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12. Electricity
Electricity has an important place in modern society. It is a controllable and convenient form of energy for a variety of uses in homes, schools, hospitals, industries and so on. What constitutes electricity? How does it flow in an electric circuit? What are the factors that control or regulate the current through an electric circuit? In this Chapter, we shall attempt to answer such questions. We shall also discuss the heating effect of electric current and its applications.
12.1 ELECTRIC CURRENT AND CIRCUIT
We are familiar with air current and water current. We know that flowing water constitute water current in rivers. Similarly, if the electric charge flows through a conductor (for example, through a metallic wire), we say that there is an electric current in the conductor. In a torch, we know that the cells (or a battery, when placed in proper order) provide flow of charges or an electric current through the torch bulb to glow. We have also seen that the torch gives light only when its switch is on. What does a switch do? A switch makes a conducting link between the cell and the bulb. A continuous and closed path of an electric current is called an electric circuit. Now, if the circuit is broken anywhere (or the switch of the torch is turned off ), the current stops flowing and the bulb does not glow.
How do we express electric current? Electric current is expressed by the amount of charge flowing through a particular area in unit time. In other words, it is the rate of flow of electric charges. In circuits using metallic wires, electrons constitute the flow of charges. However, electrons were not known at the time when the phenomenon of electricity was first observed. So, electric current was considered to be the flow of positive charges and the direction of flow of positive charges was taken to be the direction of electric current. Conventionally, in an electric circuit the direction of electric current is taken as opposite to the direction of the flow of electrons, which are negative charges.
If a net charge Q, flows across any cross-section of a conductor in time t, then the current I, through the cross-section is
The SI unit of electric charge is coulomb (C), which is equivalent to the charge contained in nearly 6 × 10 ^{18} electrons. (We know that an electron possesses a negative charge of 1.6 × 10^{–19} C .) The electric current is expressed by a unit called ampere (A), named after the French scientist, Andre-Marie Ampere (1775–1836). One ampere is constituted by the flow of one coulomb of charge per second, that is, 1 A = 1 C/1 s. Small quantities of current are expressed in milliampere (1 mA = 10^{–3} A) or in microampere (1 μA = 10^{–6} A). An instrument called ammeter measures electric current in a circuit. It is always connected in series in a circuit through which the current is to be measured. Figure 1 shows the schematic diagram of a typical electric circuit comprising a cell, an electric bulb, an ammeter and a plug key. Note that the electric current flows in the circuit from the positive terminal of the cell to the negative terminal of the cell through the bulb and ammeter.
Fig.1 A schematic diagram of an electric circuit comprising - cell, electric bulb, ammeter and plug key
A current of 0.5 A is drawn by a filament of an electric bulb for 10 minutes. Find the amount of electric charge that flows through the circuit.
We are given, I = 0.5 A; t = 10 min = 600 s.
From Eq. (1), we have
Q = It
= 0.5 A × 600 s
= 300 C
- What does an electric circuit mean?
- Define the unit of current.
- Calculate the number of electrons constituting one coulomb of charge.
12.2 ELECTRIC POTENTIAL AND POTENTIAL DIFFERENCE
What makes the electric charge to flow? Let us consider the analogy of flow of water. Charges do not flow in a copper wire by themselves, just as water in a perfectly horizontal tube does not flow. If one end of the tube is connected to a tank of water kept at a higher level, such that there is a pressure difference between the two ends of the tube, water flows out of the other end of the tube. For flow of charges in a conducting metallic wire, the gravity, of course, has no role to play; the electrons move only if there is a difference of electric pressure – called the potential difference – along the conductor. This difference of potential may be produced by a battery, consisting of one or more electric cells. The chemical action within a cell generates the potential difference across the terminals of the cell, even when no current is drawn from it. When the cell is connected to a conducting circuit element, the potential difference sets the charges in motion in the conductor and produces an electric current. In order to maintain the current in a given electric circuit, the cell has to expend its chemical energy stored in it.
