answer:
Heatingcoil

→ Which material is used to make this part?
Answer:
Nichrome:
Heating coils are made of nichrome. It is an alloy of nickel, chromium and iron.

→ What are the peculiarities of such substances?
Answer:

  • High resistivity
  • High melting points
  • Ability to remain in red hot condition for a long time without getting oxidized.
  • Thermal expansion is negligible.

Textbook Page No. 20

Question 20.
Which are the circumstances that cause high electric current, leading to the melting of fuse wire?
Answer:
Overloading or short circuit

Question 21.
How is the fuse wire connected to a circuit? In series/ parallel?
Answer:
in series

Question 22.
You know that according to Joule’s Law, more heat will be produced when electric current is increased. What happens to the fuse wire due to this?
Answer:
Fuse wire melts

Question 23.
When heat is generated, why does the fuse wire melt?
Answer:
The melting point of fuse wire is lower than that of other metals

Question 24.
When the fuse wire melts, the circuit is broken. What happens to the current in the circuit?
Answer:
The flow of current in the circuit stops simultaneously with the melting of fuse wire.

Question 25.
Why is the? fuse used in a circuit called safety fuse? Explain.
Answer:
The current in the circuit may increase due to reasons such as short circuit, overload, excess flow of current or any problems in the insulation. As the higher temperature produced in the circuit due to these reasons makes the fuse wire to melt and the flow of current stops. Thus the circuit and the appliances are protected.

Question 26.
When a fuse wire is included in a household wiring. What are the precautions to be taken?
Answer:

  • The ends of the fuse wire must be connected firmly at appropriate points.
  • The fuse wire should not project out of the carrier base.
  • Use the fuse wire of proper amperage.

Question 27.
You might have noticed the marking of 500 W on an electrical appliance. What does it indicate?
Answer:
It indicates the power of the instrument. The amount of energy consumed by an electrical appliance in unit time is its power.

Question 28.
What is the unit of power?
Answer:
Watt (W)

Textbook Page No. 22

Question 29.
If R= V/I What will be P ?
P = IR = I × ……… = ……….
Answer:
P = I2R = I2 × \(\frac { V }{ I }\) = VI

Question 30.
A current of 0.4 A flows through an electric bulb working at 230 V. What is the power of the bulb?
Answer:
V = 230 V
I = 4 A
P = VI
P = 230 × 4 = 92 W

Textbook Page No. 23

Kerala Syllabus 10th Standard Physics Solutions Chapter 1 Effects of Electric Current image 17

Question 31.
What happens if the interior of the bulb is not evacuated?
Answer:
The interior of the bulb is evacuated to pr-event the oxidation of filament. When the tungsten comes in contact with the air in the heated condition
then it undergoes oxidation.

Question 32.
Why is the bulb filled with an inert gas or nitrogen?
Answer:
To prevent the vaporization of filament.

Question 33.
What are the advantages of using tungsten as a filament?
Answer:

  • High resistivity
  • High melting point
  • High ductility
  • Ability to emit white light in the white-hot conditions

Textbook Page No. 24

Question 34.
Nichrome is not used as filament in incandescent lamps. Why?
Answer:
It can remain only in red hot condition but it can’t give light.

Question 35.
A filament lamp is lit for a short period of time. Touch it. What do you feel?
Answer:
Heat is experienced.

Question 36.
What are the other types of lamps working on electricity? List them.
Answer:

  • Discharge lamp
  • Fluorescent lamp
  • Arc lamp
  • CFL
  • LED

Textbook Page No. 25

Question 37.
What are the advantages of using discharge lamps instead of incandescent lamps?
Answer:
Loss of electricity in the form of heat is less, more life span, no shadow formation, more light is obtained, less consumption of electricity.

Question 38.
What are the factors to be considered when you select a bulb?
Answer:
Efficiency, energy consumption, low energy loss, less environmental pollution.

Question 39.
Which are the lamps that are mostly used? Why?
Answer:
LED: Low energy consumption, low energy 1 loss, less environmental pollution.

