Lesson.4 ELECTRICITY
Lesson 4 > ELECTRICITY
Formulae | |
Electric Current | I = Q/t = Charge/Time ⇒ Q = It |
Potential difference | I = W/Q = Workdone / Charge ⇒ W = vQ |
Ohm’s Law | V = IR; R = V/I |
Electrical Resisivtity (or) specific resistance | ρ = RA/L |
Conductance | G = 1/R = 1/resistance |
Conductivity | σ = 1/ρ = 1/resistivity |
Equivalent resistance in aseries combinations | R_{s} = R_{1} + R_{2} |
When ‘n’ resistors are connected in a series combinations | R_{s} = nR; |
When ‘n’ resistors are connected in parallel | R_{p} = R/n |
Total resistance in the circuit | 1/R_{p} = 1/R_{1} + 1/R_{2} + 1/R_{3} |
Series connection of parallel resistors | R_{total} = R_{P1} + R_{P2} |
Parallel connection of series resistors | 1/R_{total} = 1/R_{S1} + 1/R_{S2} |
Joule’s law of heating | H = I^{2} Rt; H = VI t |
Electric Power | P = work/time = VI/t (or) P = VI |
Electrical energy | E = power x time = VI t = VQ |
Resistance | Resistance (R) = Volatage (V) / Current (I) |
Electric Power | P = V^{2}/R (or) P = VI = I^{2} = V^{2}/R |
TEXTBOOK EVALUATION
I. Choose the correct answer
1. Which of the following is correct?
- Rate of change of charge is electrical power.
- Rate of change of charge is current.
- Rate of change of energy is current.
- Rate of change of current is charge.
Ans ; Rate of change of charge is current.
2. SI unit of resistance is
- mho
- joule
- ohm
- ohm meter
Ans ; ohm
3. In a simple circuit, why does the bulb glow when you close the switch?
- The switch produces electricity.
- Closing the switch completes the circuit.
- Closing the switch breaks the circuit.
- The bulb is getting charged.
Ans ; Closing the switch completes the circuit.
4. Kilowatt hour is the unit of
- resistivity
- conductivity
- electrical energy
- electrical power
Ans ; electrical energy
II. Fill in the blanks
1. When a circuit is open, __________ cannot pass through it.
Ans : current
2. The ratio of the potential difference to the current is known as __________.
Ans : Resistance
3. The wiring in a house consists of __________ circuits.
Ans : parallel
4. The power of an electric device is a product of __________ and __________.
Ans : votage, current
5. LED stands for __________.
Ans : Light Emitting Diode
III. State whether the following statements are true or false: If false correct the statement.
1. Ohm’s law states the relationship between power and voltage. ( False )
- Ohm’s law states the relationship between the potential difference and current.
2. MCB is used to protect house hold electrical appliances. ( True )
3. The SI unit for electric current is the coulomb. ( False )
- The SI unit for electric current is the Ampere.
4. One unit of electrical energy consumed is equal to 1000 kilowatt hour. ( False )
- One unit of electrical energy consumed is equal to 1000watt hour.
5. The effective resistance of three resistors connected in series is lesser than the lowest of the individual resistance. ( False )
- The effective resistance of three resistors connected in series is greater than the highest of the individual resistance.
IV. Match the items in column-I to the items in column-II:
Column – I | Column – II |
1) electric current | a) volt |
2) potential difference | b) ohm meter |
3) specific resistance | c) watt |
4) electrical power | d) joule |
5) electrical energy | e) ampere |
Ans ; 1 – e, 2 – a, 3 – b, 4 – c, 5 – d |
V. Assertion and reason type questions:
Mark the correct choice as
- if both the assertion and the reason are true and the reason is the correct explanation of the assertion.
- if both the assertion and the reason are true, but the reason is not the correct explanation of the assertion.
- if the assertion is true, but the reason is false.
- if the assertion is false, but the reason is true.
1. Assertion : Electric appliances with a metallic body have three wire connections.
Reason : Three pin connections reduce heating of the connecting wires.
- Ans : (c) The assertion is true, but the reason is false.