We define the electric potential difference between two points in an electric circuit carrying some current as the work done to move a unit charge from one point to the other –
Potential difference (V) between two points = Work done (W)/Charge (Q)
The SI unit of electric potential difference is volt (V), named after Alessandro Volta (1745–1827), an Italian physicist. One volt is the potential difference between two points in a current carrying conductor when 1 joule of work is done to move a charge of 1 coulomb from one point to the other.
Therefore,
1 V = 1 J C^{–1}
The potential difference is measured by means of an instrument called the voltmeter. The voltmeter is always connected in parallel across the points between which the potential difference is to be measured.
How much work is done in moving a charge of 2 C across two points having a potential difference 12 V?
The amount of charge Q, that flows between two points at potential difference V (= 12 V) is 2 C. Thus, the amount of work W, done in moving the charge [from Eq. (2)] is
W = VQ
= 12 V × 2 C
= 24 J.
- Name a device that helps to maintain a potential difference across a conductor.
- What is meant by saying that the potential difference between two points is 1 V?
- How much energy is given to each coulomb of charge passing through a 6 V battery?
12.3 CIRCUIT DIAGRAM
We know that an electric circuit, as shown in Fig.1, comprises a cell (or a battery), a plug key, electrical component(s), and connecting wires. It is often convenient to draw a schematic diagram, in which different components of the circuit are represented by the symbols conveniently used. Conventional symbols used to represent some of the most commonly used electrical components are given in Table 1.
Table 1 Symbols of some commonly used components in circuit diagrams
Sl. No. | Components | Symbols |
---|---|---|
1 | An electric cell | |
2 | A battery or a combination of cells | |
3 | Plug key or switch (open) | |
4 | Plug key or switch (closed) | |
5 | A wire joint | |
6 | Wires crossing without joining | |
7 | Electric bulb or | |
8 | A resistor of resistance R | |
9 | Variable resistance or rheostat | |
10 | Ammeter | |
11 | Voltmeter |
12.4 OHM’S LAW
Is there a relationship between the potential difference across a conductor and the current through it? Let us explore with an Activity.
In this Activity, you will find that approximately the same value for V/I is obtained in each case. Thus the V–I graph is a straight line that passes through the origin of the graph, as shown in Fig.3. Thus, V/I is a constant ratio.
In 1827, a German physicist Georg Simon Ohm (1787–1854) found out the relationship between the current I, flowing in a metallic wire and the potential difference across its terminals. He stated that the electric current flowing through a metallic wire is directly proportional to the potential difference V, across its ends provided its temperature remains the same. This is called Ohm’s law. In other words –
Fig.3 V–I graph for a nichrome wire. A straight line plot shows that as the current through a wire increases, the potential difference across the wire increases linearly – this is Ohm’s law.
In Eq. (4), R is a constant for the given metallic wire at a given temperature and is called its resistance. It is the property of a conductor to resist the flow of charges through it. Its SI unit is ohm, represented by the Greek letter Ω. According to Ohm’s law,
If the potential difference across the two ends of a conductor is 1 V and the current through it is 1 A, then the resistance R, of the conductor
is 1 Ω. That is, 1 ohm = 1 volt / 1 ampere
Also from Eq. (5) we get
It is obvious from Eq. (7) that the current through a resistor is inversely proportional to its resistance. If the resistance is doubled the current gets halved. In many practical cases it is necessary to increase or decrease the current in an electric circuit. A component used to regulate current without changing the voltage source is called variable resistance. In an electric circuit, a device called rheostat is often used to change the resistance in the circuit. We will now study about electrical resistance of a conductor with the help of following Activity.
In this Activity we observe that the current is different for different components. Why do they differ? Certain components offer an easy path for the flow of electric current while the others resist the flow. We know that motion of electrons in an electric circuit constitutes an electric current. The electrons, however, are not completely free to move within a conductor. They are restrained by the attraction of the atoms among which they move. Thus, motion of electrons through a conductor is retarded by its resistance. A component of a given size that offers a low resistance is a good conductor. A conductor having some appreciable resistance is called a resistor. A component of identical size that offers a higher resistance is a poor conductor. An insulator of the same size offers even higher resistance.
