Welcome dear students! Today we are going to learn about Electricity from Class 10 Science. 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.
[CHECKPOINT]
Let us begin with section eleven point one, Electric Current and Circuit. We are familiar with air current and water current. We know that flowing water constitutes 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.
[CHECKPOINT]
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 I equals Q divided by t. This is equation eleven point one. The SI unit of electric charge is coulomb, which is equivalent to the charge contained in nearly six times ten to the power eighteen electrons. We know that an electron possesses a negative charge of one point six times ten to the power minus nineteen coulomb. The electric current is expressed by a unit called ampere, named after the French scientist, Andre-Marie Ampere. One ampere is constituted by the flow of one coulomb of charge per second, that is, one ampere equals one coulomb per one second. Small quantities of current are expressed in milliampere, where one milliampere equals ten to the power minus three ampere, or in microampere, where one microampere equals ten to the power minus six ampere. 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 eleven point one shows the schematic diagram of a typical electric circuit comprising a cell, an electric bulb, an ammeter and a plug key. In this diagram, we can see the setup with the cell connected to the bulb, ammeter, and plug key in a single loop. 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.
[CHECKPOINT]
Let us solve Example eleven point one. A current of zero point five ampere is drawn by a filament of an electric bulb for ten minutes. Find the amount of electric charge that flows through the circuit. Solution: We are given, I equals zero point five ampere, and t equals ten minutes, which is six hundred seconds. From equation eleven point one, we have Q equals I multiplied by t. So Q equals zero point five ampere multiplied by six hundred seconds, which equals three hundred coulomb. Now, let us look at the questions following this section. Question one: What does an electric circuit mean? Question two: Define the unit of current. Question three: Calculate the number of electrons constituting one coulomb of charge. Please attempt these questions using the definitions and values provided in this section.
[CHECKPOINT]
Moving on to section eleven point two, 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 equals Work done W divided by Charge Q. V equals W divided by Q. This is equation eleven point two.
[CHECKPOINT]
The SI unit of electric potential difference is volt, named after Alessandro Volta, an Italian physicist. One volt is the potential difference between two points in a current carrying conductor when one joule of work is done to move a charge of one coulomb from one point to the other. Therefore, one volt equals one joule divided by one coulomb. One volt equals one joule per coulomb. 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. Let us solve Example eleven point two. How much work is done in moving a charge of two coulomb across two points having a potential difference twelve volt? Solution: The amount of charge Q that flows between two points at potential difference V equals twelve volt is two coulomb. Thus, the amount of work W done in moving the charge from equation eleven point two is W equals V multiplied by Q, which equals twelve volt multiplied by two coulomb, equals twenty-four joule. Now, the questions for this section. Question one: Name a device that helps to maintain a potential difference across a conductor. Question two: What is meant by saying that the potential difference between two points is one volt? Question three: How much energy is given to each coulomb of charge passing through a six volt battery? Please work these out based on the concepts discussed.
[CHECKPOINT]
Next, we will learn about section eleven point three, Circuit Diagram. We know that an electric circuit comprises a cell or a battery, a plug key, electrical components, 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. Let me describe the conventional symbols used to represent some of the most commonly used electrical components as given in Table eleven point one. First, an electric cell is represented by a long line and a short parallel line. Second, a battery or a combination of cells is shown by multiple long and short parallel lines. Third, a plug key or switch in the open position is drawn as a break in the line with a diagonal gap. Fourth, a plug key or switch in the closed position is a continuous straight line. Fifth, a wire joint is a solid dot where lines meet. Sixth, wires crossing without joining are shown as one line crossing over another without a dot. Seventh, an electric bulb is a circle with a cross inside, or a looped filament symbol. Eighth, a resistor of resistance R is a zigzag line. Ninth, a variable resistance or rheostat is a zigzag line with an arrow pointing to it. Tenth, an ammeter is a circle with the letter A inside. Eleventh, a voltmeter is a circle with the letter V inside. These symbols help us draw neat and standardized circuit diagrams.
[CHECKPOINT]
Let us proceed to section eleven point four, Ohm's Law. Is there a relationship between the potential difference across a conductor and the current through it? Let us explore with Activity eleven point one. Set up a circuit as shown in Figure eleven point two, consisting of a nichrome wire XY of length, say zero point five meter, an ammeter, a voltmeter and four cells of one point five volt each. Nichrome is an alloy of nickel, chromium, manganese, and iron metals. First use only one cell as the source in the circuit. Note the reading in the ammeter I for the current and reading of the voltmeter V for the potential difference across the nichrome wire XY in the circuit. Tabulate them. Next connect two cells in the circuit and note the respective readings of the ammeter and voltmeter for the values of current through the nichrome wire and potential difference across the nichrome wire. Repeat the above steps using three cells and then four cells in the circuit separately. Calculate the ratio of V to I for each pair of potential difference V and current I. Plot a graph between V and I, and observe the nature of the graph. In this Activity, you will find that approximately the same value for V divided by I is obtained in each case. Thus the V minus I graph is a straight line that passes through the origin of the graph. Thus, V divided by I is a constant ratio.