2. Assertion : In a simple battery circuit the point of highest potential is the positive terminal of the battery.
Reason : The current flows towards the point of the highest potential.
- Ans : (c) The assertion is true, but the reason is false.
3. Assertion : LED bulbs are far better than incandescent bulbs.
Reason : LED bulbs consume less power than incandescent bulbs.
- Ans : (a) Both the assertion and the reason are true and the reason is the correct explanation of assertion.
VI. Very short answer questions.
1. Define the unit of current.
- The SI unit of current is ampere (A).
- The current flowing through a conductor is said to be one ampere, when a charge of one coulomb flows across any cross-section of a conductor in one second.
- 1 ampere = 1 coulomb / 1second
2. What happens to the resistance, as the conductor is made thicker?
- Resistance is inversely proportional to area of cross section.
- A thicker wire has larger area of cross section and hence the resistance decreases.
- Resistance = Length / Area
- R ∝ 1/A
3. Why is tungsten metal used in bulbs, but not in fuse wires?
- Melting point of tungsten is very.
- Hence it is used in a filament bulb.
- But a fuse wire should be made up of material which has low melting point.
4. Name any two devices, which are working on the heating effect of the electric current.
- Electric heater
- Electric iron work on the heating effect of current.
VII. Short answer questions
Define electric potential and potential difference.
Electric potential:
The electric potential at a point is defined as the amount of work done in moving a unit positive charge from infinity to that point against the electric force.
Electrical potential difference:
The electric potential difference between two points is defined as the amount of work done in moving a unit positive charge from one point to another point against the electric force.
2. What is the role of the earth wire in domestic circuits?
- The earth wire provides a low resistance path to the electric current.
- The earth wire sends the current from the body of the appliance to the earth, whenever a live wire accidently touches the body of the metallic electric appliance.
- Thus the earth wire serves as a protective conductor, which saves us from electric shock.
3. State Ohm’s law.
At a constant temperature, the steady current flowing through a conductor is directly proportional to the potential difference between the two ends of the conductor.
I ∝ V
V = IR
4. Distinguish between the resistivity and conductivity of a conductor.
Resistivity | Conductivity |
1. It is the resistance of a conductor of unit length and unit area of cross section. | The reciprocal of electrical resistivity of a material is called electrical conductivity. |
2. Its unit is ohm metre | Its unit is ohm^{-1} metre^{-1} |
3. Resistivity is less for conductor than for inulators | Conductivity is more for conductors than insulators for inulators. |
4. ρ=RA/L | σ= 1/ρ |
5. What connection is used in domestic appliances and why?
Parallel connection is used in domestic appliances.
Reason:
- Each appliance will get the full voltage.
- The parallel circuit divides the current through the appliances.
- Each appliance will get the proper current depending on its resistance.
- Each of them can be put on / off independently.
VIII. Long answer questions
1. With the help of a circuit diagram derive the formula for the resultant resistance of three resistances connected: a) in series and b) in parallel.
Resistors in series:
- A series circuit connects the components one after the other to form a ‘single loop.
- A series circuit has only one loop through which current can pass.
- If the circuit is interrupted at any point in the loop, no current can pass through the circuit and hence no electric appliances connected in the circuit will work.
- Series circuits are commonly used in devices such as flashlights.
- Thus, if resistors are connected end to end, so that the same current passes through each of them, then they are said to be connected in series.
Series connection of resistors
- Let, three resistances R_{1}, R_{2}, and R_{3}, be connected in series.
- Let the current flowing through them be I.
- According to Ohm’s Law, the potential differences V_{1}, V_{2}, and V_{3} across R_{1}, R_{2}, and R_{3} respectively, are given by:
V_{1} = IR_{1}, ………………………………………… (1)
V_{2} = IR_{2}, ………………………………………… (3)
V_{3} = IR_{3}, ………………………………………… (3)
The sum of the potential differences across the ends of each resistor is given by:
V = V_{1 }+ V_{2 }+ V_{3}
Using equations (1), (2) and (3), we get
V = I R_{1 }+ I R_{2 }+ I R_{3}
- ‘The effective resistor is a single resistor, which can replace the resistors effectively, so as to allow the same current through the electric circuit.