12.5 FACTORS ON WHICH THE RESISTANCE OF A CONDUCTOR DEPENDS
It is observed that the ammeter reading decreases to one-half when the length of the wire is doubled. The ammeter reading is increased when a thicker wire of the same material and of the same length is used in the circuit. A change in ammeter reading is observed when a wire of different material of the same length and the same area of cross-section is used. On applying Ohm’s law [Eqs. (5) – (7)], we observe that the resistance of the conductor depends (i) on its length, (ii) on its area of cross-section, and (iii) on the nature of its material. Precise measurements have shown that resistance of a uniform metallic conductor is directly proportional to its length (l) and inversely proportional to the area of cross-section (A). That is,
Combining Eqs. (8) and (9) we get
R α l / A
where ρ (rho) is a constant of proportionality and is called the electrical resistivity of the material of the conductor. The SI unit of resistivity is Ω m. It is a characteristic property of the material. The metals and alloys have very low resistivity in the range of 10^{–8} Ω m to 10^{–6} Ω m. They are good conductors of electricity. Insulators like rubber and glass have resistivity of the order of 10^{12} to 10^{17} Ω m. Both the resistance and resistivity of a material vary with temperature.
Table 2 reveals that the resistivity of an alloy is generally higher than that of its constituent metals. Alloys do not oxidise (burn) readily at high temperatures. For this reason, they are commonly used in electrical heating devices, like electric iron, toasters etc. Tungsten is used almost exclusively for filaments of electric bulbs, whereas copper and aluminium are generally used for electrical transmission lines.
Table 2 Electrical resistivity* of some substances at 20°C
Material | Resistivity (Ω m) | |
---|---|---|
Conductors | Silver | 1.60 × 10^{–8} |
Copper | 1.62 × 10^{–8} | |
Aluminium | 2.63 × 10^{–8} | |
Tungstan | 5.20 × 10^{–8} | |
Nickel | 6.84 × 10^{–8} | |
Iron | 10.0 × 10^{–8} | |
Chromium | 12.9 × 10^{–8} | |
Mercury | 94.0 × 10^{–8} | |
Maganese | 1.84 × 10^{–6} | |
Alloys | Constantan (alloy of Cu andNi) | 49 × 10^{–6} |
Manganin (alloy of Cu, Mn and Ni) | 44 × 10^{–6} | |
Nichrome (alloy of Ni, Cr, Mn and Fe) | 100 × 10^{–6} | |
Insulators | Glass | 10^{10} – 10^{14} |
Hard rubber | 10^{13} – 10^{16} | |
Ebonite | 10^{15} – 10^{17} | |
Diamond | 10^{12} – 10^{13} | |
Paper (dry) | 10^{12} |
* You need not memorise these values. You can use these values for solving numerical problems.
- How much current will an electric bulb draw from a 220 V source, if the resistance of the bulb filament is 1200 Ω?
- How much current will an electric heater coil draw from a 220 V source, if the resistance of the heater coil is 100 Ω?
- We are given V = 220 V; R = 1200 Ω.
From Eq. (6), we have the current I = 220 V/1200 Ω = 0.18 A. - We are given, V = 220 V, R = 100 Ω.
From Eq. (6), we have the current I = 220 V/100 Ω = 2.2 A.
Note the difference of current drawn by an electric bulb and electric heater from the same 220 V source!
The potential difference between the terminals of an electric heater is 60 V when it draws a current of 4 A from the source. What current will the heater draw if the potential difference is increased to 120 V?
We are given, potential difference V = 60 V, current I = 4 A.
According to Ohm’s law, R = =15Ω
When the potential difference is increased to 120 V the current is given by
current = V/R = 120V/15 = 8A
The current through the heater becomes 8 A.
Resistance of a metal wire of length 1 m is 26 Ω at 20°C. If the diameter of the wire is 0.3 mm, what will be the resistivity of the metal at that temperature? Using Table 2, predict the material of the wire.
We are given the resistance R of the wire = 26 Ω, the diameter
d = 0.3 mm = 3 × 10^{-4} m, and the length l of the wire = 1 m.