Question 40.
LED.
Answer:

  • LED’s are Light Emitting Diodes.
  • As there is no filament, there is no loss j of energy in the form of heat.
  • Since there is no mercury in it, it is not harmful to environment
  • Very small
  • It requires only small amount of power.
  • No filaments

Effects of Electric Current Let Us Assess

Question 1.
Fuse wire is to be used by understanding the amperage correctly. Write down the amperage of the fuse wires that are currently available in the market.
Answer:
Fuse wire of suitable amperage should be selected. Amperage is the ratio of the power of an equipment to the voltage applied. Amperage increases with the thickness of the conductor. Thick wire takes more time to melt. Due to high current flow, the circuit may be damaged. If thin wire is used resistance is increased and hence current is reduced.
Amperage of available fuse : 0.1 A, 0.2 A, 0.5 A, 1.5 A, 3 A, 5 A.

Question 2.
0.5 A current flows through an electric heating device connected to 230 V supply.
a. the quantity of charge that flows through the circuit in 5 minutes is
i. 5 C
ii. 15 C
iii. 150 C
iv. 1500 C
b. How much is the resistance of the circuit?
c. Calculate the quantity of heat generated when current flows in the circuit for 5 minutes.
d. How much is the power of the heating device connected to the circuit if we ignore the resistance of the circuit wire?
Answer:
Kerala Syllabus 10th Standard Physics Solutions Chapter 1 Effects of Electric Current image 18
Question 3.
According to Joule’s Law the heat generated due to the flow of current is H = I2 Rt. Will the heat developed increase on increasing the resistance without changing the voltage? Explain.
Answer:
Heat decreases.
Reason H α \(\frac { 1 }{ R }\)

Question 4.
The table shows details of an electric heating device designed to work in 230 V. Complete the table by calculating the change in the heat and power on changing the voltage and resistance of the device. Analyze the table and answer the following questions.
Kerala Syllabus 10th Standard Physics Solutions Chapter 1 Effects of Electric Current image 19
a How does the voltage under which a device works affect its functioning?
b.What change happens to power on increasing the resistance without changing the voltage?
c. What change is to be brought about in the construction of household heating devices in order to increase their power?
Answer:
a. As voltage increases, power increases
b. Power decreases
c. Use thick wire, use conductor of suitable material, use conductor of small length.

Question 5.
a. Complete the table based on the amperage of the fuse wire.
Kerala Syllabus 10th Standard Physics Solutions Chapter 1 Effects of Electric Current image 20
b. The amperage of the fuse wire used in a circuit that works on 230 V is 2.2 A. If so the power of the device is
i. less than 300 W
ii. 300 W to 500 W
iii. between 500 Wand 510 W
iv. more than 510 W
Answer:
iii. between 500 W and 510 W

Question 6.
A 230 V, 115 W filament lamp works in a circuit for 10 minutes,
a. What is the current flowing through the bulb?
b. How much is the quantity of charge that flows through the bulb in 10 minutes?
Answer:
a. I = \(\frac { P }{ V }\) = \(\frac { 115 }{ 230 }\) = 0.5 A

b.  Q = I × t = 0.5 × 10 × 60 = 300 c
Question 7.
An electric heater conducts 4 A current when 60 V is applied across its terminals. What will be the current if the potential difference is 120 V?
Answer:
R = \(\frac{V}{I}=\frac{60}{4}=15 Ω
If the voltage is changed to 120 V
I = [latex]\frac{V}{R}=\frac{120}{15}\) = 8 A

Question 8.
Three resistors of 2 Ω,3 Ω and 6 Ω are given in the class.
a. What is the highest resistance that you can get using all of them?
b. What is the least resistance that you can get using all of them?
c. Can you make a resistance 4.5 Ω using these three? Draw the circuit.
Answer:
a. Highest resistance
R = R + R + R = 2 + 3 + 6 = 11 Ω

b. Least resistance
Kerala Syllabus 10th Standard Physics Solutions Chapter 1 Effects of Electric Current image 21

c. Yes
Kerala Syllabus 10th Standard Physics Solutions Chapter 1 Effects of Electric Current image 22

Question 9.
A girl has many resistors of 2 Q each. She needs a circuit of 9 Q resistance. For this draw a circuit with the minimum number of resistors.
Answer:
Kerala Syllabus 10th Standard Physics Solutions Chapter 1 Effects of Electric Current image 23

Question 10.
Kerala Syllabus 10th Standard Physics Solutions Chapter 1 Effects of Electric Current image 24
If a bulb is lit after rejoining the parts of a broken filament, what change will occur in the intensity of the light from the lamp? What will be the change in the power of the bulb?
Answer:

  • Intensity of bulb
  • Power increases.