- Let, the effective resistance of the series-combination of the resistors, be Rg. Then,
V=I R_{5}, ………………………………………… (5)
Combining equations (4) and (5), you get,
I R_{S }= I R_{1 }+ I R_{2 }+ I R_{3}
R_{S} = R_{1 }+ R_{2 }+ R_{3} ………………………………………… (6)
- Thus, you can understand that when a number of resistors are connected in series, their equivalent resistance or effective resistance is equal to the sum of the individual resistances.
- When ‘n’ resistors of equal resistance R are connected in series, the equivalent resistance is ‘n R’
i.e., R_{S}=n R
- The equivalent resistance in a series combination is greater than the highest of the individual resistances.
Resistances in Parallel:
- A parallel circuit has two or more loops through which current can pass.
- If the circuit is disconnected in one of the loops, the current can still pass through the other loop(s).
- he wiring in a house consists of parallel circuits.
Series connections of resistors
- Consider that three resistors Ry, R, and R are connected across two common points A and B.
- The potential difference across each resistance is the same and equal to the potential difference between A and B.
- This is measured using the voltmeter.
- The current I arriving at A divides into three branches I,, I, and I, passing through R,, R, and R, respectively.
According to the Ohm’s law, you have,
I_{1} = V/R_{1}, ………………………………………… (7)
I_{2} = V/R_{2}, ………………………………………… (8)
I_{3} = V/R_{3}, ………………………………………… (9)
The total current through the circuit is given by
I = I_{1 }+ I_{2 }+ I_{3}
Using equations (7), (8), and (9) you get
I_{1} = V/R_{1 }+ V/R_{2 }+ V/R_{3} ………………………………………… (10)
Let the effective resistance of the parallel combination of resistors be R_{P} Then
I = V/R_{P }………………………………………… (11)
Combining equations (10) and (11), you have
V/R_{P} = V/R_{1 }+ V/R_{2 }+ V/R_{3}
1/R_{P} = 1/R_{1 }+ 1/R_{2 }+ 1/R_{3 ………………………………………… (12)}
* Thus, when a number of resistors are connected in parallel, the sum of the reciprocals of the individual resistances is equal to the reciprocal of the effective or equivalent resistance.
* When ‘y resistors of equal resistances R are connected in parallel, the equivalent resistance is R/n.
i.e., 1/R_{P} = 1/R_{ }+ 1/R_{ }+ 1/R …….. + 1/R = n/R _{ ………………………………………… (13)}
Hence, R_{P} = R/n
The equivalent resistance in a parallel combination is less than the lowest of the individual resistances.
2. a) What is meant by electric current?
b) Name and define its unit.
c) Which instrument is used to measure the electric current? How should it be connected in a circuit?
a) Electric current:
- Electric current is often termed as ‘current’ and it is represented by the symbol T.
- It is defined as the rate of flow of charges in a conductor.
- This means that the electric current represents the amount of charges flowing in any cross section of a conductor (say a metal wire) in unit time. 9
- If a net charge ‘Q’ passes through any cross section of a conductor in time ‘t’ then the current flowing through the conductor is
b) SI unit of electric current:
- The SI unit of electric current is ampere (A).
- The current flowing through a conductor is said to be one ampere, when a charge of one coulomb flows across any cross-section of a conductor, in one second. Hence,
1 ampere = 1 coulomb / 1 second
c) Ammeter is used to measure the electric current. It should be connected in series in a circuit.
3. a) State Joule’s law of heating.
b) An alloy of nickel and chromium is used as the heating element. Why?
c) How does a fuse wire protect electrical appliances?
a) Joule’s law of heating:
Joule’s law of heating states that the heat produced in any resistor is:
- directly proportional to the square of the current passing through the resistor.
- directly proportional to the resistance of the resistor.
- directly proportional to the time for which the current is passing through the resistor.
b) (i) it has high resistivity, (ii) it has a high melting point, (iii) it is not easily oxidized.
c) The fuse wire is connected in series, in an electric circuit. When a large current passes through the circuit, the fuse wire melts due to Joule’s heating effect and hence the circuit gets disconnected. Therefore, the circuit and the electric appliances are saved from any damage. The fuse wire is made up of a material whose melting point is relatively low.