Therefore, from Eq. (10), the resistivity of the given metallic wire is
ρ = (RA/l) =(Rπd2/4l )
Substitution of values in this gives
ρ = 1.84 × 10^{–6} Ω m
The resistivity of the metal at 20°C is 1.84 × 10^{–6} Ω m. From Table 2, we see that this is the resistivity of manganese.
A 4 Ω resistance wire is doubled on it. Calculate the new resistance of the wire.
We are given, R = 4 Ω.
When a wire is doubled on it, its length would become half and area of cross-section would double. That is, a wire of length l and area of cross-section A becomes of length l/2 and area of cross section 2A. From Eq. (10), we have
R = ρ(l/A)
R_{1} = ρ((l/A) / 2A)
where R1 is the new resistance.
Therefore, R_{1}/R = ρ((l/A)/2A) / ρ(l/A) = 1/4
Or, R_{1} = R/4 = 4Ω/4 = 1Ω
The new resistance of the wire is 1 Ω.
- On what factors does the resistance of a conductor depend?
- Will current flow more easily through a thick wire or a thin wire of the same material, when connected to the same source? Why?
- Let the resistance of an electrical component remains constant while the potential difference across the two ends of the component decreases to half of its former value. What change will occur in the current through it?
- Why are coils of electric toasters and electric irons made of an alloy rather than a pure metal?
- Use the data in Table 2 to answer the following –
- Which among iron and mercury is a better conductor?
- Which material is the best conductor?
12.6 RESISTANCE OF A SYSTEM OF RESISTORS
In preceding sections, we learnt about some simple electric circuits. We have noticed how the current through a conductor depends upon its resistance and the potential difference across its ends. In various electrical gadgets, we often use resistors in various combinations. We now therefore intend to see how Ohm’s law can be applied to combinations of resistors.
There are two methods of joining the resistors together. Figure 6 shows an electric circuit in which three resistors having resistances R_{1}, R_{2} and R_{3}, respectively, are joined end to end. Here the resistors are said to be connected in series.
Fig.6 Resistors in series
Figure 7 shows a combination of resistors in which three resistors are connected together between points X and Y. Here, the resistors are said to be connected in parallel.
Fig.7 Resistors in parallel
12.6.1 Resistors in Series
What happens to the value of current when a number of resistors are connected in series in a circuit? What would be their equivalent resistance? Let us try to understand these with the help of the following activities.
You will observe that the value of the current in the ammeter is the same, independent of its position in the electric circuit. It means that in a series combination of resistors the current is the same in every part of the circuit or the same current through each resistor.
You will observe that the potential difference V is equal to the sum of potential differences V_{1}, V_{2}, and V_{3}. That is the total potential difference across a combination of resistors in series is equal to the sum of potential difference across the individual resistors. That is,
In the electric circuit shown in Fig.8, let I be the current through the circuit. The current through each resistor is also I. It is possible to replace the three resistors joined in series by an equivalent single resistor of resistance R, such that the potential difference V across it, and the current I through the circuit remains the same. Applying the Ohm’s law to the entire circuit, we have
On applying Ohm’s law to the three resistors separately, we further have
From Eq. (11),
I R = I R_{1} + I R_{2} + I R_{3}
Or
We can conclude that when several resistors are joined in series, the resistance of the combination R_{s} equals the sum of their individual resistances, R_{1}, R_{2}, R_{3}, and is thus greater than any individual resistance.
An electric lamp, whose resistance is 20 Ω, and a conductor of 4 Ω resistance are connected to a 6 V battery (Fig. 9). Calculate
- the total resistance of the circuit,
- the current through the circuit, and
- the potential difference across the electric lamp and conductor.
Fig.9 An electric lamp connected in series with a resistor of 4 Ω to a 6 V battery
The resistance of electric lamp, R_{1} = 20 Ω,
The resistance of the conductor connected in series, R_{2} = 4 Ω.
Then the total resistance in the circuit
R = R_{1} + R_{2}
R_{s} = 20 Ω + 4 Ω = 24 Ω.
The total potential difference across the two terminals of the battery
V = 6 V.
Now by Ohm’s law, the current through the circuit is given by
I = V/R_{s}
= 6 V/24 Ω
= 0.25 A.