Question 11.
Which of the following does not indicate the power of a circuit?
a. I28
b. VI
c. 1R2
d. V2/R
Answer:
c. 1R2

Question 12.
How much will be the power of a 220 V, 100 W electric bulb working at 110 V?
a. 100 W
b. 75 W
c. 50 W
d. 25 W
Answer:
d. 25 W

Question 13.
Which of the following should be connected in parallel to a device in a circuit?
a. voltmeter
b. ammeter
c. galvanometer
Answer:
a. voltmeter

Question 14.
When a 12 V battery is connected to resistor, 2.5 mA current flows through the circuit. If so what is the resistance of the resistor?
Answer:
Kerala Syllabus 10th Standard Physics Solutions Chapter 1 Effects of Electric Current image 25
Question 15.
If 0.2Ω, 0.3Ω, 0.4 Ω, 0.5 Ω, and 12Ω resistors are connected to a 9 V battery in parallel, what will be the current through the 12 Ω resistor?
Answer:
I = \(\frac{V}{R}=\frac{9}{12}\) = 0.75 A

Question 16.
How many resistors of 176 Ω should be connected in parallel to get 5A current from 220 V supply?
a. 2
b. 3
c. 6
d. 4
Answer:
4

Question 17.
Depict a figure showing the arrangement of three resistors in a circuit to get an effective resistance of
(i) 9 Ω
(ii) 4 Ω
Answer:
i.
Kerala Syllabus 10th Standard Physics Solutions Chapter 1 Effects of Electric Current image 26
ii.
Kerala Syllabus 10th Standard Physics Solutions Chapter 1 Effects of Electric Current image 27

Effects of Electric Current Extended Activities

Question 1.
Analyse and describe the working of a microwave oven.
Answer:
Microwave oven is a device which is used for heating effect of electric current. It produces heat without a heating coil. Eddy current is used in the working of microwave oven. The heat is generated as a result of microwave radiations. The water molecules are stimulated using ‘magnet one’ and thus attains high temperature. Metal utensils are avoided and the substances without water content are not heated by this.

Question 2.
How does an arc lamp help in rescue operations?
Answer:
The intense light of arc lamps are used in search lights and rescue operations during night time. It is used to rescue the victims of natural calamities, boat accidents.

Question 3.
With the help of teachers and the Internet find out the following
a. What is the percentage of nickel, chromium and iron in Nichrome?
b. How much is the melting point of nichrome in degree Celsius?
c. How much is the resistivity of Nichrome?
d Does the result of your observation justify the use of nichrome as a heating element?
Answer:
a. 61 % Ni, 15% Cr, 24% Fe
b. 1350°C
c. (1.0 – 1.5)x 10 – 6 Ωm
d. 1. High resistivity
2. High melting points
3. Ability to remain in red hot condition for a long time without getting oxidized. :
4. Thermal expansion is negligible.

Question 4.
Analyse the merits and demerits of the following lamps and find out which is best in the group. Justify your answers.
a. filament lamp
b. fluorescent lamp
c. arc lamp dCFL
e. LED bulb
Answer:
a. Filament lamp:
Demerits: Loss of energy as heat, low light, low light.
Merits: Work in both DC and AC.

b. Fluorescent lamp:
Demerits: Cause environmental pollution,
Merit: Produce more light

c. Arc lamp:
Demerits: Carbon rods are to be changed frequently.
Merit: Intense light is produced.

d. CFL:
Merit: Produce more light at low power.
Demerits: Cause environmental pollution,

e. LED Bulb :
Merit: No environmental pollution, low energy consumption, long life. The best lamp is LED.