4, Explain about domestic electric circuits. (circuit diagram not required)
- The electricity produced in power stations is distributed to all the domestic and industrial consumers through overhead and underground cables.
- In our homes, electricity is distributed through the domestic electric circuits wired by the electricians.
- ‘The first stage of the domestic circuit is to bring the power supply to the main-box from a distribution panel, such as a transformer.
Main Box Contains:
i) Fuse Box:
- ‘The fuse box contains either a fuse wire or a miniature circuit breaker (MCB).
- The function of the fuse wire or a MCB is to protect the house hold electrical appliances from overloading due to excess current.
- It has a spring attached to the switch, which is attracted by an electromagnet when an excess current passes through the circuit.
- An MCB is a switching device, which can be activated automatically as well as manually.
- Hence, the circuit is broken and the protection of the appliance is ensured.
ii) Meter:
- The meter is used to record the consumption of electrical energy.
Insulated Wire: - The electricity is brought to houses by two insulated wires.
- Out of these two wires, one wire has a red insulation and is called the ‘live wire’.
- The other wire has a black insulation and is called the ‘neutral wire’.
- Both, the live wire and the neutral wire enter into a box where the main fuse is connected with the live wire.
- After the electricity meter, these wires enter into the main switch, which is used to discontinue the electricity supply whenever required.
- After the main switch, these wires are connected to live wires of two separate circuits.
5A rating circuit:
- Out of these two circuits, one circuit is of a5 A rating, which is used to run the electric appliances with a lower power rating, such as tube lights, bulbs and fans.
15 A rating circuit:
- The other circuit is of a 15 A rating, which is used to run electric appliances with a high power rating, such as air-conditioners, refrigerators, electric iron and heaters.
- It should be noted that all the circuits in a house are connected in parallel, so that the disconnection of one circuit does not affect the other circuit.
- The electricity supplied to your house is actually an alternating current having an electric potential of 220 V.
- One more advantage of the parallel connection of circuits is that each electric appliance gets an equal voltage.
5. a) What are the advantages of LED TV over the normal TV?
b) List the merits of LED bulb.
a) Advantages of LED television:
- It has brighter picture quality.
- It is thinner in size.
- Tt uses less power and consumes very less energy.
- Its life span is more.
- It is more reliable.
b) Merits of a LED bulb:
- As there is no filament, there is no loss of energy in the form of heat. It is cooler than the incandescent bulb
- In comparison with the fluorescent light, the LED bulbs have significantly low power requirement.
- It is not harmful to the environment.
- A wide range of colours is possible here.
- It is cost-efficient and energy efficient.
- Mercury and other toxic materials are not required.
IX. Numerical Problems:
1. An electric iron consumes energy at the rate of 420 W when heating is at the maximum rate and 180 W when heating is at the minimum rate. The applied voltage is 220 V. What is the current in each case?
Given:
Energy consumed when heating is maximum = 420 W at a given rate
i.e., P_{1} = 420 W
Energy consumed when heating is minimum at a given rate P_{2} = 180 W
Applied voltage = 220 V
Current in each case =?
Formula used: |
Power = Voltage x Current |
P
case (i) I_{1} case (ii) I_{2} |
= V x I
= P_{1}/V = 420/220 = 1.9 A = P_{2}/V = 180/220 = 1.81 A |
2. A 100 watt electric bulb is used for 5 hours daily and four 60 watt bulbs are used for 5 hours daily. Calculate the energy consumed (in kWh) in the month of January
Formula used: |
E = P x t |
Given:
Power of the first electric bulb = 100 W = 100/ 1000 = 0.1 kW
Time = 5 hours
Power of the second electric bulb = 60 watt = 60/1000 = 0.06 kW
Total number of bulbs = 4
Time = 5 hours.
Energy consumed in the month of January = ?