Applying Ohm’s law to the electric lamp and conductor separately,we get potential difference across the electric lamp,
V_{1} = 20 Ω × 0.25 A
= 5 V;
and, that across the conductor, V_{2} = 4 Ω × 0.25 A
= 1 V.
Suppose that we like to replace the series combination of electric lamp and conductor by a single and equivalent resistor. Its resistance must be such that a potential difference of 6 V across the battery terminals will cause a current of 0.25 A in the circuit. The resistance
R of this equivalent resistor would be
R = V/I
= 6 V/ 0.25 A
= 24 Ω.
This is the total resistance of the series circuit; it is equal to the sum of the two resistances.
- Draw a schematic diagram of a circuit consisting of a battery of three cells of 2 V each, a 5 Ω resistor, an 8 Ω resistor, and a 12 Ω resistor, and a plug key, all connected in series.
- Redraw the circuit of Question 1, putting in an ammeter to measure the current through the resistors and a voltmeter to measure the potential difference across the 12 Ω resistor. What would be the readings in the ammeter and the voltmeter?
12.6.2 Resistors in Parallel
Now, let us consider the arrangement of three resistors joined in parallel with a combination of cells (or a battery), as shown in Fig.7.
It is observed that the total current I, is equal to the sum of the separate currents through each branch of the combination.
Let R_{p} be the equivalent resistance of the parallel combination of resistors. By applying Ohm’s law to the parallel combination of resistors, we have
On applying Ohm’s law to each resistor, we have
From Eqs. (15) to (17), we have
V/R_{p} = V/R_{1} + V/R_{2} + V/R_{3}
Or
Thus, we may conclude that the reciprocal of the equivalent resistance of a group of resistances joined in parallel is equal to the sum of the reciprocals of the individual resistances.
In the circuit diagram given in Fig.10, suppose the resistors R_{1}, R_{2} and R_{3} have the values 5 Ω, 10 Ω, 30 Ω, respectively, which have been connected to a battery of 12 V. Calculate
- the current through each resistor,
- the total current in the circuit, and
- the total circuit resistance.
R_{1} = 5 Ω, R_{2} = 10 Ω, and R_{3} = 30 Ω.
Potential difference across the battery, V = 12 V.
This is also the potential difference across each of the individual
resistor; therefore, to calculate the current in the resistors, we use
Ohm’s law.
The current I_{1}, through R_{1} = V/ R_{1}
I_{1} = 12 V/5 Ω = 2.4 A.
The current I_{2}, through R_{2} = V/ R_{2}
I_{2} = 12 V/10 Ω = 1.2 A.
The current I_{3}, through R_{3} = V/R_{3}
I_{3} = 12 V/30 Ω = 0.4 A.
The total current in the circuit,
I = I_{1} + I_{2} + I_{3}
= (2.4 + 1.2 + 0.4) A
= 4 A
The total resistance Rp, is given by [Eq. (18)]
1/R_{p} = 1/5 + 1/10 + 1/30 = 1/3
Thus, R_{p} = 3 Ω.
If in Fig.12, R_{1} = 10 Ω, R_{2} = 40 Ω, R_{3} = 30 Ω, R_{4} = 20 Ω, R_{5} = 60 Ω, and a 12 V battery is connected to the arrangement. Calculate
- the total resistance in the circuit, and
- the total current flowing in the circuit.
Fig.12 An electric circuit showing the combination of series and parallel resistors
Suppose we replace the parallel resistors R_{1} and R_{2} by an equivalent resistor of resistance, R′. Similarly we replace the parallel resistors R_{3}, R_{4} and R_{5} by an equivalent single resistor of resistance R″. Then using Eq. (18), we have
1/ R′ = 1/10 + 1/40 = 5/40; that is R′ = 8 Ω.
Similarly, 1/ R″ = 1/30 + 1/20 + 1/60 = 6/60;
that is, R″ = 10 Ω.
Thus, the total resistance, R = R′ + R″ = 18 Ω.
To calculate the current, we use Ohm’s law, and get
I = V/R = 12 V/18 Ω = 0.67 A.
