STD 10 Physics: Chapter 01 Effects of Electric Current - Questions and Answers
Questions and Answers for Std X Physics: Chapter 1 വൈദ്യുത പ്രവാഹത്തിന്റെ ഫലങ്ങൾ | Physics (English Medium) Chapter 01 Effects of Electric Current
Effects of Electric Current: Questions And Answers
Electric energy can easily be converted to other forms of energy using suitable devices. That is why we prefer electric energy to any other sources of energy. It is given a list of familiar devices in which electric energy is converted to required form of energy and effects of current made in use.
That is, I = Q/t
If current in a circuit is I ampere, the quantity of charge flows through the conductor in t second is, Q = It In order to have an electric current through a conductor, a potential difference is to be maintained across its ends. The unit of potential difference is volt.
The potential difference between two points will be one volt if one joule of work is done to move one-coulomb charge from one point to the other.
The process by which heat is developed in a conductor on passing a current through it is known as Joule heating or Ohmic heating.
Heat generated by Joule heating will be, H = VIt
According to Ohm's law we have, V = IR.
Then heat is also equal to, H = (IR)It = I²Rt
Joule's Law:
The heat generated (H) in a current-carrying conductor is directly proportional to the product of the square of the current (I) in the conductor, the resistance of the conductor (R ) and the time (t) of the flow of current.
That is, H = I²Rt
According to Joule’s Law, Heat generated in a current-carrying conductor can also be calculated by H = VIt and H = V²t/R
It is noted that the time should be taken in seconds.
Factors influencing Joule Heating Effect.
* Current passing through the conductor. (I) : Heat increases with Voltage.
* Resistance of the conductor. R:
i. If the voltage is the same, heat increases with a decrease in Resistance.
ii. If the current is the same, heat increases with an increase in Resistance.
* Time through which current passes. (t)
PRACTICE QUESTIONS & ANSWERS
a. Heat generated in a current-carrying conductor is known as ......
b. What will be the change in the heat if the resistance in the circuit is halved (R/2)?
c. What about it if the voltage is halved?
Answer:
a. Joule heating effect.
b. Since H = V²t/R, when resistance is halved heat will be doubled.
c. Since H = V²t/R, heat becomes 1/4th when V is halved.
2. Basheer is demonstrating an experiment to find out the relation between current, resistance, time and heat.
a. What might be the device he used to regulate the current?
b. Draw the symbol of the device.
c. What will be the change in heat if the current through a conductor is tripled?
Answer:
a. Rheostat.
c. Since the heat is proportional to the squire of current,
heat will be 3x3 =9 times greater.
3. 3A current is passed through an Iron box that works on 230V. Calculate the heat produced in it in ½ hour.
Answer: V = 230V, I = 3A t = ½ hr = ½ x 60x60 = 1800s.
H = VIt = 230x3x1800 = 1242000 joule.
4. Light, heat, sound and electricity are various forms of energy. Give one major feature of electricity over other forms of energy.
Answer: Electric energy can easily be converted to other forms using a suitable device.
5. A copper wire and nichrome wire of the same length and cross-section area are connected in two circuits as shown.
b. Find out the wire in which more heat is generated.
Answer:
a. As the resistance of a copper wire is less than that of a nichrome wire, more current will flow through the circuit.1
b. For the same voltage heat is inversely proportional to resistance. (H=V²t/R). So more heat will be produced in copper as its resistance is small.
6. 0.4A current passes through a bulb when it works on 230V.
a. Calculate the charge flowing through the circuit in 2 minutes.
Answer:
a. Charge Q = Ixt = 0.4x2x60 = 48 coulomb
7. The coil of a heater is cut into equal halves and one is used in the heater.
a. What happened to the resistance of the coil?
b. What will be the change in heat produced?
Answer:
a. Resistance is halved. (Because the resistance of a conductor is directly proportional to its length)
b. We have Heat H=V²t/R. Since heat is inversely proportional to resistance, heat will be doubled when resistance is halved.
8. Which of the following device is to be connected in parallel in a circuit?
Voltmeter, Ammeter, Galvanometer.