Effects of Electric Current Orukkam Questions and Answers

Question 1.
Complete the table based on the effects of electric current and energy change.
Kerala Syllabus 10th Standard Physics Solutions Chapter 1 Effects of Electric Current image 28
Answer:
a. Light energy
b. Mechanical effect
c. Heat energy
d. Chemical effect

→ Given below questions are related with heating of an electric iron box. Answer them.
a. Which is the part that produces heat in an electric iron?
b. Which nature of this part is made use in the above situation?
c. What is the relation between intensity of electric current and heat energy generated?
d. What are the factors that affect the heat generated in such heating appliances?
e. What is the relation connecting these factors with the heat generated?
f. What is this law known as?
g. Name a device that works on this law used for ensuring safety in electric circuits?
Answer:
a. Heating coil
b. High resistivity, High melting point
c. As electric current increases, Heat energy increases.
d. Electric current, Resistance, time
e. H = I2 Rt
f. Joules law
g. Safety fuse

Question 2.
Incandescent lamp, discharge lamp, C.F.L, LED lamp, arc lamp are given for observation (Otherwise, make use of their pictures)
Answer the following questions after observing them.
a. Which metal is used to make filament of an incandescent lamp? What are the advantages of using this metal as a filament?
b. Name the lamps which belong to the group of discharge lamp.
c. The color of the light depends on the gas inside the discharge lamp. Which gases are to be filled for getting white light and yellow light?
d. Which lamps are harmful to environment because of the presence of mercury in it?
e. Which lamp is used in rescue work during night time and used in searchlights?
Answer:
a. Tungsten. Advantages of tungsten are high ductility, high resistivity, high melting point
b. Fluorescent lamp, CFL, LED, Arc lamp.
c. For white light – Mercury
For yellow light- Sodium Vapour
d. Fluorescent lamps
e. Arc Lamp

Effects of Electric Current SCERT Questions and Answers

Question 1.
Observe the circuit diagram given below and answer the following questions.
Kerala Syllabus 10th Standard Physics Solutions Chapter 1 Effects of Electric Current image 29
a. Which are the instruments labeled as P and Q in the diagram?
b. If you replace the copper wire AB with a Nichrome wire of same length and area of cross-section.
i. What change would you notice in the reading on the device Q? Why?
ii. What will happen to the heat produced in the conductor? Explain with reference to Joule’s Law.
Answer:
a. P- Rheostat.
Q- Ammeter.

b. i. Reading will decrease. Due to higher resistance current decreases
ii. Because of Nichrome’s high resistivity, the current in the circuit will decrease. According to Joules law H= I2 Rt, decrease in the amount of current will reduce the amount of heat.

Question 2.
Write down the names of four types of lamps working on the lighting effects of electricity.
Answer:
Discharge lamp, Fluorescent lamp, LED, Arc lamp.

Question 3.
Two types of lamps are given below.
1. Discharge lamp
2. Filament lamp.
a. If nitrogen gas is filled in each lamp, what change will happen to their working?
b. Why is it said that the use of filament lamp must be controlled?
Answer:
a. Nitrogen filled discharge lamp will give red light, but in a filament lamp nitrogen is filled to reduce the evaporation of the filament.

b. In a filament lamp, major part of the electrical energy we supply is converted into heat energy.

Question 4.
Kerala Syllabus 10th Standard Physics Solutions Chapter 1 Effects of Electric Current image 30
a. Identify the device shown in the picture.
b. How are such devices used in rescue ope-rations?
Answer:
a. Arc lamp

b. Light intensity in an Arc lamp is very high compared to other lamps, so Arc lamps are helpful in rescue operations even in adverse climatic conditions.