We have seen that in a series circuit the current is constant throughout the electric circuit. Thus it is obviously impracticable to connect an electric bulb and an electric heater in series, because they need currents of widely different values to operate properly (see Example.3). Another major disadvantage of a series circuit is that when one component fails the circuit is broken and none of the components works. If you have used ‘fairy lights’ to decorate buildings on festivals, on marriage celebrations etc., you might have seen the electrician spending lot of time in trouble-locating and replacing the ‘dead’ bulb – each has to be tested to find which has fused or gone. On the other hand, a parallel circuit divides the current through the electrical gadgets. The total resistance in a parallel circuit is decreased as per Eq. (18). This is helpful particularly when each gadget has different resistance and requires different current to operate properly.
- Judge the equivalent resistance when the following are connected in parallel – (a) 1 Ω and 10^{6} Ω, (b) 1 Ω and 10^{3} Ω, and 10^{6} Ω.
- An electric lamp of 100 Ω, a toaster of resistance 50 Ω, and a water filter of resistance 500 Ω are connected in parallel to a 220 V source. What is the resistance of an electric iron connected to the same source that takes as much current as all three appliances, and what is the current through it?
- What are the advantages of connecting electrical devices in parallel with the battery instead of connecting them in series?
- How can three resistors of resistances 2 Ω, 3 Ω, and 6 Ω be connected to give a total resistance of (a) 4 Ω, (b) 1 Ω?
- What is (a) the highest, (b) the lowest total resistance that can be secured by combinations of four coils of resistance 4 Ω, 8 Ω, 12 Ω, 24 Ω?
12.7 HEATING EFFECT OF ELECTRIC CURRENT
We know that a battery or a cell is a source of electrical energy. The chemical reaction within the cell generates the potential difference between its two terminals that sets the electrons in motion to flow the current through a resistor or a system of resistors connected to the battery. We have also seen, in Section 2, that to maintain the current, the source has to keep expending its energy. Where does this energy go? A part of the source energy in maintaining the current may be consumed into useful work (like in rotating the blades of an electric fan). Rest of the source energy may be expended in heat to raise the temperature of gadget. We often observe this in our everyday life. For example, an electric fan becomes warm if used continuously for longer time etc. On the other hand, if the electric circuit is purely resistive, that is, a configuration of resistors only connected to a battery; the source energy continually gets dissipated entirely in the form of heat. This is known as the heating effect of electric current. This effect is utilised in devices such as electric heater, electric iron etc.
Consider a current I flowing through a resistor of resistance R. Let the potential difference across it be V (Fig. 13). Let t be the time during which a charge Q flows across. The work done in moving the charge Q through a potential difference V is VQ. Therefore, the source must supply energy equal to VQ in time t. Hence the power input to the circuit by the source is
Or the energy supplied to the circuit by the source in time t is P × t, that is, VIt. What happens to this energy expended by the source? This energy gets dissipated in the resistor as heat. Thus for a steady current I, the amount of heat H produced in time t is
Applying Ohm’s law [Eq. (5)], we get
This is known as Joule’s law of heating. The law implies that heat produced in a resistor is (i) directly proportional to the square of current for a given resistance, (ii) directly proportional to resistance for a given current, and (iii) directly proportional to the time for which the current flows through the resistor. In practical situations, when an electric appliance is connected to a known voltage source, Eq. (21) is used after calculating the current through it, using the relation I = V/R.
Fig.13 A steady current in a purely resistive electric circuit
An electric iron consumes energy at a rate of 840 W when heating is at the maximum rate and 360 W when the heating is at the minimum. The voltage is 220 V. What are the current and the resistance in each case?
From Eq. (19), we know that the power input is
P = V I
Thus the current I = P/V
(a) When heating is at the maximum rate,
I = 840 W/220 V = 3.82 A;
and the resistance of the electric iron is
R = V/I = 220 V/3.82 A = 57.60 Ω.
(b) When heating is at the minimum rate,
I = 360 W/220 V = 1.64 A;
and the resistance of the electric iron is
R = V/I = 220 V/1.64 A = 134.15 Ω.
