-6<\/sup> A.<\/p>\n<\/p>\n
Question 13.
\nHow is an ammeter connected in a circuit?
\nAnswer:
\nAn ammeter is always connected in series in a circuit.<\/p>\n
Question 14.
\nDraw a schematic diagram of a typical electric circuit comprising a cell, an electric bulb, an ammeter and a plug key.
\nAnswer:
\n<\/p>\n
Question 15.
\nIn any circuit, state the direction of electric current and the direction of electron current.
\nAnswer:
\nIn any circuit, electric current flows from the positive terminal of the cell towards the negative terminal. Electron current, however, flows from the negative terminal towards the positive terminal of the cell. This means that the direction opposite to the flow of electrons is taken as the direction of conventional current.<\/p>\n
Question 16.
\nDefine electrical potential at a point. Give an equation for the same.
\nAnswer:
\nThe electric potential at a point is defined as the work done in moving a unit positive charge from infinity to that point.<\/p>\n
If W is the work done in moving a charge Q from infinity to a point, then, the electrical potential V at that point is given by<\/p>\n
Electrical potential = \\(\\frac{\\text { Work done }}{\\text { Charge moved }}\\)
\nV = \\(\\frac{\\mathrm{W}}{\\mathrm{Q}}\\)<\/p>\n
Question 17.
\nDefine and explain potential difference. Give an equation for the same.
\nAnswer:
\nJust as liquids flow from a higher level to a lower level, electricity also flows through a conductor from a point of higher electric potential to a point of lower electric potential. This means that a potential difference is necessary for the flow of current through a conductor.<\/p>\n
The potential difference between any two points is defined as the work done in moving a unit positive charge from one of those points to the other. It is also called \u2018voltage\u2019.
\n
\nConsider a charge Q moving from point B to point A. Let us say that the work done during the process is W. Let Vbe the potential difference between the two points. Now,
\nPotential difference = \\(\\frac{\\text { Work done }}{\\text { Charge moved }}\\)
\nV = \\(\\frac{\\mathrm{W}}{\\mathrm{Q}}\\)<\/p>\n
<\/p>\n
Question 18.
\nWhat is the S.I. unit of \u2018electric potential\u2019 and \u2018potential difference\u2019?
\nAnswer:
\nBoth electrical potential and potential difference are expressed by the same unit namely \u2018volt\u2019. It is represented by the symbol V.<\/p>\n
Question 19.
\nDefine 1 V of electric potential
\nAnswer:
\nThe electric potential at a point is said to be 1 volt if 1 joule of work is done in moving a charge of positive 1 coulomb from infinity to that point.
\n1 volt = \\(\\frac{1 \\text { joule }}{1 \\text { coulomb }}\\)
\n1 V = 1 JC-1<\/sup>.<\/p>\nQuestion 20.
\nDefine 1 V of potential difference.<\/p>\n
OR<\/p>\n
What is meant by saving that the potential difference between two points is 1 V?
\nAnswer:
\nThe potential difference between two points is said to be 1 volt if 1 joule of work is done in moving a charge of +1 C from one of those points to the other.
\n1 V = \\(\\frac{1 \\text { joule }}{1 \\text { coulomb }}\\)
\n1 V = 1 JC-1<\/sup>.<\/p>\nQuestion 21.
\nWhat is the device used to measure the potential difference across a conductor? How is a voltmeter connected in the circuit to measure the potential difference between two points?<\/p>\n
OR<\/p>\n
Name the device used to measure potential difference across a conductor. How should this device be connected in a circuit?
\nAnswer:
\nThe potential difference across a conductor is measured by using a device called voltmeter. A voltmeter is always connected in parallel in the circuit across the points between which the potential difference is to be measured.<\/p>\n
<\/p>\n
Question 22.
\nName a device that helps to maintain a potential difference across a conductor.
\nAnswer:
\nWe can set up a potential difference by using an electric cell or a battery.<\/p>\n
Question 23.
\nHow does an electric cell help to maintain a potential difference across a conductor?
\nAnswer:
\nThe chemical reaction within a cell generates potential difference across the terminals of the cell. When the cell is connected to a circuit, the potential difference sets the charges in motion through the conductor and produces an electric current. The cell maintains the current in the circuit by expending the chemical energy stored in it.<\/p>\n
Question 24.
\nIn the circuit given below, the potential difference across the conductor AB is to be measured. Show by a diagram how you would connect a voltmeter to measure the potential difference across AB.
\n
\nAnswer:
\n
\nA voltmeter should be connected in parallel between the points across which we want to measure the potential difference. Here, we desire to measure the potential difference across AB. The connection of the voltmeter is shown in the circuit diagram given below:<\/p>\n
Question 25.
\nHow much work is done in moving a charge of 2 C across two points having a potential difference 12 V?
\nAnswer:
\nGiven: Charge moved, Q = 2 C; Potential difference, V= 12 V; Work done, W = ?
\nW= VQ
\n= 12 V \u00d7 2 C = 24 J.<\/p>\n
Question 26.
\nThe work done in moving 5 C of charge between two points in an electric field is 30 J. Calculate the potential difference between the two points.
\nAnswer:
\nGiven: Charge moved, Q = 5 C, Work done W= 30 J, Potential difference V= ?
\nV = \\(\\frac{w}{Q}\\)
\n= \\(\\frac{30 \\mathrm{J}}{5 \\mathrm{C}}\\) = 6 JC-1<\/sup>
\n= 6V<\/p>\n<\/p>\n
Question 27.