Answer: Voltmeter.
9. A wire AB of 20Ω resistance is included in a circuit as shown in the figure.
b.If the wire is folded into half and included in the same circuit, what will be the
current?
c. Calculate the heat produced in the circuit in one second then.
Answer:
a. R= 20 Ω V = 10V I=V/R = 10/20= 0.5A
b. When it is folded, length is halved and thickness is doubled. So resistance becomes ¼ of the previous value.
Therefore new resistance R¹ = 20/4 = 5Ω. Then current I¹= V/R¹= 10/5 = 2A.
c. Heat produced in one second,H = VIt = 10x2x1 = 20J.
10. 230V is applied to a coil of resistance 92 Ω.
a. Find the current in the circuit.
b. Calculate the heat produced in it when it works for 10 minutes.
c.If the coil is folded into half and applied to the same voltage, how much heat will be produced at the same time?
Answer:
a. I = V/R = 230/92 = 2.5A.
b. H =VIt = 230x2.5x14x60 =483000 joule.
c. When it is folded, resistance becomes 1/4th.
So the new resistance R¹ = 92/4 = 23Ω .
Then heat = V²xt/R¹ = 230x230x14x60/23 = 1932000 J
11. 0.2 A current flows through a resistor of resistance 100 Ω for 2 minutes.
a. Calculate the heat generated.
b. What will be the heat if resistance is changed to 200 Ω keeping I and t remaining the same?
c. What will be the heat if the current is doubled keeping R and t remain the same?
Answer:
a. H = I²Rt = 0.2x0.2x100x2x60 = 480 J
b. H = 0.2x0.2x200x2x60 = 960 J
c. H = 0.4x0.4x100x2x60 = 1920 J. When current is doubled, the heat is quadrupled.
12. Three amperes current flows through an iron box working under 230 V for half an hour. Calculate the heat generated in the Iron box.
Answer: H = VIt = 230x3x30x60 = 1242000 J
13. A heating appliance is connected to a 230V supply line. If the current is 2A, calculate the heat produced in 5 minutes.
Answer:
d. V = 230V, I = 2A, t = 5 minute = 5x60 = 300 seconds.
Heat,H = VIt = 230x2x300 = 138000J.
Arrangement of Resistors
Resistors can be connected in series and parallel modes.
Series Connection.
Series connection of two resistors (R₁&R₂) is shown. When resistors are connected in this mode, effective resistance increases.
The effective resistance of this circuit is, R = R₁ + R₂
That is, the effective resistance of the series combination is equal to the sum of the resistance of all resistors.
Note: When 'n' resistors of equal resistance r Ω are connected in series,
the effective resistance will be ‘nr’.
Parallel Connection.
The parallel connections of resistors is shown in the figure.
Then, 1/R = 1/R₁ + 1/R2₂ Or R = R₁.R₂/(R₁+R₂)
Note: When 'n' resistors of equal resistance r Ω are connected in parallel, the effective resistance, R = r/n.
1. It is given 5Ω, 20Ω resistors and a 10V battery.
b. What is the effective resistance of this circuit?
c. What is the current in the circuit?
d. What will be the effective resistance if they are connected in parallel.
e. What will be the current then?
Answer:
b. Effective resistance, R = R₁ + R₂ = 5+20 = 25 Ω
c. Current, I = V/R = 10/25 = 0.4 A
d. Effective resistance, RP = R₁.R₂/(R₁+R₂) = 5x20/(5+20) = 100/25 = 4 Ω
e. Current, I = V/R = 10/4 = 2.5A
2. See the circuit,
b. What is the effective resistance in the circuit?
c. High voltage is dropped across ………. (100Ω/200Ω)
d. More heat will be generated in …….. (100Ω/200Ω)
e. Identify the resistor through which a large current passes.
f. If the potential difference between 100 Ω is 10V, how much work is done by the battery to move one-coulomb charge from A to B?