[CHECKPOINT]
In eighteen twenty seven, a German physicist Georg Simon Ohm found out the relationship between the current I flowing in a metallic wire and the potential difference across its terminals. The potential difference, V, across the ends of a given metallic wire in an electric circuit is directly proportional to the current flowing through it, provided its temperature remains the same. This is called Ohm's law. In other words, V is directly proportional to I, or V divided by I equals constant equals R, or V equals I R. This is equation eleven point four and eleven point five. In equation eleven point four, 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 capital omega. According to Ohm's law, R equals V divided by I. This is equation eleven point six. If the potential difference across the two ends of a conductor is one volt and the current through it is one ampere, then the resistance R of the conductor is one ohm. That is, one ohm equals one volt divided by one ampere. Also from equation eleven point five we get I equals V divided by R. This is equation eleven point seven. It is obvious from equation eleven point seven 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.
[CHECKPOINT]
We will now study about electrical resistance of a conductor with the help of Activity eleven point two. Take a nichrome wire, a torch bulb, a ten watt bulb and an ammeter with zero to five ampere range, a plug key and some connecting wires. Set up the circuit by connecting four dry cells of one point five volt each in series with the ammeter leaving a gap XY in the circuit, as shown in Figure eleven point four. In this diagram, the battery, ammeter, and plug key are connected in series, with a gap labeled XY where we will place different components. Complete the circuit by connecting the nichrome wire in the gap XY. Plug the key. Note down the ammeter reading. Take out the key from the plug. Note: Always take out the key from the plug after measuring the current through the circuit. Replace the nichrome wire with the torch bulb in the circuit and find the current through it by measuring the reading of the ammeter. Now repeat the above step with the ten watt bulb in the gap XY. Are the ammeter readings different for different components connected in the gap XY? What do the above observations indicate? You may repeat this Activity by keeping any material component in the gap. Observe the ammeter readings in each case. Analyse the observations. 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.
[CHECKPOINT]
Moving to section eleven point five, Factors on which the Resistance of a Conductor Depends. Let us perform Activity eleven point three. Complete an electric circuit consisting of a cell, an ammeter, a nichrome wire of length l, say marked one, and a plug key, as shown in Figure eleven point five. In this setup, the cell, ammeter, and nichrome wire are in series. Now, plug the key. Note the current in the ammeter. Replace the nichrome wire by another nichrome wire of same thickness but twice the length, that is two l, marked two in the figure. Note the ammeter reading. Now replace the wire by a thicker nichrome wire, of the same length l, marked three. A thicker wire has a larger cross-sectional area. Again note down the current through the circuit. Instead of taking a nichrome wire, connect a copper wire, marked four in the figure, in the circuit. Let the wire be of the same length and same area of cross-section as that of the first nichrome wire, marked one. Note the value of the current. Notice the difference in the current in all cases. Does the current depend on the length of the conductor? Does the current depend on the area of cross-section of the wire used? 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, we observe that the resistance of the conductor depends on its length, on its area of cross-section, and on the nature of its material.
[CHECKPOINT]
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, R is directly proportional to l, and R is directly proportional to one divided by A. Combining these, we get R is directly proportional to l divided by A, or R equals rho l divided by A. This is equation eleven point eight, eleven point nine, and eleven point ten. Here, rho is a constant of proportionality and is called the electrical resistivity of the material of the conductor. The SI unit of resistivity is ohm meter. It is a characteristic property of the material. The metals and alloys have very low resistivity in the range of ten to the power minus eight ohm meter to ten to the power minus six ohm meter. They are good conductors of electricity. Insulators like rubber and glass have resistivity of the order of ten to the power twelve to ten to the power seventeen ohm meter. Both the resistance and resistivity of a material vary with temperature. Table eleven point two reveals that the resistivity of an alloy is generally higher than that of its constituent metals. Alloys do not oxidise 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. Let me read the values from Table eleven point two for electrical resistivity of some substances at twenty degrees Celsius. For conductors: Silver is one point six zero times ten to the power minus eight ohm meter. Copper is one point six two times ten to the power minus eight ohm meter. Aluminium is two point six three times ten to the power minus eight ohm meter. Tungsten is five point two zero times ten to the power minus eight ohm meter. Nickel is six point eight four times ten to the power minus eight ohm meter. Iron is ten point zero times ten to the power minus eight ohm meter. Chromium is twelve point nine times ten to the power minus eight ohm meter. Mercury is ninety-four point zero times ten to the power minus eight ohm meter. For alloys: Manganese is one point eight four times ten to the power minus six ohm meter. Constantan, an alloy of copper and nickel, is forty-nine times ten to the power minus six ohm meter. Manganin, an alloy of copper, manganese and nickel, is forty-four times ten to the power minus six ohm meter. Nichrome, an alloy of nickel, chromium, manganese and iron, is one hundred times ten to the power minus six ohm meter. For insulators: Glass is ten to the power ten to ten to the power fourteen ohm meter. Hard rubber is ten to the power thirteen to ten to the power sixteen ohm meter. Ebonite is ten to the power fifteen to ten to the power seventeen ohm meter. Diamond is ten to the power twelve to ten to the power thirteen ohm meter. Paper dry is ten to the power twelve ohm meter. You need not memorise these values. You can use these values for solving numerical problems.