Energy = Power x time
Energy consumed by the first bulb in a day = 0.1 x 5 = 0.5 kWh
Energy consumed by the four 60 W bulb in a day = 0.06 x 4x 5=1.2 kWh
Total energy consumed by both the bulbs = 0.5 + 1.2 = 1.7 kWh
Total energy consumed in the month of january = 31 x 1.7 = 52.7 kWh
3. A torch bulb is rated at 3 V and 600 mA. Calculate it’s
a) power
b) resistance
c) energy consumed if it is used for 4 hour.
Formula used: |
P = V x I, R = V/I, E = P x t |
Given
V = 3V, I = 300mA = 300 x 10^{-3} A, Time = 4 hours
Power = ?, resistance = ?, energy = ?
a). Power (P) = V x I
P = 3 x 600 x 10^{-3} = 1.8W
b). Resistance (R) = V/I [V = IR]
R = 3 /600 x 10^{-3} = 5Ω
c). Energy consumed in 4 hours
E = p x t
= 1.8 x 4 = 7.2 wh
4 A piece of wire having a resistance R is cut into five equal parts.
a) How will the resistance of each part of the wire change compared with the original resistance?
b) If the five parts of the wire are placed in parallel, how will the resistance of the combination change?
c) What will be ratio of the effective resistance in series connection to that of the parallel connection?
a) Wire is cut into 5 equal parts. Since all dimensions are same, resistance of each wire is equal and has a value = R/5
b) Formula for finding the effective resistance when connected in parallel is
1/R_{p} = 1/R_{1 }+ 1/R_{2} + 1/R_{3} + 1/R_{4} + 1/R_{5}
Formula used: |
Resistors in parallel – 1/R_{p} = 1/R_{1 }+ 1/R_{2} ……………………….. + 1/R_{n} |
Here, R_{1 }= R_{2} = R_{3} = R_{4} = R_{5 }= R/5
1/R_{p} = 5/R_{ }+ 5/R + 5/R + 5/R + 5/R = 25/R
R_{p }= R/25 Ω
c) If the resistors are connected in series, then the effective resistance will be
Formula used: |
Resistors in series – R_{S} = R_{1 }+ R_{2} ……………………….. + R_{n} |
R_{p}” = R/5_{ }+ R/5 + R/5 + R/5 + R/5
R_{p}“= 5R/R = R
Ratio of effective resistance in series connection to that of the parallel connection is
R_{S}/R_{p }= R/R/25 = 25
X. HOTS:
1. Two resistors when connected in parallel give the resultant resistance of 2 ohm; but when connected in series the effective resistance becomes 9 ohm. Calculate the value of each resistance.
Formula used: |
1/R_{p} = 1/R_{1 }+ 1/R_{2} ……………………….. + 1/R_{n} |
R_{S} = R_{1 }+ R_{2} |
Given :
R_{S} = 2Ω, R_{S} = 9Ω
1/R_{p} = 1/R_{1 }+ 1/R_{2}
R_{S} = R_{1 }+ R_{2}
1/R_{p} = R_{1 }+ R_{2 }/ R_{1 }R_{2}
R_{p} = R_{1 }R_{2 }/ R_{1 }+ R_{2}
R_{p} = 2Ω; R_{1 }+ R_{2 }= 9Ω
2 = R_{1 }R_{2} /9 ; R_{1 }R_{2 }= 18Ω
On solving,
R_{1 }= 3Ω (or) R_{1 }= 6Ω
R_{2 }= 6Ω (or) R_{2 }= 3Ω
2. How many electrons are passing per second in a circuit in which there is a
current of 5 A?
Given ;
I = 5A, t = 1s
Number of electrons?
Formula used: |
q = It |
n = q/e |
q = It = 5 x 1 = 5C
q = ne where n is the number of electrons; e is the charge of an electron which is equal to 1.6x 10^{-19}C
n = q/e = 5/1.6×10^{-19} = 31.25 x 10^{18} electrons
3. A piece of wire of resistance 10 ohm is drawn out so that its length is increased to three times its original length. Calculate the new resistance.
Given :
R = 10Ω; original lenth = l, new length l’ = 3l
Area will decrease by 3 times.
Formula used: |
New resistance Value R’ = ρ l’/A’ |
R^{1} = ρ l^{1}/A^{1} = ρ 3l/A/3 = 9pl/A = 9R = 9 (10) = 90Ω
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