\nHow much energy is given to each coulomb of charge passing through a 6 V battery?
\nAnswer:
\nGiven: Charge moved, Q = 1 C; Potential difference, V = 6 V, Energy given, W = ?
\nThe energy given is nothing but the work done in moving the charge.
\nW = VQ
\n= 6V \u00d7 1C = 6 J.
\nThe energy given to each coulomb is 6 J.<\/p>\n
Question 28.
\nName the law that gives the relationship between current flowing through a conductor and the potential difference across it.
\nAnswer:
\nThe law that gives the relationship between current flowing through a conductor and the potential difference across it is Ohm’s law.<\/p>\n
Question 29.
\nState Ohm’s law. Express it in the form of an equation.
\nAnswer:
\nOhm’s law states that \u2018at constant temperature, the potential difference across a conductor is directly proportional to the current flowing through it\u2019.
\n
\nConsider a conductor AB. Let V be the potential difference between these points and I be the current flowing through it. Now, according to Ohm’s law,
\nPotential difference \u221d Current
\nV \u221d I
\nV = RI where R is a constant called \u2018resistance\u2019 of the conductor.
\n\u2234 \\(\\frac{\\mathrm{V}}{\\mathrm{I}}\\) = Constant.<\/p>\n
<\/p>\n
Question 30.
\nWhat is electrical resistance? Mention its S.I unit.<\/p>\n
OR<\/p>\n
What is resistance of a conductor?
\nAnswer:
\nAll conductors oppose the flow of electricity through them. The opposition offered by a conductor to the flow of current through it is called its electrical resistance. The S.I unit of electrical resistance is called \u2018ohm\u2019. It is represented by the symbol \u03a9.<\/p>\n
Question 31.
\nDefine 1 \u03a9 of resistance.
\nAnswer:
\nThe resistance of a conductor is said to be 1 ? if 1 A of current flows through it when a potential difference of IV is applied across it.
\n1 \u03a9 = \\(\\frac{1 \\mathrm{V}}{1 \\mathrm{A}}\\)
\n1 \u03a9 = 1 VA-1<\/sup><\/p>\nQuestion 32.
\nDraw a simple circuit diagram showing the arrangement of apparatus for the verification of Ohm’s law.<\/p>\n
OR<\/p>\n
Describe an experiment to show the relationship between potential difference and current.<\/p>\n
OR<\/p>\n
How do you verify Ohm\u2019s law?
\nAnswer:
\n
\nTake an electrical conductor XY made of nichrome. Connect it in series with a battery, a key, a rheostat and an ammeter. Connect a voltmeter parallel to the given resistor. Close the key.<\/p>\n
Adjust the rheostat for a small value of current. Note the current and the potential difference by noting down the readings in the ammeter and voltmeter respectively. Go on increasing the value of current in equal but small steps.<\/p>\n
Note down the corresponding value of potential difference (V) for each value of current (I). Find the value of V\/I in each case. Tabulate the readings as follows:
\n
\nThe value of V\/I is found to be a constant. This shows that the potential difference across a conductor is directly proportional to the current flowing through it.
\nThe graph of V against I:
\n
\nPlot a graph of potential difference against current. The graph of V against I is found to be a straight line passing through the origin. This again shows that the potential difference is directly proportional to the current flowing through it. This verifies Ohm\u2019s law.<\/p>\n
<\/p>\n
Question 33.
\nGowri, while studying the variation of current through a conductor with potential difference, prepared the following table. However, she has left a few blanks in the table. Complete the table using your knowledge of Ohm’s law.
\nAnswer:
\n
\n<\/p>\n
Question 34.
\nA student measured the potential difference across a nichrome wire for various values of current flowing through it. The values are given in the table below:
\nAnswer:
\n
\nDraw a graph of potential difference against current. Interpret the graph. Find the value of the nichrome wire used in this activity.
\nAnswer:
\n
\nThe graph of potential difference (V) against current (I) is a straight line passing through the origin. This shows that the potential difference across the conductor is directly proportional to the current flowing through it.<\/p>\n
In other words, the value of \\(\\frac{\\mathrm{V}}{\\mathrm{I}}\\) is a constant. This constant value is nothing but the resistance of the given nichrome wire.
\nResistance, R = \\(\\frac{\\mathrm{V}}{\\mathrm{I}}=\\frac{0.4 \\mathrm{V}}{0.1 \\mathrm{A}}\\) = 4 \u03a9.<\/p>\n
<\/p>\n
Question 35.
\nName the device used to provide variable resistance in a circuit.
\nAnswer:
\nIn an electric circuit, a device called rheostat is often used to provide continuously variable resistance. This device can be used to either increase or decrease the resistance of the circuit.<\/p>\n
Question 36.
\nHow do you show that the current flowing through an electric component depends upon its resistance?
\nAnswer:
\nTake a nichrome wire, a torch bulb, a 10 W bulb and an ammeter (0-5 A range), a plug key and some connecting wires. Set up the circuit by connecting four dry cells of 1.5 V each in series with the ammeter leaving a gap XY in the circuit, as shown in the figure below:
\n
\nConnect the nichrome wire in the gap XY. Close the plug key. Note down the current in the circuit from the reading of the ammeter. Repeat the activity by replacing the nichrome wire with the torch bulb and the 10 W bulb one after the other.<\/p>\n
Note down the ammeter reading in each case. Compare the current in the circuit in all the three cases. This is due to the fact that the resistance of each of three gadgets connected in the gap is different. From this we can conclude that the current through an electric component depends on its resistance.<\/p>\n
<\/p>\n
Question 37.