Question 5.
Observe the diagram and answer the questions below.
Kerala Syllabus 10th Standard Physics Solutions Chapter 1 Effects of Electric Current image 31
a. Which bulbs will glow when S1 switched on?
b. Which bulbs will low when S1 and S1 are switched on?
c. What change will happen to the circuit when S3 is switched on?
d. Calculate the amperage of the fuse to be used in the circuit,
e. Describe short circuit and overloading with the help of the given circuit
Answer:
a. B1
b. Bulb will not glow, (fuse burns out due to overloading)
c. The circuit breaks as fuse burns out due to short circuit.
d. Amperage
\(\frac{200}{100}=2 A(\text { more than } 2 \mathrm{A})\)
e. Short circuit happens when two wires in the mains in contact without the presen-ce of a resistance in between. Over loading is connecting appliances with more power in the circuit, that it can bear.

Question 6.
A 800W electrical device is designed to work in 200V. What will be its power if the device is working in 100 V?
Answer:
Resistance of the appliance
Kerala Syllabus 10th Standard Physics Solutions Chapter 1 Effects of Electric Current image 32
Question 7.
Bulbs marked 200V and 500W are shown in the picture.
Kerala Syllabus 10th Standard Physics Solutions Chapter 1 Effects of Electric Current image 33
a. Calculate the resistance of each bulb in the circuit.
b. What is the power with which the bulb in circuit 1 glows?
c. What is the power of the bulb in circuit 2? Why?
d. How is the power and resistance of an electrical device related to each other when the voltage is the same?
Answer:
Resistance of B1 = \(\frac { V2 }{ P }\) = \(\frac { 200 × 200 }{ 50 }\) = 800 Ω
Resistance of B2=800 Ω

b. Will work in 50 W

c. Since the 50 W bulbs are in series current will be equal and voltage will behalf. Since P=VI, Power also will become half.

d. P = \(\frac { V2 }{ R }\) (No change in resistance, Power will change)

Question 8.
Electric bulbs are connected in a 240V supply line are shown in the figure.
Kerala Syllabus 10th Standard Physics Solutions Chapter 1 Effects of Electric Current image 34
a. What is the total wattage of appliances used in the circuit?
b. Calculate the amperage of the fuse to be used in the circuit.
Answer:
a. P = 20 W+ 20 W + 20 W = 60W

Kerala Syllabus 10th Standard Physics Solutions Chapter 1 Effects of Electric Current image 35
Question 9.
Given below are the steps in the working of a fluorescent lamp. Arrange them in the proper order.
a. Ultraviolet rays are produced
b. Visible light is emitted.
c. Fastly moving electrons collide with the unionized molecules of mercury.
d. Due to electric current thorium oxide coated heating element becomes red hot.
Answer:
(d),
(c),
(a),
(b)

Question 10.
Correct the mistakes, if any:
a. Amperage decreases in proportion to the decrease in the area of cross-section of the conductor.
b. Connecting appliances in a circuit beyond its power capacity is short circuit.
c. It is to reduce the heat loss that electric lamps are filled with inert gases.
Answer:
b. Overloading is connecting appliances with more power in the circuit, than it can bear.
c. Inert gases are filled in filament lamps to reduce the rate of evaporation.

Question 11.
Power of an electric heater working in 230 V is 1000 W. Calculate the heat produced if the current passes for 5 minutes through the circuit.
Answer:
Heat = P × t
= 1000 × 5 × 60
= 300000 J

Question 12.
An electric bulb has marking 110 V, 100 W on it.
a. How much energy is used per second by the circuit?
b. What is the resistance of electric bulb?
Answer:
Kerala Syllabus 10th Standard Physics Solutions Chapter 1 Effects of Electric Current image 36

Question 13.
Fill up the blanks by finding the suitable relationship.
Kerala Syllabus 10th Standard Physics Solutions Chapter 1 Effects of Electric Current image 37
Answer:
A= Fuse wire
B= High melting point

Question 14.
If the resistance of two soldering irons working in 250 V are 500 Ω and 750 Ω.
a. Calculate which one of these will carry more current.
b. Find out which soldering iron has more power.
c. Calculate the heat produced in 5 minutes in the soldering iron having resistance 750 Ω.
Answer:
Kerala Syllabus 10th Standard Physics Solutions Chapter 1 Effects of Electric Current image 38
b. Soldering iron has less resistance
Kerala Syllabus 10th Standard Physics Solutions Chapter 1 Effects of Electric Current image 39