100 J of heat are produced each second in a 4 Ω resistance. Find the potential difference across the resistor.
H = 100 J, R = 4 Ω, t = 1 s, V = ?
From Eq. (21) we have the current through the resistor as
I=√(H/Rt)
=√[100 J/(4 Ω × 1 s)]
=5A
Thus the potential difference across the resistor, V [from Eq. (5)] is
V = IR
= 5A × 4 Ω
= 20 V.
- Why does the cord of an electric heater not glow while the heating element does?
- Compute the heat generated while transferring 96000 coulomb of charge in one hour through a potential difference of 50 V.
- An electric iron of resistance 20 Ω takes a current of 5 A. Calculate the heat developed in 30 s.
12.7.1 Practical Applications of Heating Effect of Electric Current
The generation of heat in a conductor is an inevitable consequence of electric current. In many cases, it is undesirable as it converts useful electrical energy into heat. In electric circuits, the unavoidable heating can increase the temperature of the components and alter their properties. However, heating effect of electric current has many useful applications. The electric laundry iron, electric toaster, electric oven, electric kettle and electric heater are some of the familiar devices based on Joule’s heating.
The electric heating is also used to produce light, as in an electric bulb. Here, the filament must retain as much of the heat generated as is possible, so that it gets very hot and emits light. It must not melt at such high temperature. A strong metal with high melting point such as tungsten (melting point 3380°C) is used for making bulb filaments. The filament should be thermally isolated as much as possible, using insulating support, etc. The bulbs are usually filled with chemically inactive nitrogen and argon gases to prolong the life of filament. Most of the power consumed by the filament appears as heat, but a small part of it is in the form of light radiated.
Another common application of Joule’s heating is the fuse used in electric circuits. It protects circuits and appliances by stopping the flow of any unduly high electric current. The fuse is placed in series with the device. It consists of a piece of wire made of a metal or an alloy of appropriate melting point, for example aluminium, copper, iron, lead etc. If a current larger than the specified value flows through the circuit, the temperature of the fuse wire increases. This melts the fuse wire and breaks the circuit. The fuse wire is usually encased in a cartridge of porcelain or similar material with metal ends. The fuses used for domestic purposes are rated as 1 A, 2 A, 3 A, 5 A, 10 A, etc. For an electric iron which consumes 1 kW electric power when operated at 220 V, a current of (1000/220) A, that is, 4.54 A will flow in the circuit. In this case, a 5 A fuse must be used.
12.8 ELECTRIC POWER
You have studied in your earlier Class that the rate of doing work is power. This is also the rate of consumption of energy.
Equation (21) gives the rate at which electric energy is dissipated or consumed in an electric circuit. This is also termed as electric power.
The power P is given by
P = VI
The SI unit of electric power is watt (W). It is the power consumed by a device that carries 1 A of current when operated at a potential difference of 1 V. Thus,
The unit ‘watt’ is very small. Therefore, in actual practice we use a much larger unit called ‘kilowatt’. It is equal to 1000 watts. Since electrical energy is the product of power and time, the unit of electric energy is, therefore, watt hour (W h). One watt hour is the energy consumed when 1 watt of power is used for 1 hour. The commercial unit of electric energy is kilowatt hour (kW h), commonly known as ‘unit’.
1 kW h = 1000 watt × 3600 second
= 3.6 × 10^{6} watt second
= 3.6 × 10^{6} joule (J)
An electric bulb is connected to a 220 V generator. The current is 0.50 A. What is the power of the bulb?
P = VI
= 220 V × 0.50 A
= 110 J/s
= 110 W.
An electric refrigerator rated 400 W operates 8 hour/day. What is the cost of the energy to operate it for 30 days at Rs 3.00 per kW h?
The total energy consumed by the refrigerator in 30 days would be
400 W × 8.0 hour/day × 30 days = 96000 W h
= 96 kW h
Thus the cost of energy to operate the refrigerator for 30 days is
96 kW h × Rs 3.00 per kW h = Rs 288.00
- What determines the rate at which energy is delivered by a current?
- An electric motor takes 5 A from a 220 V line. Determine the power of the motor and the energy consumed in 2 h.