\nThe values of current I flowing in a given resistor for the corresponding values of potential difference V across the resistor are given below:
\n
\nPlot a graph between V and I and calculate the resistance of the resistor.
\nAnswer:
\nIf we plot the values of current along the Y-axis and the potential difference (voltage) along X-axis for the values given, we get the following graph:
\n
\nResistance of resistor (R)
\n<\/p>\n
Question 38.
\nWhy does the current flowing through a circuit depend on the resistance of the circuit? How does the current through a device vary with its resistance?
\nAnswer:
\nCertain components offer an easy path for the flow of electric current while some others resist the flow. We know that motion of electrons in an electric circuit constitutes an electric current.<\/p>\n
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, its resistance retards motion of electrons through a conductor. The current flowing through a device is inversely proportional to its resistance.<\/p>\n
Question 39.
\nLet the resistance of an electrical component remain 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?
\nAnswer:
\nWe know that the current flowing through a conductor is directly proportional to the potential difference across it. Therefore, when the potential difference across the component is reduced to half the initial value, the current will also get halved.<\/p>\n
<\/p>\n
Question 40.
\nDistinguish between a good conductor, a resistor, a poor conductor and an insulator on the basis of resistance.
\nAnswer:
\nA material, which offers low resistance to the flow of electrons or electric current in an electric circuit, is known as a good conductor.<\/p>\n
A component in an electric circuit, which offers resistance to the flow of electrons constituting electric current, is known as resistor.<\/p>\n
A material, which offers higher resistance to the flow of electrons or electric current in an electric circuit, is known as poor conductor.<\/p>\n
A material, which offers very high resistance to the flow of electrons or electric current in an electric circuit, is known as insulator. Electric current does not flow through an insulator.<\/p>\n
Question 41.
\nWhat potential difference must be applied across a 10 \u03a9 wire in order that a current of 2.5 A flows through it?
\nAnswer:
\nGiven: Current I = 2.5 A; Resistance R= 10 \u03a9; Potential difference V = ?
\nV = RI (Ohm’s law)
\n= 10 \u03a9 \u00d7 2.5 A = 25 V.<\/p>\n
Question 42.
\nWhen a 12 V battery is connected across an unknown resistor, there is a current of 2.5 mA in the circuit. Find the value of the resistance of the resistor.
\nAnswer:
\nGiven: Potential difference, V = 12 V, Current, I = 2.5 mA = 0.0025 A, Resistance, R = ?
\nAccording to Ohm’s law, R = \\(\\frac{v}{1}\\)
\n= \\(=\\frac{12 \\mathrm{V}}{0.0025 \\mathrm{A}}\\) = 4800 \u03a9.<\/p>\n
Question 43.<\/p>\n
\n- How much current will an electric bulb draw from a 220 V source, if the resistance of the bulb filament is 1200 \u03a9?<\/li>\n
- How much current will an electric heater coil draw from a 220 V source, if the resistance of the heater coil is 100 \u03a9?<\/li>\n<\/ol>\n
Answer:
\n1. Given: Voltage, V= 220 V; Resistance, R = 1200 \u03a9; Current, I = ?
\n<\/p>\n
2. Given: Voltage, V= 220 V; Resistance, R = 100 \u03a9; Current, I = ?
\n<\/p>\n
<\/p>\n
Question 44.
\nThe 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?
\nAnswer:
\nWe are given, potential difference V= 60 V, current 7 = 4A.
\nV 60 V
\nAccording to Ohm’s law, R = \\(\\frac{V}{I}=\\frac{60 \\mathrm{V}}{4 \\mathrm{A}}\\) = 15 \u03a9.
\nWhen the potential difference is increased to 120 V the current is given by
\nCurrent = \\(\\frac{V}{R}=\\frac{120 \\mathrm{V}}{15 \\Omega}\\) = 8 A.
\nThe current through the heater becomes 8A.<\/p>\n
Question 45.
\nOn what factors does the resistance of a conductor depend?
\nAnswer:
\nThe resistance of a conductor depends on the following factors:<\/p>\n
\n- Length of the conductor,<\/li>\n
- area of cross-section of the conductor, and<\/li>\n
- nature of the material of the conductor.<\/li>\n<\/ol>\n
Question 46.
\nThe resistance of a conductor varies with its length. How do you establish this fact?
\nAnswer:
\nTake three nichrome wires L1<\/sub>, L2<\/sub> and L3<\/sub> of same thickness but of lengths in the ratio 1:2:3 respectively. Complete an electric circuit consisting of a cell, an ammeter, nichrome wire L1<\/sub>\u00a0and a plug key, as shown in the figure below.
\n
\nNow, close the plug key. Note the current in the ammeter. Replace the nichrome wire L1<\/sub> with nichrome wire L2<\/sub>. Note the ammeter reading. Now replace this wire with nichrome wire L3<\/sub>. Note down the ammeter reading again.<\/p>\nCompare the values of the current through L1<\/sub>, L2<\/sub> and L3<\/sub>. The current is found to be in the ratio 3:2:1 respectively. This shows that the resistance of L2<\/sub> and L3<\/sub> is twice and thrice that of L1<\/sub> respectively. This establishes that the resistance of a conductor is directly proportional to its length.<\/p>\nQuestion 47.
\nHow do you show experimentally that the resistance of a conductor varies with its area of cross section?
\nAnswer:
\nTake three nichrome wires L1<\/sub>, L2<\/sub> and L3<\/sub> of same length but of area of cross section in the ratio 1:2:3 respectively. Complete an electric circuit consisting of a cell, an ammeter, nichrome wire L1<\/sub>\u00a0and a plug key, as shown in the figure.