Question 15.
Two bulbs of 500 W and 100 W are connected parallel in a circuit of 250 V.
a. Which bulb will have more brightness?
b. Through which bulb is current greater?
c. Find out the resistance of filaments in each bulb.
d. Which of these two bulbs will glow with more brightness if they are connected in series? Explain the reason.
Answer:
a. 500 W Bulb

b. Bulb which has power 500 W (P = VI).

c. 500 W bulb
R = \(\frac { V2 }{ P }\) = \(\frac { 250 × 250 }{ 500 }\) = 125 Ω
100 W Bulb
R = \(\frac { V2 }{ P }\) = \(\frac { 100 × 100 }{ 250 }\) = 40 Ω

d. 100 W bulb — Current will be the same in series, High resistance bulb will shine more brightly.

Question 16.
Fill in the blanks suitably using the relationship in the first pair,
a. Filament: High melting point
Fuse : ………………
b. Nitrogen filled discharge lamp : Red
………………………… : Blue
Answer:
a. Low melting point

b. Hydrogen used in discharge lamp

Question 17.
Choose and write the facts related to LED lamps from the given list.
a. Require only a small quantity of power.
b. UV rays are produced due to electric discharge.
c. Not harmful to environment since there is no mercury.
d. Intense light is produced when high voltage is applied.
Answer:
a, c

Effects of Electric Current Exam Oriented Questions and Answers

Very Short Answer Type Questions (Score 1)

Question 1.
Using the relation from the first pair, complete the other.
Filament: Tungsten Heating coil:
Answer:
Nichrome

Question 2.
Which of the given material has the highest resistivity?
a. Nichrome
b.Copper
c. Aluminium
d. Silver
Answer:
Nichrome

Question 3.
What is the color of the light emitted by discharge lamp filled with nitrogen when it works ?
Answer:
Red

Question 4.
Using the relation from the first pair, complete the other.
Fuse wire : low melting point
Tungsten: …………..
Answer:
High melting point

Question 5.
Which among the following is the special characteristics of a material to be used as a heating coil in a electric heating device?
a. Low melting point
b. High resistivity
c. Large area of cross section
d. Low resistance
Answer:
High resistivity

Question 6.
Which material is used as filament of bulb?
Answer:
Tungsten

Question 7.
Find out the odd one from the following also write the reason.
[long glass tube, Mercury vapor, Fluorescent coating, Thin tungsten filament]
Answer:
Thin tungsten filament. Others are parts of a fluorescent lamp.

Question 8.
If we apply 230 V for a device having 230 V and 400 W, what will happen the power of the device ?
(increases, decreases, doesn’t change)
Answer:
decreases

Question 9.
In incandescent lamp nichrome is not used as filament. Choose the reason from the following
a. Glows brightly
b. It emit white light when it is in hot condition. ,
c. It can remain only in red hot condition but it can’t give light.
Answer:
c

Question 10.
Find out the odd one which is not suitable for the advantages of LED.
a. It requires only a small quantity of power.
b. It is not harmful to environment.
c. Ultraviolet fays are formed due to discharge.
Answer:
c

Short Answer Type Questions (Score 2)

Question 11.
A 400 W electrical device is designed to work in 100 V. What will be its power if the device is working in 100 V?
Answer:
Resistance of the appliance
R = \(\frac { V2 }{ P }\) = \(\frac { 100 × 100 }{ 400 }\) = 25 Ω
Power when connected in 100 V
P = \(\frac { V2 }{ R }\) = \(\frac { 100 × 100 }{ 25 }\) = 400 Ω
Answer:
b,
d,

Question 12.
An electric bulb marked 500 W, 250 V is connected to a 250 V supply.
a. Find the electric current through the circuit.
b. Calculate the resistance of the bulb.
Answer:
Kerala Syllabus 10th Standard Physics Solutions Chapter 1 Effects of Electric Current image 40
Question 13.
What are the advantages of fluorescent lamps over incandescent lamps.
Answer:

  1. Saves electrical energy to a greater extent.
  2. The inconvenience caused by shadow is minimized.
  3. The life of fluorescent lamp is about 5 times that of incandescent lamps.