[CHECKPOINT]
Let us solve Example eleven point three. Part a: How much current will an electric bulb draw from a two hundred twenty volt source, if the resistance of the bulb filament is one thousand two hundred ohm? Part b: How much current will an electric heater coil draw from a two hundred twenty volt source, if the resistance of the heater coil is one hundred ohm? Solution for part a: We are given V equals two hundred twenty volt, R equals one thousand two hundred ohm. From Ohm's law equation, we have the current I equals two hundred twenty volt divided by one thousand two hundred ohm, which equals zero point one eight ampere. Solution for part b: We are given V equals two hundred twenty volt, R equals one hundred ohm. From Ohm's law equation, we have the current I equals two hundred twenty volt divided by one hundred ohm, which equals two point two ampere. Note the difference of current drawn by an electric bulb and electric heater from the same two hundred twenty volt source! Example eleven point four: The potential difference between the terminals of an electric heater is sixty volt when it draws a current of four ampere from the source. What current will the heater draw if the potential difference is increased to one hundred twenty volt? Solution: We are given potential difference V equals sixty volt, current I equals four ampere. According to Ohm's law, R equals V divided by I equals sixty volt divided by four ampere, which equals fifteen ohm. When the potential difference is increased to one hundred twenty volt the current is given by current equals V divided by R equals one hundred twenty volt divided by fifteen ohm, which equals eight ampere. The current through the heater becomes eight ampere.
[CHECKPOINT]
Example eleven point five: Resistance of a metal wire of length one meter is twenty-six ohm at twenty degrees Celsius. If the diameter of the wire is zero point three millimeter, what will be the resistivity of the metal at that temperature? Using Table eleven point two, predict the material of the wire. Solution: We are given the resistance R of the wire equals twenty-six ohm, the diameter d equals zero point three millimeter, which is three times ten to the power minus four meter, and the length l of the wire equals one meter. Therefore, from equation eleven point ten, the resistivity of the given metallic wire is rho equals R multiplied by A divided by l. Since A equals pi d squared divided by four, rho equals R multiplied by pi d squared divided by four l. Substitution of values in this gives rho equals one point eight four times ten to the power minus six ohm meter. The resistivity of the metal at twenty degrees Celsius is one point eight four times ten to the power minus six ohm meter. From Table eleven point two, we see that this is the resistivity of manganese. Example eleven point six: A wire of given material having length l and area of cross-section A has a resistance of four ohm. What would be the resistance of another wire of the same material having length l divided by two and area of cross-section two A? Solution: For first wire, R one equals rho l divided by A, which equals four ohm. Now for second wire, R two equals rho multiplied by l divided by two, divided by two A, which equals one fourth rho l divided by A. So R two equals one fourth R one, which equals one ohm. The resistance of the new wire is one ohm.
[CHECKPOINT]
Now let us look at the questions for this section. Question one: On what factors does the resistance of a conductor depend? Question two: Will current flow more easily through a thick wire or a thin wire of the same material, when connected to the same source? Why? Question three: 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? Question four: Why are coils of electric toasters and electric irons made of an alloy rather than a pure metal? Question five: Use the data in Table eleven point two to answer the following. Part a: Which among iron and mercury is a better conductor? Part b: Which material is the best conductor? Please attempt these using the principles of resistance and resistivity discussed above.
[CHECKPOINT]
Moving to section eleven point six, 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 eleven point six shows an electric circuit in which three resistors having resistances R one, R two and R three, respectively, are joined end to end. Here the resistors are said to be connected in series. Figure eleven point seven 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. Let us study section eleven point six point one, 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 Activity eleven point four. Join three resistors of different values in series. Connect them with a battery, an ammeter and a plug key, as shown in Figure eleven point six. You may use the resistors of values like one ohm, two ohm, three ohm etc., and a battery of six volt for performing this Activity. Plug the key. Note the ammeter reading. Change the position of ammeter to anywhere in between the resistors. Note the ammeter reading each time. Do you find any change in the value of current through the ammeter? 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.