\n
\nNow, close the plug key. Note the current in the ammeter. Replace the nichrome wire L1<\/sub> with nichrome wire L2<\/sub>. Note the ammeter reading. Now replace this wire with nichrome wire L3<\/sub>.<\/p>\nNote down the ammeter reading again. Compare the values of the current through L1<\/sub>, L2<\/sub> and L3<\/sub>. The current is found to be in the ratio 1:2:3 respectively.<\/p>\nThis shows that the resistance of L2<\/sub> and L3<\/sub> is half and one-third that of L1<\/sub> respectively. This establishes that the resistance of a conductor is inversely proportional to its area of cross section.<\/p>\n<\/p>\n
Question 48.
\nHow does the resistance of a wire vary with its area of cross-section?
\nAnswer:
\nResistance of a wire is inversely proportional to its area of cross-section. This means that the resistance of a wire decreases in the same ratio as the increase in the area of its cross-section and vice- versa. Thus, if the wire is thick (large area of cross-section), then resistance is less. If the wire is thin (less area of cross-section), then resistance is large.<\/p>\n
Question 49.
\nWill current flow more easily through a thick wire or a thin wire of the same material, when connected to the same source? Why?
\nAnswer:
\nThe resistance of a conductor is inversely proportional to the area of cross-section of the conductor. The current will flow more easily through the thicker wire because the thicker wire has more area of cross-section hence has lesser resistance to the flow of current.<\/p>\n
Question 50.
\nHow does the resistance of a conductor vary with (a) its length, and (b) area of cross-section?
\nAnswer:<\/p>\n
\n- The resistance of a conductor is directly proportional to its length.<\/li>\n
- The resistance of a conductor is inversely proportional to its area of cross-section.<\/li>\n<\/ol>\n
Question 51.
\nExplain the factors that affect electrical resistance.
\nAnswer:
\n1. The electrical resistance of a conductor is directly proportional to its length. This means that the resistance of a conductor increases in the same ratio as the increase in its length.<\/p>\n
2. The resistance of a conductor is inversely proportional to its area of cross-section. This implies that the resistance decreases in the same ratio as the increase in its area of cross-section.<\/p>\n
3. The resistance of a conductor depends bn its temperature. The resistance of most of the electrical conductors increases with the increase in temperature.<\/p>\n
4. Finally, the resistance also depends on the material of the conductor. This means identical conductors made of different materials have different resistances.<\/p>\n
<\/p>\n
Question 52.
\nDefine \u2018electrical resistivity\u2019 of the material of a conductor. Give an expression for the resistivity of a material.
\nAnswer:
\nThe electrical resistivity of the material of a conductor is defined as the resistance of a conductor made of that material of length 1 m and area of cross-section 1 m2<\/sup>.<\/p>\nConsider a conductor of length l and area of cross-section A. Let its resistance be R. We know that the resistance of a conductor is directly proportional to its length t and inversely proportional to its area of cross-section A.
\n
\nwhere \u03c1 is a constant called \u2018resistivity\u2019 of the material.
\nResistivity, \u03c1 = \\(\\frac{\\mathrm{RA}}{l}\\).<\/p>\n
Question 53.
\nWhat is the S.I unit of resistivity? What is the other name for resistivity of a material?
\nAnswer:
\nThe S.I. unit of resistivity is called \u2018ohm metre\u2019. Its symbol is \u2018\u03a9. m\u2019. The resistivity of a material is also called its \u2018specific electrical resistance\u2019.<\/p>\n
Question 54.
\nYou are given three copper wires of length l, 2l and 3l having area of cross-section A, 2A and 3A. Which of these has highest resistivity?
\nAnswer:
\nResistivity of a conductor depends only on the material it is made of and not on its dimensions. Therefore, all the three wires have equal resistivity, as all of them are made of the same material namely copper.<\/p>\n
Question 55.
\nA wire of resistivity ? is pulled to double its length. What will be its new resistivity?
\nAnswer:
\nSince resistivity of a conductor depends only on the material it is made of and not on its dimensions, the resistivity \u03c1 remains the same.<\/p>\n
<\/p>\n
Question 56.
\nThe electrical resistivity of copper is 1.62 \u00d7 10-8<\/sup>? m. What is the meaning of this statement?
\nAnswer:
\nThe statement above means that a copper wire of length 1 m and area of cross-section 1 m2<\/sup> has resistance of 1.62 \u00d7 10-18<\/sup> \u03a9.<\/p>\nQuestion 57.