Answer:
a. Series.
b. 300Ω (R = R₁+R₂)
c. 200Ω (When resistors are connected in series more voltage is dropped across the high resistor)
d. 200Ω (When resistors are connected in series more heat is generated in a high resistor)
e. Same current passes through both resistors. (When resistors are connected in series same current passes through all the resistors)
f. 10J (If potential difference between two point is V volt, V joule of work is to be done to move one-coulomb charge from one point to other).
3. See the circuit.
b. What will be the total resistance of the circuit when the switch kept on?
c. What will be the ammeter reading then?
Answer:
a. Ammeter reading (Current I) = V/R = 10/5 = 2A
b. When the switch is ON, the two resistors are in parallel.
Effective resistance R = R₁.R₂/(R₁+R₂) = 20x5/(20+5) = 100/25 = 4 Ω
c. Then the Current I = V/R = 10/4 = 2.5 A
4. Draw the diagrams of connecting 3Ω resistors and 6Ω resistors, so as to get maximum resistance and minimum resistance. Also, calculate the effective resistance in each case.
Answer: Maximum resistance is obtained when they are connected in series.
And the effective value R = R₁ + R₂ = 3+6 = 9 Ω
Minimum value is obtained by connecting them in parallel. Effective value in parallel combination, R = R₁.R₂/(R₁+R₂) = 3x6/(3+6) =18/9 = 2 Ω
5. In the circuit, PQ and RS are a nichrome wire and copper wire of the same length and thickness which are immersed in equal amounts of water at the same temperature.
b. Of the wires PQ and RS which one gets more current?
c. Identify the beaker where the water gets heated more when
the switch is kept ON for a few minutes.
d. Explain the reason for the difference in the rising in the temperature of the water.
Answer:
a. Series.
b. Same amount of current passes through both wires.
c. Water in beaker A ( water in which the nichrome wire is immersed)heats up more.
d. When current passes through resistors in series connection, more heat will be generated in the highest resistor.
6. Insert the following statements in the table given.
* When the number of resistors increases current also increases.
* When number of resistors increases effective resistance decreases.
* Same amount of current passes through all the resistors.
* Potential difference is same for all the resistors.
* High resistor gets heated more.
* Applied voltage will be split among the resistors.
* Effective resistance is minimum.
7. If 2Ω, 3Ω, 4Ω, 5 Ω resistors are connected in parallel, the effective resistance will be...
14 Ω / greater than 2Ω / less than 2Ω / none of these
Answer: Less than 2Ω
Explanation: When a number of resistors are connected in parallel, the effective resistance will be less than the least one among them.
8. See the circuit.
Answer: The brightness will increase. Because when the switch is ON, the second resistor is also included in the circuit. Since they are connected in parallel the effective resistance decreases and hence increases the current
in the circuit.
9. 20 resistors of 2Ω each are connected in parallel. Calculate the effective resistance.
Answer: Effective resistance R = r/n = 2/20 = 0.1 Ω
10. Three resistors of 2Ω, 3Ω, 6Ω 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 the resistance of 4.5Ω using these three. Draw the circuit.
Answer:
a. Highest resistance is obtained when they are connected in series.
Hence R = R₁ + R₂ + R₃ = 2+3+6 = 11 Ω
b. Least resistance is obtained when they are connected in parallel.
Then 1/R = 1/R₁ +1/R₂ + 1/R₃ = ½ + ⅓ + ⅙ = 18/18 = 1 Ω Or R = 1 Ω
c. If they are arranged as shown in the figure, 4.5 Ω can be made available.
11. A boy has many resistors of 2Ω each. He needs a circuit of 9 Ω resistance. For this draw a circuit with the minimum number of resistors.
Answer:
Answer: As the resistors are connected in parallel, all the resistors will get 9V.
So current through 12 Ω resistor, I= V/R = 9/12 = 3/4 A = 0.75A
13. How many resistors of 176 Ω should be connected in parallel to get 5A current from a 220V supply?
Answer: The required net resistance R = V/I = 220/5 = 44 Ω
When four 176 Ω resistors are connected in parallel, we get 44 Ω effective resistance.