[CHECKPOINT]
Now Activity eleven point five. In Activity eleven point four, insert a voltmeter across the ends X and Y of the series combination of three resistors, as shown in Figure eleven point six. Plug the key in the circuit and note the voltmeter reading. It gives the potential difference across the series combination of resistors. Let it be V. Now measure the potential difference across the two terminals of the battery. Compare the two values. Take out the plug key and disconnect the voltmeter. Now insert the voltmeter across the ends X and P of the first resistor, as shown in Figure eleven point eight. In this diagram, the voltmeter is connected in parallel across the first resistor only. Plug the key and measure the potential difference across the first resistor. Let it be V one. Similarly, measure the potential difference across the other two resistors, separately. Let these values be V two and V three, respectively. Deduce a relationship between V, V one, V two and V three. You will observe that the potential difference V is equal to the sum of potential differences V one, V two, and V three. 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, V equals V one plus V two plus V three. This is equation eleven point eleven. In the electric circuit shown in Figure eleven point eight, 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 Ohm's law to the entire circuit, we have V equals I R. This is equation eleven point twelve.
[CHECKPOINT]
On applying Ohm's law to the three resistors separately, we further have V one equals I R one, V two equals I R two, and V three equals I R three. From equation eleven point eleven, I R equals I R one plus I R two plus I R three, or R sub s equals R one plus R two plus R three. This is equation eleven point fourteen. We can conclude that when several resistors are joined in series, the resistance of the combination R sub s equals the sum of their individual resistances, R one, R two, R three, and is thus greater than any individual resistance. Let us solve Example eleven point seven. An electric lamp, whose resistance is twenty ohm, and a conductor of four ohm resistance are connected to a six volt battery. Calculate a) the total resistance of the circuit, b) the current through the circuit, and c) the potential difference across the electric lamp and conductor. Solution: The resistance of electric lamp, R one equals twenty ohm. The resistance of the conductor connected in series, R two equals four ohm. Then the total resistance in the circuit R sub s equals R one plus R two equals twenty ohm plus four ohm equals twenty-four ohm. The total potential difference across the two terminals of the battery V equals six volt. Now by Ohm's law, the current through the circuit is given by I equals V divided by R sub s equals six volt divided by twenty-four ohm equals zero point two five ampere. Applying Ohm's law to the electric lamp and conductor separately, we get potential difference across the electric lamp, V one equals twenty ohm multiplied by zero point two five ampere equals five volt, and that across the conductor, V two equals four ohm multiplied by zero point two five ampere equals one volt. 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 six volt across the battery terminals will cause a current of zero point two five ampere in the circuit. The resistance R of this equivalent resistor would be R equals V divided by I equals six volt divided by zero point two five ampere equals twenty-four ohm. This is the total resistance of the series circuit; it is equal to the sum of the two resistances.
[CHECKPOINT]
Let us look at the questions for this section. Question one: Draw a schematic diagram of a circuit consisting of a battery of three cells of two volt each, a five ohm resistor, an eight ohm resistor, and a twelve ohm resistor, and a plug key, all connected in series. Question two: Redraw the circuit of Question one, putting in an ammeter to measure the current through the resistors and a voltmeter to measure the potential difference across the twelve ohm resistor. What would be the readings in the ammeter and the voltmeter? Please practice drawing these diagrams and calculating the values using the series circuit rules.
[CHECKPOINT]
Now we move to section eleven point six point two, Resistors in Parallel. Let us consider the arrangement of three resistors joined in parallel with a combination of cells, as shown in Figure eleven point seven. Activity eleven point six: Make a parallel combination, XY, of three resistors having resistances R one, R two, and R three, respectively. Connect it with a battery, a plug key and an ammeter, as shown in Figure eleven point ten. In this diagram, the three resistors are connected side by side between two common points X and Y, and this combination is connected in series with the battery, plug key, and ammeter. Also connect a voltmeter in parallel with the combination of resistors. Plug the key and note the ammeter reading. Let the current be I. Also take the voltmeter reading. It gives the potential difference V, across the combination. The potential difference across each resistor is also V. This can be checked by connecting the voltmeter across each individual resistor, as shown in Figure eleven point eleven. Take out the plug from the key. Remove the ammeter and voltmeter from the circuit. Insert the ammeter in series with the resistor R one, as shown in Figure eleven point eleven. Note the ammeter reading, I one. Similarly, measure the currents through R two and R three. Let these be I two and I three, respectively. What is the relationship between I, I one, I two and I three? It is observed that the total current I is equal to the sum of the separate currents through each branch of the combination. I equals I one plus I two plus I three. This is equation eleven point fifteen.