\nThe resistivity of some materials is siven in the table below:<\/p>\n
\n\n\nMaterial<\/td>\n | Resistivity (\u2126 m)<\/td>\n<\/tr>\n |
\n1. Copper<\/td>\n | 1.62 x 10-8<\/sup><\/td>\n<\/tr>\n\n2. Iron<\/td>\n | 10.0 x 10-8<\/sup><\/td>\n<\/tr>\n\n3. Mercury<\/td>\n | 94.0 x 10-8<\/sup><\/td>\n<\/tr>\n\n4. Silver<\/td>\n | 1.60 x 10-8<\/sup><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n\n- Which among iron and mercury is a better conductor?<\/li>\n
- Which material is the best conductor?<\/li>\n<\/ol>\n
Answer:<\/p>\n \n- Among iron and mercury, the better conductor is iron. This is because the resistivity of iron is lower than that of mercury.<\/li>\n
- Among the given materials, silver is the best conductor of electricity. This is because silver has the lowest value of resistivity.<\/li>\n<\/ol>\n
Question 58. \nResistance of a metal wire of length 1 m is 26 \u03a9 at 20\u00b0C. If the diameter of the wire is 0.3 mm, what will be the resistivity of the metal at that temperature? \nAnswer: \nGiven: Resistance, R = 26 \u03a9; Diameter, d = 0.3 mm = 3 \u00d7 10-4<\/sup>m, Length of the wire, l = 1 m. \n<\/p>\n<\/p>\n Question 59. \nA wire of given material having length l and area of cross-section A has a resistance of 4 \u03a9. What would be the resistance of another wire of the same material having length l\/2 and area of cross-section 2 A ? \nAnswer: \nFor the first wire: R1<\/sub> = \u03c1 \\(\\frac{l}{A}\\) = 4 \u03a9 \nNow for the second wire \nR2<\/sub> = \\(\\frac{1 \/ 2}{2 A}=\\frac{1}{4} \\rho \\frac{l}{A}\\) \nR2<\/sub> = \\(\\frac{1}{4}\\) R \nR1<\/sub> = 1 \u03a9. \nThe resistance of the new wire is 1 \u03a9.<\/p>\nQuestion 60. \nA copper wire has diameter 0.5 mm and resistivity of 1.6 \u00d7 10-8<\/sup> \u03a9 m. What will be the length of this wire to make its resistance 10 \u03a9? How much does the resistance change if the diameter is doubled? \nAnswer: \nArea of cross-section of the wire, A = p \\(\\left(\\frac{d}{2}\\right)^{2}\\) \nDiameter = 0.5 mm = 0.0005 m \nResistance, R = 10 \u03a9 \n \n= 122.72 m \n\u2234 Length of the wire = 122.72 m. \nIf the diameter of the wire is doubled, new diameter = 2 \u00d7 0.5 = 1 mm = 0.001 m. \nLet new resistance be R’. \nR’ = \u03c1 \\(\\frac{l}{A}\\) \n \n= 250.2 \u00d7 10-2<\/sup> \n= 2.5 \u03a9 \nTherefore, the length of the wire is 122.7 m and the new resistance is 2.5 \u03a9. \nThus, the resistance of the conductor will become \\(\\frac{1}{4}\\) th its initial value when the diameter is doubled.<\/p>\n<\/p>\n Question 61. \nDistinguish between resistance and resistivity of a conductor. \nAnswer:<\/p>\n \n\n\nResistance<\/td>\n | Resistivity<\/td>\n<\/tr>\n | \n1. Resistance of a conductor is the obstruction offered by the conductor for the flow of current through it.<\/td>\n | Resistivity of a material is the obstruction offered by a conductor of that material of length 1 m and area of cross-section 1 m2<\/sup>.<\/td>\n<\/tr>\n\n2. It depends on the material, length and area of cross-section of the conductor.<\/td>\n | It does not depend on the length and area of cross-section of the conductor.<\/td>\n<\/tr>\n | \n3. The S.I. unit of resistance is called \u2018ohm\u2019 (\u2126).<\/td>\n | The S.I. unit of resistivity is called \u2018ohm-metre\u2019 (\u2126m).<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n Question 62. \nWhen do we say that a group of resistances are connected in series? Show by a diagram the arrangement of three resistances in series. \nAnswer: \nA group of resistances are said to be in series if they are connected from end to end so that each of them carries the same current. \nThree resistances R1<\/sub>, R2<\/sub> and R3<\/sub> are connected in series as shown below: \n<\/p>\nQuestion 63. \nDraw the diagram of the electric circuit in which the resistors are connected in series.<\/p>\n OR<\/p>\n Three resistances R1<\/sub>, R2<\/sub> and R3<\/sub> are connected in series. This combination is connected in series with a battery, a key and an ammeter. A voltmeter is connected to measure the potential difference across the combination. Draw a suitable circuit diagram for this. \nAnswer: \n<\/p>\nQuestion 67. \nWhat is the equivalent resistance of a group of resistances connected in series? Give a suitable equation for it. \nAnswer: \nThe equivalent resistance of a group of resistances connected in series is equal to the sum of all the individual resistances.<\/p>\n Consider a number of resistances R1<\/sub>, R2<\/sub>, R3<\/sub>………Rn<\/sub> connected in series. Let R be their equivalent resistance. Now, \nR = R1<\/sub> + R2<\/sub> + R3<\/sub> + ……… + Rn<\/sub>.<\/p>\n<\/p>\n Question 65. \nObtain an equation for the combined resistance of a group of resistances in series. \nAnswer: \n \nA group of resistances are said to be in series if they are connected from end to end so that they carry the same current through each of them. Consider three resistances R1<\/sub>, R2<\/sub> and R3<\/sub> connected in series.