14. Depict a figure showing the arrangement of three resistors in a circuit to get an effective resistance of (i). 9 Ω and (ii). 4 Ω
Answer:
Heating Coil: It is in a heating coil, heat produces in heating appliances like iron boxes, heaters etc... Heating coils are made of nichrome, an alloy of nickel, chromium, and iron.
Features of Nichrome: Ability to remain red hot for a long time, high resistivity, high melting point.
Short circuit and Overloading.
Short circuits and overloading are the two situations that lead to excess current in an electric circuit.
A short circuit is a direct contact between positive& negative terminals of a battery or two wires from the mains without the presence of resistance in between them.
In the circuit shown here, if switch K is turned on, it will be short-circuited. Because now the negative and positive terminals of the battery are in direct contact.
Safety fuse
A safety fuse is a device which protects electric circuits and appliances from danger due to excess current flows through the circuit. It works on the heating effect of electric current. The main part of a safety fuse is a fuse wire, which is made of an alloy of lead and Tin having a low melting point. It is connected in series to the circuit. When the current exceeds the permissible limit, the heat generated in the fuse wire becomes excessive and hence it melts and breaks the circuit.
Amperage: Amperage is the sufficient current required for the proper functioning of an electric appliance. It is the ratio of the power of the appliance to the working voltage. (In the case of a conducting wire or fuse wire, amperage is the maximum current that can withstand)
Amperage = Wattage/Voltage.
The amperage of a conducting wire depends on its thickness.
That is amperage increases with the thickness of the wire.
As the intensity of current differs from one appliance/circuit to another, fuse wire of appropriate amperage is to be selected.
Fuses are safety devices. But their improper handling may lead to adverse effects. The following precautions are to be taken while using fuses.
* The ends of the fuse wire must be connected firmly at appropriate points for avoiding loose contact.
* Fuse wire should not project out of the carrier base.
* Use fuse wire having required amperage/ thickness according to the load of the circuit.
* Use fuse wires made of suitable material having a low melting point.
PRACTICE QUESTIONS & ANSWERS
1. Amperage indicates the current bearing capacity of a conductor / electric device. What is the relation between amperage and thickness of a wire?
Answer: a. Amperage increases with thickness.
2. Analyse the circuit and suggest the amperage of the fuse wire required for this circuit.
= total power/voltage= (250+1500+1000+750)/250
= 3500/250 =14A.
So the required amperage of the fuse wire is 14A.
3. Iron box is a heating appliance.
a. Which substance is used to make a heating coil?
b. What are the constituents of this substance?
c. Give two features of this substance.
Answer:
a. Nichrome.
b. Nickel, Chromium, Iron.
c. High melting point, High resistivity, ability to remain hot red for a long time, low vaporisation.
4. Safety fuse is a device that protects circuits and appliances from danger due to excess current flow through the circuit. a. Which effect of current is used in the safety fuse?
b. How is the fuse connected in a circuit? (in parallel/series)
c. What must be the major feature of the substance used to make fuse wire?
d. Briefly explain how does a safety fuse make sure the safety of the circuit and appliance.
e. What is your opinion about using thick wire as fuse wire?
Answer:
a. Heating effect.
b. In series
c. Low melting point.
d. Melting point of the fuse wire is low. When excess current flows through the circuit due to a short circuit or overload, the fuse gets heated. As its melting point is low, it melts and the circuit is broken.
e. If we use thick wire, it may not melt and break while the excess flow of current. So it is not good to use thick wire as fuse wire.
5. The following are the electrical appliances in a house.
One 1000W Iron box, One 750 W Mixie, One 2000W heater, and five 60W bulbs. If all are designed to work on 230V, find an amperage of fuse wire suitable to this house.
Answer: Total power of the appliances = 1000+750+2000+5x60= 4050W
Total current required = 4050/230 = 17.6 A =18 A.
Therefore Amperage of the fuse wire is 18A.
6. Which of the following feature is correct in respect of a fuse wire?
a. High melting point.
b. Low melting point.
c. High resistance.
Answer: Low melting point.
Electric Power
The amount of energy consumed/released by an electrical appliance in unit time (in one second) is its power.