[CHECKPOINT]
Let R sub p be the equivalent resistance of the parallel combination of resistors. By applying Ohm's law to the parallel combination of resistors, we have I equals V divided by R sub p. This is equation eleven point sixteen. On applying Ohm's law to each resistor, we have I one equals V divided by R one, I two equals V divided by R two, and I three equals V divided by R three. This is equation eleven point seventeen. From equations eleven point fifteen to eleven point seventeen, we have V divided by R sub p equals V divided by R one plus V divided by R two plus V divided by R three, or one divided by R sub p equals one divided by R one plus one divided by R two plus one divided by R three. This is equation eleven point eighteen. 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. Let us solve Example eleven point eight. In the circuit diagram given in Figure eleven point ten, suppose the resistors R one, R two and R three have the values five ohm, ten ohm, thirty ohm, respectively, which have been connected to a battery of twelve volt. Calculate a) the current through each resistor, b) the total current in the circuit, and c) the total circuit resistance. Solution: R one equals five ohm, R two equals ten ohm, and R three equals thirty ohm. Potential difference across the battery, V equals twelve volt. 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 one through R one equals twelve volt divided by five ohm equals two point four ampere. The current I two through R two equals twelve volt divided by ten ohm equals one point two ampere. The current I three through R three equals twelve volt divided by thirty ohm equals zero point four ampere. The total current in the circuit, I equals I one plus I two plus I three equals two point four plus one point two plus zero point four ampere, which equals four ampere. The total resistance R sub p is given by equation eleven point eighteen: one divided by R sub p equals one divided by five plus one divided by ten plus one divided by thirty, which equals one third. Thus, R sub p equals three ohm.
[CHECKPOINT]
Example eleven point nine: If in Figure eleven point twelve, R one equals ten ohm, R two equals forty ohm, R three equals thirty ohm, R four equals twenty ohm, R five equals sixty ohm, and a twelve volt battery is connected to the arrangement. Calculate a) the total resistance in the circuit, and b) the total current flowing in the circuit. Solution: Suppose we replace the parallel resistors R one and R two by an equivalent resistor of resistance R prime. Similarly we replace the parallel resistors R three, R four and R five by an equivalent single resistor of resistance R double prime. Then using equation eleven point eighteen, we have one divided by R prime equals one divided by ten plus one divided by forty equals five divided by forty, that is R prime equals eight ohm. Similarly, one divided by R double prime equals one divided by thirty plus one divided by twenty plus one divided by sixty equals six divided by sixty, that is R double prime equals ten ohm. Thus, the total resistance, R equals R prime plus R double prime equals eighteen ohm. To calculate the current, we use Ohm's law, and get I equals V divided by R equals twelve volt divided by eighteen ohm equals zero point six seven ampere. 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. 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 equation eleven point eighteen. This is helpful particularly when each gadget has different resistance and requires different current to operate properly.
[CHECKPOINT]
Let us look at the questions for this section. Question one: Judge the equivalent resistance when the following are connected in parallel: a) one ohm and ten to the power six ohm, b) one ohm and ten to the power three ohm, and ten to the power six ohm. Question two: An electric lamp of one hundred ohm, a toaster of resistance fifty ohm, and a water filter of resistance five hundred ohm are connected in parallel to a two hundred twenty volt 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? Question three: What are the advantages of connecting electrical devices in parallel with the battery instead of connecting them in series? Question four: How can three resistors of resistances two ohm, three ohm, and six ohm be connected to give a total resistance of a) four ohm, b) one ohm? Question five: What is a) the highest, b) the lowest total resistance that can be secured by combinations of four coils of resistance four ohm, eight ohm, twelve ohm, twenty-four ohm? Please solve these to strengthen your understanding of parallel and series combinations.
[CHECKPOINT]
Now let us move to section eleven point seven, 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 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, as shown in Figure eleven point thirteen. 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 V Q. Therefore, the source must supply energy equal to V Q in time t. Hence the power input to the circuit by the source is P equals V multiplied by Q divided by t, which equals V I. This is equation eleven point nineteen. Or the energy supplied to the circuit by the source in time t is P multiplied by t, that is, V I t. 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 H equals V I t. This is equation eleven point twenty.