<\/p>\nLet V1<\/sub> be the potential difference across R1<\/sub>, V2<\/sub> be the potential difference across R2<\/sub> and V3<\/sub> be the potential difference across R3<\/sub>.<\/p>\nLet I be the current flowing through each of these resistors. Let V be the potential difference across the combination and R be the total resistance of the system. \nTotal potential difference V = V1<\/sub> + V2<\/sub> + V3<\/sub>. \nBut, V = IR, V1<\/sub> = IR1<\/sub>, V2<\/sub> = IR2<\/sub>, and V3<\/sub> = IR3<\/sub>. \nSubstituting these values in the equation above, we get, \nIR = IR1<\/sub> + IR2<\/sub> + IR3<\/sub> \n\u2234 Rs<\/sub> = R1<\/sub> + R2<\/sub> + R3<\/sub> \nThis shows that the combined resistance of a group of resistances connected in series is equal to the sum of all the individual resistances in the group.<\/p>\nQuestion 66. \nHow do you show that the same current flows through each resistor when a group of resistors are connected in series? \nAnswer: \n \nConnect three resistors, say of 1 \u03a9, 2 \u03a9, 3 \u03a9, in series. Connect this combination with a battery of say 6 V, an ammeter and a plug key in series. Close the circuit. Note the ammeter reading. Change the position of the ammeter to anywhere in between the resistors and note the reading of the ammeter each time.<\/p>\n Each time the ammeter shows the same current. This shows that, whenever a number of resistors are connected in series, the current flowing through each part of the circuit is same. This is because the current has only one path to flow.<\/p>\n <\/p>\n Question 67. \nWhen do we say that a group of resistances are in parallel arrangement? Show by a diagram the arrangement of three resistances in parallel arrangement. \nAnswer: \nA group of resistances are said to be in parallel arrangement when one end of each resistance is connected to one common terminal and the other end of each resistance is connected to another common terminal. In such an arrangement, the potential difference across each resistance would be the same as all of them have common end points. Three resistances R1<\/sub>, R2<\/sub> and R3<\/sub> are connected, as shown below, in parallel arrangement: \n<\/p>\nQuestion 68. \nDraw the diagram of the electric circuit in which the resistors R1<\/sub>, R2<\/sub> and R3<\/sub> are connected in parallel including ammeter and voltmeter and mark the direction of current. \nAnswer: \n<\/p>\nQuestion 69. \nWhat is the equivalent resistance of a group of resistances connected in parallel? Give a suitable equation for it. \nAnswer: \nThe reciprocal of the equivalent resistance of a group of resistances in parallel is equal to the sum of the reciprocals of all the individual resistances.<\/p>\n Consider a number of resistances R1<\/sub>, R2<\/sub>, R3<\/sub> ……….. Rn<\/sub> connected in parallel. Let R be their equivalent \nresistance. Now, \n<\/p>\n<\/p>\n Question 70. \nObtain an equation for the combined resistance of a group of resistances in parallel \nAnswer: \n \nA group of resistances are said to be in parallel if one end of each resistance is connected to one common terminal and the other end is connected to another common terminal so that each resistor has the same potential difference across it.<\/p>\n Consider three resistances R1<\/sub>, R2<\/sub> and R3<\/sub> connected in parallel. Let I1<\/sub> be the current through R1<\/sub>, I2<\/sub> be the current through R2<\/sub> and I3<\/sub> be the current through R3<\/sub>. Let I be the total current through the system, V be the potential difference across each resistor and R be the total resistance of the system. \nTotal current I = I1<\/sub> + I2<\/sub> + I3<\/sub>. \nSubstituting these values in the equation above, we get, \n \nThus, the reciprocal of the combined resistance of a group of resistances in parallel is equal to the sum of the reciprocals of all the individual resistances.<\/p>\nQuestion 71. \nWhy is the series arrangement not used for domestic circuits?<\/p>\n OR<\/p>\n Mention the disadvantages of connecting electrical appliances in series in domestic wiring. \nAnswer: \nIn domestic circuits if gadgets are connected in series, voltage is divided. Each component of a series circuit receives a small voltage so that the amount of current decreases and the device becomes hot and does not work properly.<\/p>\n Secondly, when gadgets are in series arrangement, they cannot be operated independently. If one of them does not work, the circuit is broken and the other components also do not work. Hence, series arrangement is not used in domestic electric circuits.<\/p>\n <\/p>\n Question 72. \nThree resistors of 3 \u03a9, 4 \u03a9 and 5 \u03a9 are connected in<\/p>\n \n- series and in<\/li>\n
- parallel Find the equivalent resistance in each case.<\/li>\n<\/ol>\n