Question 14.
Complete the table suitably.

Gas-filled in the discharge lamp Colour
Hydrogen (a)
(b) Orange-red
Nitrogen (c)
(d) Green

Answer:
a. Blue
b. Neon
c. Red
d. Chlorine

Short Answer Type Questions (Score 3)

Question 15.
In figure, Beakers A and B contain 100 ml water. PQ is a nichrome wire and RS is a copper wire of same length and diameter.
Kerala Syllabus 10th Standard Physics Solutions Chapter 1 Effects of Electric Current image 41
a. Water in which beaker will be at higher temperatures? What is the reason?
b. If the current is doubled using the rheostat, what happens to the quantity of heat produced in the wire PQ?
Answer:
a. Beaker A, since the combination is series, the current remain same. As nichrome has high resistance, more heat is produced in it. (H = I2 Rt)

b. Heat produced becomes 4 times.
Heat H = I2 Rt, Heat increases as current is squared.

Question 16.
Classify the following as those suitable for fluorescent lamps and for arc lamps.
a. The main part is carbon rods.
b. The heating coil is coated with thorium oxide.
c. used in searchlights.
d. more harmful for the environment.
e. more intense light.
f. ultraviolet rays are produced.
Answer:
Arc lamps: a, c, e
Fluorescent lamps : b, d, f

Question 17.
Find out the relation and complete the missing parts.
i. tungsten: ………….. (A) ……………. : high melting point
ii. alloy of suitable : fuse wire: ……….(B)……….. metals
iii (C)…….. heating coils: high melting point
Answer:
A. Filament
B. Low melting point
C. Nichrome

Question 18.
The filament of a filament lamp is broken. It is rejoined and is lit again.
a. What happens to the intensity of light from it? Describe the reason behind.
b. Incandescent lamps are filled with nitrogen at low pressure. What is the advantage of doing so?
c. You are given two filament lamps of resistance 75012 and 100012. Of these, which has more power?
Answer:
a. Increases. Because resistance decreases and current increases when length decreasing.

b. Nitrogen doesn’t expand over less change in temperature. At normal temperature it behave as a inert gas. It is plenty in atmosphere.

c. 750 Ω

Questions 19.
Choose the appropriate items from the box.
Nichrome, Tungsten, Fuse Wire, Nitrogen
a. Which is used as heating coil ?
b. Which is an alloy of tin and lead?
c. Which is used as filament ?
Answer:
a. Nichrome
b. Fuse wire
c. Tungsten

Short Answer Type Questions (Score 4)

Question 20.
a. What is meant by amperage ?
b. An appliance of power 690 W is used in a branch circuit. If the voltage is 230 V,what is its amperage?
Answer:
a. Amperage is the ratio of the power of an equipment to the voltage applied.
b. A = \(\frac { P }{ V }\) = \(\frac { 690 }{ 230 }\) = 3 A

Question 21.
Analyse the table and fill it suitably.
Kerala Syllabus 10th Standard Physics Solutions Chapter 1 Effects of Electric Current image 42
Answer:
(1) 9 times,
(2) Series,
(3) 1/16 times

Question 22.
An electric heater working at 230 V produces 1000 J energy in one second.
a. What is the power of the lamp ?
b. Calculate the resistance of the heating coil used in it.
c. Calculate the heat generated by it when it works for 5 minute.
Answer:
Kerala Syllabus 10th Standard Physics Solutions Chapter 1 Effects of Electric Current image 43
Question 23.
Write down the reasons for the following statements.
a. Don’t throw unwanted fluorescent tubes.
b. Nichrome is not used as filament in incandescent lamps.
c. Fuse wire is melt when electric current increased.
d. Why is the incandescent filled with an inert gas or nitrogen?
Answer:
a. It contains mercury which causes several environmental problems.
b. It can remain only in red hot condition but it can’t give light.
c. When current flows as the higher temperature produced in the circuit. Fuse wire has low melting point compared to other, so it melts.
d. To prevent the vaporization of filament.

Memory Map:

Kerala Syllabus 10th Standard Physics Solutions Chapter 1 Effects of Electric Current image 44

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