[CHECKPOINT]
Applying Ohm's law, we get H equals I squared R t. This is equation eleven point twenty-one. This is known as Joule's law of heating. The law implies that heat produced in a resistor is directly proportional to the square of current for a given resistance, directly proportional to resistance for a given current, and 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, equation eleven point twenty-one is used after calculating the current through it, using the relation I equals V divided by R. Let us solve Example eleven point ten. An electric iron consumes energy at a rate of eight hundred forty watt when heating is at the maximum rate and three hundred sixty watt when the heating is at the minimum. The voltage is two hundred twenty volt. What are the current and the resistance in each case? Solution: From equation eleven point nineteen, we know that the power input is P equals V I. Thus the current I equals P divided by V. When heating is at the maximum rate, I equals eight hundred forty watt divided by two hundred twenty volt equals three point eight two ampere, and the resistance of the electric iron is R equals V divided by I equals two hundred twenty volt divided by three point eight two ampere equals fifty-seven point six zero ohm. When heating is at the minimum rate, I equals three hundred sixty watt divided by two hundred twenty volt equals one point six four ampere, and the resistance of the electric iron is R equals two hundred twenty volt divided by one point six four ampere equals one hundred thirty-four point one five ohm.
[CHECKPOINT]
Example eleven point eleven: One hundred joule of heat is produced each second in a four ohm resistance. Find the potential difference across the resistor. Solution: H equals one hundred joule, R equals four ohm, t equals one second, V equals question mark. From equation eleven point twenty-one we have the current through the resistor as I equals square root of H divided by R t equals square root of one hundred joule divided by four ohm multiplied by one second equals five ampere. Thus the potential difference across the resistor, V from equation eleven point five is V equals I R equals five ampere multiplied by four ohm equals twenty volt. Let us look at the questions for this section. Question one: Why does the cord of an electric heater not glow while the heating element does? Question two: Compute the heat generated while transferring ninety-six thousand coulomb of charge in one hour through a potential difference of fifty volt. Question three: An electric iron of resistance twenty ohm takes a current of five ampere. Calculate the heat developed in thirty seconds. Please attempt these numerical and conceptual questions.
[CHECKPOINT]
Now section eleven point seven point one, 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 three thousand three hundred eighty degrees Celsius, 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 one ampere, two ampere, three ampere, five ampere, ten ampere, etc. For an electric iron which consumes one kilowatt electric power when operated at two hundred twenty volt, a current of one thousand divided by two hundred twenty ampere, that is four point five four ampere will flow in the circuit. In this case, a five ampere fuse must be used.
[CHECKPOINT]
Let us proceed to section eleven point eight, 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 eleven point twenty-one 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 equals V I, or P equals I squared R equals V squared divided by R. This is equation eleven point twenty-two. The SI unit of electric power is watt. It is the power consumed by a device that carries one ampere of current when operated at a potential difference of one volt. Thus, one watt equals one volt multiplied by one ampere. The unit watt is very small. Therefore, in actual practice we use a much larger unit called kilowatt. It is equal to one thousand watts. Since electrical energy is the product of power and time, the unit of electric energy is, therefore, watt hour. One watt hour is the energy consumed when one watt of power is used for one hour. The commercial unit of electric energy is kilowatt hour, commonly known as unit. One kilowatt hour equals one thousand watt multiplied by three thousand six hundred second equals three point six times ten to the power six watt second equals three point six times ten to the power six joule. More to Know! Many people think that electrons are consumed in an electric circuit. This is wrong! We pay the electricity board or electric company to provide energy to move electrons through the electric gadgets like electric bulb, fan and engines. We pay for the energy that we use.
[CHECKPOINT]
Let us solve Example eleven point twelve. An electric bulb is connected to a two hundred twenty volt generator. The current is zero point five zero ampere. What is the power of the bulb? Solution: P equals V I equals two hundred twenty volt multiplied by zero point five zero ampere equals one hundred ten joule per second equals one hundred ten watt. Example eleven point thirteen: An electric refrigerator rated four hundred watt operates eight hour per day. What is the cost of the energy to operate it for thirty days at three rupees per kilowatt hour? Solution: The total energy consumed by the refrigerator in thirty days would be four hundred watt multiplied by eight point zero hour per day multiplied by thirty days equals ninety-six thousand watt hour equals ninety-six kilowatt hour. Thus the cost of energy to operate the refrigerator for thirty days is ninety-six kilowatt hour multiplied by three rupees per kilowatt hour equals two hundred eighty-eight rupees. Let us look at the questions for this section. Question one: What determines the rate at which energy is delivered by a current? Question two: An electric motor takes five ampere from a two hundred twenty volt line. Determine the power of the motor and the energy consumed in two hours. Please solve these to apply the power and energy formulas.