Answer: \nR1<\/sub> = 3 \u03a9, R2<\/sub> = 4 \u03a9, R3<\/sub> = 5 \u03a9, combined resistance R = ?<\/p>\n1. When resistances are connected in series, the equivalent resistance R is given by \nR = R1<\/sub> + R2<\/sub> + R3<\/sub> \nR = 3 \u03a9 + 4 \u03a9 + 5 \u03a9 \n= 12 \u03a9<\/p>\n2. When resistances are in parallel arrangement, the equivalent resistance R is given by \n \n<\/p>\n Question 73. \nIn the circuit diagram given below, suppose the resistors R1<\/sub>, R2<\/sub> and R3<\/sub> have values 5 \u03a9, 10 \u03a9, 30 \u03a9? respectively, which are connected to a battery of 12 V. Calculate<\/p>\n\n- The current through each resistor,<\/li>\n
- The total current in the circuit, and<\/li>\n
- The total circuit resistance.<\/li>\n<\/ol>\n
Answer: \n \n1. Current through 5 \u03a9 resistor = \\(\\frac{V}{R}=\\frac{12 \\mathrm{V}}{5 \\Omega}\\) = 2.4 A \nCurrent through 10 \u03a9 resistor = \\(\\frac{V}{R}=\\frac{12 \\mathrm{V}}{10 \\Omega}\\) = 1.2 A \nCurrent through 30 \u03a9 resistor = \\(\\frac{V}{R}=\\frac{12 \\mathrm{V}}{30 \\Omega}\\) = 0.4 A<\/p>\n 2. Total current through the circuit, I = 2.4 A + 1.2 A + 0.4 A = 4 A<\/p>\n 3. Total resistance of the circuit, R = \\(\\frac{V}{I}=\\frac{12 \\mathrm{V}}{4 \\mathrm{A}}\\) = 3 \u03a9.<\/p>\n <\/p>\n Question 74. \nIf, in the figure given below, R1<\/sub> = 10 \u03a9, R2<\/sub> = 40 \u03a9, R3<\/sub> = 30 \u03a9, R4<\/sub> = 20 \u03a9, R5<\/sub> = 60 \u03a9, and a 12 V battery is connected to the arrangement Calculate<\/p>\n\n- The total resistance in the circuit, and<\/li>\n
- The total current flowing in the circuit<\/li>\n<\/ol>\n
Answer: \n \n1. Suppose we replace the parallel resistors R1<\/sub> and R2<\/sub> by an equivalent resistor of resistance R’ and resistors R3<\/sub>, R4<\/sub> and R5<\/sub> by an equivalent single resistor R”. Then, \n \nNow, R’ and R” are in series. Let their combined resistance be R. Now, R is the total resistance of the circuit. \nR = R\u2019 + R” \nR = 8 \u03a9 + 10 \u03a9 = 18 \u03a9.<\/p>\n2. Current through the circuit, I = \\(\\frac{V}{R}=\\frac{12 \\mathrm{V}}{18 \\Omega}\\) = 0.67 A.<\/p>\n Question 75. \nFour resistors 5 \u03a9, 6 \u03a9, 4 \u03a9 and 8 \u03a9 are connected in<\/p>\n \n- series and in<\/li>\n
- parallel. Find the equivalent resistance in each case.<\/li>\n<\/ol>\n
Answer: \nR1<\/sub> = 5 \u03a9, R2<\/sub> = 6 \u03a9, R3<\/sub> = 4 \u03a9, R4<\/sub>= 8\u03a9 combined resistance R = ?<\/p>\n1. When the resistances are connected in series, the equivalent resistance R is given by \nR = R1<\/sub> + R2<\/sub> + R3<\/sub> + R4<\/sub> \nR = 5 \u03a9 + 6 \u03a9 + 4 \u03a9 + 8 \u03a9 = 23 \u03a9.<\/p>\n2. When the resistances are in parallel arrangement, the equivalent resistance R is given by \n<\/p>\n <\/p>\n Question 76. \nWhy are coils of electric toasters and electric irons made of an alloy rather than a pure metal?<\/p>\n OR<\/p>\n Why are the conductors of electric heating devices, such as bread-toasters and electric irons, made of an alloy rather than a pure metal? \nAnswer: \nCoils of electric toasters and electric irons are made of an alloy rather than a pure metal for the following reasons:<\/p>\n \n- Alloys have higher resistivity than pure metals. This enables them to generate more heat.<\/li>\n
- Alloys, unlike pure metals, generally do not melt easily even at higher temperatures.<\/li>\n<\/ol>\n
Question 77. \nDraw a schematic diagram of a circuit consisting of a battery of three cells of 2 V each, a 5 \u03a9 resistor, an 8 \u03a9 resistor, and a 12 \u03a9 resistor, and a plus key, all connected in series. \nAnswer: \n<\/p>\n Question 78. \nDraw a schematic diagram of a circuit consisting of a battery of three cells of 2 V each, a 5 \u03a9 resistor, an 8 \u03a9 resistor, and a 12 \u03a9 resistor, and a plus key, all connected in series. Put in an ammeter to measure the current through the resistors and a voltmeter to measure the potential difference across the 12 \u03a9 resistor. What would be the readinss in the ammeter and the voltmeter? \nAnswer: \n \nThe resistors of 5 \u03a9, 8 \u03a9 and 12 \u03a9 are connected in series. \nThe resistance of the circuit = 5 \u03a9 + 8 \u03a9 + 12 \u03a9 = 25 \u03a9. \nCurrent through the circuit (Ammeter reading) = I = \\(\\frac{6 \\mathrm{V}}{25 \\Omega}\\) = 0.24 A \nTherefore, ammeter reading = 0.24 A. \nVoltmeter reading (P.D across 12 \u03a9) = 12 \u03a9 \u00d7 0.24 A = 2.88 V. \nTherefore, voltmeter reading = 2.88 V.<\/p>\n <\/p>\n Question 79. \nA battery of 9 V is connected in series with resistors of 0.2 \u03a9, 0.3 \u03a9, 0.4 \u03a9, 0.5 \u03a9 and 12 \u03a9, respectively. How much current would flow through the 12 \u03a9 resistor? \nAnswer: \n \nTotal resistance of the circuit, R = 0.2 \u03a9 + 0.3 \u03a9 + 0.4 \u03a9 + 0.5 \u03a9 + 12 \u03a9 = 13.4 \u03a9 \nCurrent through the 12 \u03a9 resistor, I = \\(\\frac{V}{R}=\\frac{9 V}{13.4 \\Omega}\\) = 0.67 A \nEach resistor in the circuit carries the same current. Therefore, the current flowing through the 12 \u03a9 resistor is 0.67 A.<\/p>\n Question 80. \nAn electric lamp, whose resistance is 20 \u03a9, and a conductor of 4 \u03a9 resistance are connected to a 6 V battery as shown in the figure below. Calculate<\/p>\n \n- The total resistance of the circuit,<\/li>\n