The unit of power is joule/second and is called 'watt'.
Power P = H/t = I²Rt/t = I²R
But using the relations I = V/R & R = V/I, power can also be expressed as
P = V²/R = VI
Note i) If voltage is constant, power is inversely proportional to the resistance.
That is, Pα (1/R)
ii) If the current is constant, power is directly proportional to the resistance. That is, PαR
• An electric heater of power 460 W is worked at the voltage 230V.
a. What is its amperage?
b. Find the resistance of the heating coil.
Answer:
a. Amperage, I = P/V = 460/230 = 2 A
b. Resistance R = V²/P = 230x230/460 = 115 Ω
Lighting Effect of electric current.
Incandescent lamp (Filament lamp): Incandescent lamp works on the lighting effect of electric current. In this, a filament is supported by two copper wires. The filament is made of pure metal tungsten that can become white-hot and emit white light when a current is passed through it. In order to avoid oxidation of the filament, the bulb is evacuated. And it is filled with inert gas or nitrogen gas for preventing vaporisation of the filament.
Note: The word 'incandescent” means glowing with heat.
Features of tungsten as a filament.
* Emit white light when it is being heated* High melting point * High resistivity
* High ductility.
Merits and demerits of Incandescent lamps.
Merits: Low cost, It doesn't make pollution.
Demerit: Major part of (above 60%) electric energy given to the lamp is lost as heat.
Discharge lamps: In discharge lamps light is produced as a result of the discharge of electricity through gases filled in a tube at low pressure. Discharge lamps are more efficient than incandescent lamps. Sodium vapour lamps, fluorescent lamps, CFL, Arc lamps etc are discharge lamps.
When voltage is applied between the electrodes of these lamps, the atoms of the gas inside the tube get excited. When the excited atom comes back to its original state, the light will be radiated.
Energy loss in most of the discharge lamps is very small. That is why we widely use discharge lamps like fluorescent lamps, CFLs etc.
LED Lamps: The full form of LED is Light Emitting Diode. It is an electronic device which converts electric energy into light energy by consuming low power. LED lamps are more efficient than any other lamps.
Advantages of LED Lamps:
i) Work at low power.
ii)No energy loss in the form of heat as there is no filament. (high efficiency)
iii) High longevity
iv. Not harmful to the environment
As LED lamps are more efficient, their production and use are to be encouraged.
PRACTICE QUESTIONS & ANSWERS
1. When 1.5V is applied to a piece of tungsten filament, it burns into ashes. But it doesn't happen so even at 230V when the filament is inside the bulb. Why?
Answer: Tungsten is a metal than can oxidise at high temperatures. That is why filament burns away even when a small voltage is applied in the presence of air. But there is no oxygen inside the bulb and hence the filament doesn't oxidise even at high voltage.
2. Gas atoms in the discharge lamps are excited to high energy levels. Why do the excited atoms give out energy?
Answer: For coming back to their original state to attain stability.
3. “The use of LED lamps is to be encouraged” Why?
Answer:
i. Its efficiency is high.
ii. It doesn't make environmental pollution.
4. Ramesh got a bulb of broken filament. He gently joined the ends of the filament and worked it. Then it is seen that its power is increased. Can you explain why?
Answer: When filament was rejoined, its length would have been reduced. It may lead to reducing the resistance of the filament. As power is inversely proportional to resistance at constant voltage, power is increased.
5. What happens to the power of a device/conductor due to the change in applied voltage.
Answer: We have P = V²/R. So, power is directly proportional to the square of the voltage. That is, if a voltage is doubled, power will be quadrupled. Or If voltage is halved, power will be one fourth.
6. It is marked as 100W, 220V in an electric appliance. What will be its power if the applied voltage is 110V.
Answer: Resistance of the appliance R = V²/P = 220x220/100 = 484 Ω.
Then the power at 110V = V²/R = 110x110/484 = 25 W
Another method: Power is directly proportional to the square of the voltage. Here the voltage is halved. So power will be ¼th of the marked power. That is, new power = 100/4 = 25W
7. Two bulbs are marked as 40W, 240V and 100W,240V. Which one has greater resistance?
Answer: Resistance of 40W bulb , R₄₀ = V²/P = 240x240/40 = 1440 Ω.