[CHECKPOINT]
Let us quickly review what you have learnt. A stream of electrons moving through a conductor constitutes an electric current. Conventionally, the direction of current is taken opposite to the direction of flow of electrons. The SI unit of electric current is ampere. To set the electrons in motion in an electric circuit, we use a cell or a battery. A cell generates a potential difference across its terminals. It is measured in volts. Resistance is a property that resists the flow of electrons in a conductor. It controls the magnitude of the current. The SI unit of resistance is ohm. Ohm's law states that the potential difference across the ends of a resistor is directly proportional to the current through it, provided its temperature remains the same. The resistance of a conductor depends directly on its length, inversely on its area of cross-section, and also on the material of the conductor. The equivalent resistance of several resistors in series is equal to the sum of their individual resistances. A set of resistors connected in parallel has an equivalent resistance R sub p given by one divided by R sub p equals one divided by R one plus one divided by R two plus one divided by R three and so on. The electrical energy dissipated in a resistor is given by W equals V multiplied by I multiplied by t. The unit of power is watt. One watt of power is consumed when one ampere of current flows at a potential difference of one volt. The commercial unit of electrical energy is kilowatt hour. One kilowatt hour equals three point six million joule.
[CHECKPOINT]
Now, let us solve the exercises completely. Exercise one: A piece of wire of resistance R is cut into five equal parts. These parts are then connected in parallel. If the equivalent resistance of this combination is R prime, then the ratio R divided by R prime is? Answer: Each part has resistance R divided by five. For five equal resistors in parallel, one divided by R prime equals five divided by (R divided by five) equals twenty-five divided by R. So R prime equals R divided by twenty-five. Therefore, R divided by R prime equals twenty-five. The correct option is d. Exercise two: Which of the following terms does not represent electrical power in a circuit? Answer: Power is V I, I squared R, and V squared divided by R. I R squared is not a valid power formula. The correct option is b. Exercise three: An electric bulb is rated two hundred twenty volt and one hundred watt. When it is operated on one hundred ten volt, the power consumed will be? Answer: First find resistance: R equals V squared divided by P equals two hundred twenty squared divided by one hundred equals four hundred eighty-four ohm. At one hundred ten volt, P equals V squared divided by R equals one hundred ten squared divided by four hundred eighty-four equals twenty-five watt. The correct option is d. Exercise four: Two conducting wires of the same material and of equal lengths and equal diameters are first connected in series and then parallel in a circuit across the same potential difference. The ratio of heat produced in series and parallel combinations would be? Answer: Let each wire resistance be R. Series resistance is two R. Parallel resistance is R divided by two. Heat is proportional to V squared divided by R. Ratio of heat series to parallel equals (V squared divided by two R) divided by (V squared divided by R divided by two) equals one fourth. So ratio is one to four. The correct option is c.
[CHECKPOINT]
Exercise five: How is a voltmeter connected in the circuit to measure the potential difference between two points? Answer: A voltmeter is always connected in parallel across the two points between which the potential difference is to be measured. Exercise six: A copper wire has diameter zero point five millimeter and resistivity of one point six times ten to the power minus eight ohm meter. What will be the length of this wire to make its resistance ten ohm? How much does the resistance change if the diameter is doubled? Answer: Area A equals pi d squared divided by four. d equals zero point five times ten to the power minus three meter. A equals three point one four times zero point two five times ten to the power minus six divided by four equals one point nine six times ten to the power minus seven meter squared. Using R equals rho l divided by A, l equals R A divided by rho equals ten multiplied by one point nine six times ten to the power minus seven divided by one point six times ten to the power minus eight equals one hundred twenty-two point five meter. If diameter is doubled, area becomes four times larger. Since R is inversely proportional to A, resistance becomes one fourth of original, so it decreases to two point five ohm. Exercise seven: The values of current I flowing in a given resistor for the corresponding values of potential difference V across the resistor are given. Plot a graph between V and I and calculate the resistance of that resistor. Answer: The data points are (zero point five, one point six), (one point zero, three point four), (two point zero, six point seven), (three point zero, ten point two), (four point zero, thirteen point two). The graph is a straight line passing through origin. Slope equals V divided by I. Taking average, resistance is approximately three point three to three point four ohm. Specifically, using last point, R equals thirteen point two divided by four equals three point three ohm.