- The current through the circuit, and<\/li>\n
- The potential difference across the electric lamp and conductor.<\/li>\n<\/ol>\n
Answer: \n \nGiven: Resistance of the electric lamp, R1<\/sub> = 20 \u03a9, Resistance of the conductor connected in series, R2<\/sub> = 4 \u03a9, Total resistance of the circuit, R = ?, Current, I = ?, Potential difference across the lamp VL<\/sub> = ?, Potential difference across the 4 \u03a9 resistor = ?<\/p>\n1. The total resistance in the circuit, R = R1<\/sub> + R2<\/sub> \n= 20 \u03a9 + 4 \u03a9 = 24 \u03a9.<\/p>\n2. Current through the circuit, I = \\(\\frac{V}{R}=\\frac{6 V}{24 \\Omega}\\) = 0.25 A.<\/p>\n 3. Potential difference across the lamp, VL<\/sub> = 20 \u03a9 \u00d7 0.25 A = 5 V \nPotential difference across the conductor = 4 \u03a9 \u00d7 0.25 A = 1 V.<\/p>\n<\/p>\n Question 81. \nJudee the equivalent resistance when the following are connected in parallel:<\/p>\n \n- 1 \u03a9 and 106<\/sup> \u03a9.<\/li>\n
- 1 \u03a9, 103<\/sup> \u03a9 and 106<\/sup> \u03a9.<\/li>\n<\/ol>\n
Answer: \n1. When 1 \u03a9 and 106<\/sup> \u03a9 are connected in parallel, the equivalent resistance R is given by: \n<\/p>\n2. When 1 \u03a9, 103<\/sup> \u03a9 and 106<\/sup> \u03a9 are connected in parallel, the equivalent resistance R is given by: \n<\/p>\nQuestion 82. \nHow can three resistors of resistances 2 \u03a9, 3 \u03a9, and 6 \u03a9 be connected to give a total resistance of<\/p>\n \n- 4 \u03a9<\/li>\n
- 1 \u03a9?<\/li>\n<\/ol>\n
Answer: \n1. We must connect 3 \u03a9 and 6 \u03a9 resistors in parallel. \nThe resulting resistance would be \\(=\\frac{3 \\Omega \\times 6 \\Omega}{3 \\Omega+6 \\Omega}\\) = 2 \u03a9 \nNow, the combination above must be connected in series with the 2 \u03a9 wire. \nThe resulting resistance would be = 2 \u03a9 + 2 \u03a9 = 4 \u03a9. \n<\/p>\n 2. To get 1 \u03a9, we must connect 2 \u03a9, 3 \u03a9, and 6 \u03a9 resistors in parallel, as shown below: \n \nLet the combined resistance be R. Now, \n<\/p>\n <\/p>\n Question 83. \nWhat is<\/p>\n \n- The highest,<\/li>\n
- The lowest total resistance that can be secured by combinations of four coils of resistance 4 \u03a9, 8 \u03a9, 12 \u03a9, 24 \u03a9?<\/li>\n<\/ol>\n
Answer: \n1. The highest resistance can be secured by combining all the given four resistors in series as shown below: \n \nThe highest total resistance of the system = R1<\/sub> + R2<\/sub> + R3<\/sub> + R4<\/sub> \n= 4 \u03a9 + 8 \u03a9 + 12 \u03a9 + 24 \u03a9 \n= 48 \u03a9.<\/p>\n2. The lowest total resistance is secured when the four resistors are in parallel arrangement as shown in the figure below. \n \nLet the combined resistance be R. Now, \n \n<\/p>\n <\/p>\n Question 84. \nShow how you would connect three resistors, each of resistance 6 \u03a9, so that the combination has a resistance of<\/p>\n \n- 9 \u03a9,<\/li>\n
- 4 \u03a9.<\/li>\n<\/ol>\n
Answer: \n1. Two resistors in parallel and the combination is in series with the third: \n \nCombined resistance of two 6 \u03a9 resistors in parallel = \\(\\frac{6 \\Omega \\times 6 \\Omega}{6 \\Omega+6 \\Omega}\\) = 3 \u03a9 \nWhen this combination is in series with the other 6 ? resistor, the total resistance = 6 \u03a9 + 3 \u03a9 = 9 \u03a9.<\/p>\n 2. Two resistors in series and the combination in series with the third: \n \nCombined resistance of two 6 ? resistors in series = 6 \u03a9 + 6 \u03a9 = 12 \u03a9. \nWhen this combination is in parallel with the other 6 \u03a9, \nTotal resistance = \n<\/p>\n Question 85. \nHow many 176 \u03a9 resistors (in parallel) are required to carry 5 A on a 220 V line? \nAnswer: \nLet the number of 176 \u03a9 resistors required to be connected in parallel to carry 5 A current on a 220 V line be n. \n \nThus, four resistors of 176 \u03a9 (in parallel) are required to carry 5 A on a 220 V line.<\/p>\n <\/p>\n Question 86. \nAn electric lamp of 100 ?, a toaster of resistance 50 ?, and a water filter of resistance 500 \u03a9 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? \nAnswer: \nGiven: Resistance of electric lamp, R1<\/sub> = 100 \u03a9, Resistance of toaster, R2<\/sub> = 50 \u03a9, Resistance of water filter, R3<\/sub> = 500 \u03a9, Potential difference of the source, V = 220 V. \nSince all the resistances are in parallel, their equivalent resistance R is given by: \n \nCurrent drawn by the three appliances, I \n \nThe resistance of electric iron which draws same current, R \n \nTherefore, the resistance of the electric iron is 31.25 \u03a9 and the current flowing through it is 7.04 A.<\/p>\nQuestion 87. | | | | | |