Resistance of 100W bulb , R₁₀₀ = V²/P = 240x240/100 = 576Ω.
8. Two bulbs are connected in series as shown in the figure.
b. If the current passes through the bulb L₁ is 0.45A, what will be the current
through the bulb L₂.
c. Of these which will show more brightness?
d.Why does this bulb emit more light?
e. If they were connected parallel, which bulb would give more light. Why?
Answer:
a. Resistance of L₁, R₁= V²/P = 200x200/40 = 1000 Ω
Resistance of L₂, R₂= V²/P = 200x200/50 = 800 Ω
b. As the bulbs are connected in series, the current will be the same through both bulbs. That is 0.45 A itself through the bulb L₂.
c. L₁ will glow brighter.
d. Because the resistance of L₁ is greater than that of L₂.
e. The bulb L₂ will give more light. Because when they are connected parallel, more current will pass through a low resistance.
9. Filament lamps are also called incandescent lamps.
a. What is the meaning of the word "incandescent"
b. Name the substance used for making filament?
c. What are the features of this substance?
d. What are the advantages of filling the bulbs with nitrogen after removing air from them?
e. What is the major limitation of filament lamps?
f. Name three light sources that can be used as substitutes for incandescent lamps.
Answer:
a. glowing with heat.
b. Tungsten
c. ability to emit white light on being heated, high melting point, high resistivity, high ductility.
d. prevent oxidation and vapourisation of filament.
e. Major portion (above 60%) of electrical energy consumed is lost in the form of heat.
f. discharge lamp, fluorescent lamp, CF lamp, LED lamp.
10. What is your opinion about making filament using nichrome?
Answer: It doesn't work. Because i) Nichrome cannot emit white light when it gets heated.
ii) Nichrome almost completely converts the consumed electric energy into heat energy.
11. A bulb was designed to get a specific power when it works on 250V. When this bulb is worked at 100V, its power is 16W. Then what was its desired power?
Answer: We have P = V²/R Then the resistance of the bulb, R = V²/P = 100x100/16 =625Ω
So the desired power of the bulb P¹=V²/R =250x250/625 = 100W.
12. Power of an electric Iron is 1000W. When it works at its marked power what will be heat energy released per second?
Answer: Power is the energy consumed or released per second. So this iron box will release 1000 J of heat energy in one second.
13. An electric appliance designed to work at 230V has 690 Ω resistance. Find its power.
Answer: Power P = V²R = 230x230/690 = 76.7 W
14. Whether a 1kW heater or a 100W bulb which works on 230V possess more resistance? Why?
Answer: We have P = V²/R OR R= V²/P
Resistance of heater RH = 230x230/1000 = 52.9 Ω
Resistance of bulb RB = 230x230/100 = 529 Ω
a. What is indicated in it?
b. Calculate the energy produced if it works at 230V for 5 minutes.
Answer:
a. 750 W is its power and 230 V is the required voltage to consume this power.
b. Heat produced in 5 minutes= power x time = 750x 5x60= 225000 J.
15. A 40 W fluorescent lamp and 40 W incandescent lamp are used in a house.
Which lamp consumes more current?
Answer: a. As the power of both lamps is the same, the same current will be drawn by each.
16. A wire AB of 20Ω resistance is included in a circuit as shown in the figure.
b. If the wire is folded into half and included in the same circuit, what will be
current?
c. Calculate the heat produced in the circuit in one second then.
d. What is the power then?
Answer:
a. R= 20 Ω V = 10V I=V/R = 10/20= 0.5A
b. When it is folded, length is halved and thickness is doubled. So resistance becomes ¼ of the previous value. Therefore new resistance R¹ = 20/4 = 5 Ω.
Then current I¹= V/R¹ = 10/5 - 2A
c. Heat produced in one second, H = VIt = 10x2x1 = 20J.
d. Power is the energy released or consumed per second. So the power of the wire is 20W.