[CHECKPOINT]
Exercise eight: When a twelve volt battery is connected across an unknown resistor, there is a current of two point five milliampere in the circuit. Find the value of the resistance of the resistor. Answer: I equals two point five times ten to the power minus three ampere. R equals V divided by I equals twelve divided by zero point zero zero two five equals four thousand eight hundred ohm or four point eight kilo ohm. Exercise nine: A battery of nine volt is connected in series with resistors of zero point two ohm, zero point three ohm, zero point four ohm, zero point five ohm and twelve ohm, respectively. How much current would flow through the twelve ohm resistor? Answer: Total resistance R equals zero point two plus zero point three plus zero point four plus zero point five plus twelve equals thirteen point four ohm. Current I equals V divided by R equals nine divided by thirteen point four equals zero point six seven ampere. In series, current is same everywhere, so current through twelve ohm resistor is zero point six seven ampere. Exercise ten: How many one hundred seventy-six ohm resistors in parallel are required to carry five ampere on a two hundred twenty volt line? Answer: Total resistance needed R equals V divided by I equals two hundred twenty divided by five equals forty-four ohm. For n identical resistors in parallel, R equals R one divided by n. So forty-four equals one hundred seventy-six divided by n. n equals one hundred seventy-six divided by forty-four equals four. Four resistors are required. Exercise eleven: Show how you would connect three resistors, each of resistance six ohm, so that the combination has a resistance of i) nine ohm, ii) four ohm. Answer: For nine ohm, connect two six ohm resistors in parallel to get three ohm, then connect this in series with the third six ohm resistor. Three plus six equals nine ohm. For four ohm, connect two six ohm resistors in series to get twelve ohm, then connect this in parallel with the third six ohm resistor. One divided by R equals one divided by twelve plus one divided by six equals three divided by twelve equals one fourth. So R equals four ohm.
[CHECKPOINT]
Exercise twelve: Several electric bulbs designed to be used on a two hundred twenty volt electric supply line, are rated ten watt. How many lamps can be connected in parallel with each other across the two wires of two hundred twenty volt line if the maximum allowable current is five ampere? Answer: Total power allowed P equals V I equals two hundred twenty multiplied by five equals one thousand one hundred watt. Each bulb is ten watt. Number of bulbs equals one thousand one hundred divided by ten equals one hundred ten. Exercise thirteen: A hot plate of an electric oven connected to a two hundred twenty volt line has two resistance coils A and B, each of twenty-four ohm resistance, which may be used separately, in series, or in parallel. What are the currents in the three cases? Answer: Used separately: I equals V divided by R equals two hundred twenty divided by twenty-four equals nine point one seven ampere. In series: Total R equals forty-eight ohm. I equals two hundred twenty divided by forty-eight equals four point five eight ampere. In parallel: Total R equals twenty-four divided by two equals twelve ohm. I equals two hundred twenty divided by twelve equals eighteen point three three ampere. Exercise fourteen: Compare the power used in the two ohm resistor in each of the following circuits: i) a six volt battery in series with one ohm and two ohm resistors, and ii) a four volt battery in parallel with twelve ohm and two ohm resistors. Answer: Case i: Total R equals three ohm. I equals six divided by three equals two ampere. Power in two ohm resistor equals I squared R equals four multiplied by two equals eight watt. Case ii: Voltage across two ohm resistor is four volt. Power equals V squared divided by R equals sixteen divided by two equals eight watt. Power is same in both cases.
[CHECKPOINT]
Exercise fifteen: Two lamps, one rated one hundred watt at two hundred twenty volt, and the other sixty watt at two hundred twenty volt, are connected in parallel to electric mains supply. What current is drawn from the line if the supply voltage is two hundred twenty volt? Answer: Total power equals one hundred plus sixty equals one hundred sixty watt. Current I equals P divided by V equals one hundred sixty divided by two hundred twenty equals zero point seven three ampere. Exercise sixteen: Which uses more energy, a two hundred fifty watt TV set in one hour, or a one thousand two hundred watt toaster in ten minutes? Answer: Energy for TV equals two hundred fifty watt multiplied by one hour equals two hundred fifty watt hour. Energy for toaster equals one thousand two hundred watt multiplied by one sixth hour equals two hundred watt hour. The TV uses more energy. Exercise seventeen: An electric heater of resistance eight ohm draws fifteen ampere from the service mains two hours. Calculate the rate at which heat is developed in the heater. Answer: Rate of heat development is power. P equals I squared R equals fifteen squared multiplied by eight equals two hundred twenty-five multiplied by eight equals one thousand eight hundred watt. Exercise eighteen: Explain the following. a) Why is tungsten used almost exclusively for filament of electric lamps? Answer: Tungsten has a very high melting point and high resistivity, so it can get very hot and emit light without melting. b) Why are the conductors of electric heating devices made of an alloy rather than a pure metal? Answer: Alloys have higher resistivity and do not oxidise easily at high temperatures, making them durable and efficient for heating. c) Why is the series arrangement not used for domestic circuits? Answer: In series, voltage divides, devices get insufficient voltage, and if one fails, all stop working. Parallel provides full voltage and independent operation. d) How does the resistance of a wire vary with its area of cross-section? Answer: Resistance is inversely proportional to the area of cross-section. If area increases, resistance decreases. e) Why are copper and aluminium wires usually employed for electricity transmission? Answer: They have very low resistivity, are highly conductive, ductile, and relatively inexpensive, making them ideal for long-distance transmission.
Thank you for listening! Keep revising and practicing. Goodbye! [CHAPTER_COMPLETE]