17. …..... is an electronic device that works on the lighting effect of current. Answer: LED Lamp.
18. Which of the following is not expressing power.
a. I²Rt. b. V²/R c. I²R. d. VI
Answer: a. I²Rt.
19. The marking on an electrical appliance is 800W, 200V.
a. If it works on 100 V, what will be the consumed power?
b. What is the power when it works on 50V?
Answer:
a. Resistance of the appliance, R = V²/P = 200x200/800 = 50Ω
Power when it is worked on 100V, P = V²R= 100x100/50 = 200W.
b. Power when it is worked on 50V, P = V²R = 50x50/50 = 50W
More Questions and Answers
1. What advantages of nichrome are made use of in electric heating appliances?
Answer: High resistivity, high melting point, the ability to remain in red hot condition, negligible thermal expansion
2. If the same current flows through nichrome and copper wire of the same length, which one gets more heated?
Answer: Nichrome, It has high resistance
3. State Joule’s law
Answer: The quantity of heat produced when current is passed through a conductor is directly proportional to the product of the square, current, resistance and time for which the current flows
4. Write a short note 1. overloading 2. short circuit
Answer: If the power of all appliances in a circuit is more than that the circuit can withstand, the circuit is said to be overloaded
Direct contact between positive and negative terminals of battery or phase and neutral wires of mains supply without a resistance
5. How is a safety fuse provide safety to the circuit?
Answer: Whenever excess current passes through the circuit due to overloading or short circuit the fuse wire gets heated and melts off and breaks the circuits. Fuse wire is made up of an alloy of tin and lead have a low melting point hence it melts suddenly due to do the heat developed in the circuit
6. How is the fuse wire connected in a circuit
Answer:
1. The ends of the fuse wire are to be fixed firmly at appropriate points
2. The fuse wire should not project out of the carrier base
3. Use fuse wire with the proper amperage
4. Connect fuse wire in series in the phase line
5. Do not use copper wire or thick wires as fuse wire
7. What are the properties of fuse wire?
Answer: Alloy of tin and lead, low melting point
8. What are the properties of tungsten that make it suitable for being used as a filament?
Answer: High resistivity, high melting point, high ductility, ability to emit white light in the white-hot condition
9. Why is the glass bulb of an incandescent lamp filled with inert gas or nitrogen?
Answer: Vaporisation of the filament can be reduced by filling inert gas or nitrogen, it increases the lifespan of the bulb
10. Explain the working of the discharge lamp
Answer: Discharge lamps are glass tubes fitted with two electrodes. They emit light as a result of the discharge of electricity through a gas-filled tube. when a high potential difference is applied the gas molecules get excited. Excited atoms come back to their original state for attaining stability. During this process, energy stored in them will be radiated as light
11. What are the advantages of using fluorescent lamps instead of incandescent lamps?
Answer: Loss of electricity in the form of heat is less, more lifespan, no Shadow formation, less consumption of electricity, more light than the same power filament lamp
12. What are the advantages of LED lamps?
Answer: There is no loss of energy in the form of heat because there is no filament, No mercury vapour in LED lamps hence it is not harmful to the environment, It requires a small amount of electric power, more bright light as compared to other types of lamps with the same power, long lifespan
13. What are the disadvantages of CFL?
Answer: it contains Mercury it is harmful and creates environmental pollution
14. What are the parts of LED lamp
Answer: base unit, heat sink, base plate, back conductor screws, power supply board, printed circuit board (LED chipboard), diffuser cup
15. How can LED bulbs be disposed of scientifically?
Answer: segregate the plastic, electronic, and metal components of LED bulbs and transfer them to their respective disposal unit
16. When an incandescent lamp is lighted with its broken filament joined together, we get more light output why?
Answer: The length of the tungsten filament decreases, and the resistance also decreases .more current flows through the filament then the heat and power increase.
17. what is the method of connecting resistors? List its differences in series and parallel
Answer:
Answer: Heat increases by four times
19. Volt meter connects parallel to the resistance and Ammeter connects series to the resistance
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 Chapter 01 Effects of Electric Current){kind=link}
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