In-text Questions (Page 202)<\/span><\/p>\nQuestion 1.
\nName a device that helps to maintain a potential difference across a conductor.
\nAnswer:
\nBattery or Cell.<\/p>\n
Question 2.
\nWhat is meant by saying that the potential difference between two points is 1V?
\nAnswer:
\nThe potential difference between two points in a current carrying conductor is said to be one volt when one joule of work is done to move a charge of one coulomb from one point to the other.<\/p>\n
Question 3.
\nHow much energy is given to each coulomb of charge passing through a 6 V battery?
\nAnswer:
\nWe know that, the amount of energy or work is given by the formula.
\nW = VQ
\n= 6 V \u00d7 1 C
\n= 6 V – C = 6 J<\/p>\n
<\/p>\n
In-text Questions (Page 209)<\/span><\/p>\nQuestion 1.
\nOn what factors does the resistance of a conductor depend?
\nAnswer:
\nThe resistance of a conductor depends upon two factors.
\n(i) Length of the conductor ; Resistance of a conductor is directly proportional to its length i.e.,
\nR \u221d 1
\n(ii) Area of cross section of the conductor :
\nThe resistance of conductor is directly proportional to its area of cross section.<\/p>\n
Question 2.
\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:
\nCurrent will flow more easily through a thick wire because the resistance of a wire is inversely proportional to its area of cross section. Thick wire has a large surface area than a thin wire.<\/p>\n
Question 3.
\nLet 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?
\nAnswer:
\nLet ‘R’ be the resistance, and ‘V’ be the potential difference than ‘I’ current flow through the wire, so from Ohm’s Law
\nV = iR ……….(i)
\nWhen potential difference is half of its initial value than, from Ohm’s Law
\n\\(\\frac{\\mathrm{V}}{2}\\) = i1<\/sub> R …….(ii)
\nNow from equation (i) and (ii) we get
\n
\nThe current become half of its initial value (i).<\/p>\nQuestion 4.
\nWhy are coils of electric toasters and electric irons made of an alloy rather than a pure metal?
\nAnswer:
\nThe resistivity of an alloy is generally higher than that of its constituent metals but alloys do not oxidise readily at high temperature so they are commonly used in elecricity toasters and electric irons.<\/p>\n
<\/p>\n
Question 5.
\nUse the data in Table 12.2 to answer the following-
\n(a) Which among iron and mercury is a better conductor?
\n(b) Which material is the best conductor?
\nAnswer:
\n(a) Iron because its resistivity is lower than mercury.
\n(b) Silver is the best conductor because it has lowest resistivity.
\nTable 12.2 s Electrical resistivity of some substances at 20\u00b0C
\n
\n<\/p>\n
In-text Questions (Page 213)<\/span><\/p>\nQuestion 1.
\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 plug key, all connected in series.
\nAnswer:
\n<\/p>\n
Question 2.
\nRedraw the circuit of Question 1, putting in an ammeter to measure the current through the resistors and a voltmeter to measure the potential difference across the 12 \u03a9 resistor. What would be the readings in the ammeter and the voltmeter?
\nAnswer:
\n
\nAccording to Ohm’s Law
\nV = IR
\nsince R = R1<\/sub> + R2<\/sub> + R3<\/sub>
\n= 5\u03a9 + 8\u03a9 + 12\u03a9
\nR= 25 W
\nAnd V = 6 volt
\n6 volt = 1 \u00d7 25 Ohm
\nI = \\(\\frac{6 \\text { volt }}{25 \\mathrm{Ohm}}\\) = 0.24 Ampere
\nV1<\/sup> = -MR
\nV1<\/sup> = 0.24 \\(\\frac{\\text { volt }}{\\mathrm{Ohm}}\\) \u00d7 12 ohm
\nV1<\/sup> = 2.88 volt
\nSo the reading in the ammeter = 0.24 Ampere
\nAnd reading of the volmeter = 2.88 volt.<\/p>\n<\/p>\n
In-text Questions (Page 216)<\/span><\/p>\nQuestion 1.
\nJudge the equivalent resistance when the following are connected in parallel.
\n(A) 1 \u03a9 and 106<\/sup> \u03a9
\n(b) 1 \u03a9 and 103<\/sup> \u03a9, and 106<\/sup> \u03a9.
\nAnswer:
\n(a) Let R is the equivalent resistance
\n<\/p>\n(b) Let ‘R’ is the equivalent resistance.
\n<\/p>\n
Question 2.
\nAn electric lamp of 100 \u03a9, a toaster of resistance 50 \u03a9?, 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:
\n
\n<\/p>\n
Question 3.
\nWhat are the advantages of connecting electrical devices in parallel with the battery instead of connecting them in series?
\nAnswer:
\nIn series the current is constant throughout the electric circuit Thus, it is obviously impracticable to connect an electric bulb and and an electric heater in series because they require currents of widely different values to operate properly. Another major disadvantages of a series circuit is that when one component fails the circuit is broken and none of the components works. On the other hand, a parallel circuit divides the current through the electrical gadgets. The total resistance in a parallel circuit is decreased. This is helpful particularly when each gadget has different resistance and require different current to operate properly.<\/p>\n
<\/p>\n
Question 4.
\nHow can three resistors of resistances 2 \u03a9, 3 \u03a9, and 6 \u03a9 be connected to give a total resistance of
\n(a) 4 \u03a9
\n(b) 1 \u03a9?
\nAnswer:
\nIf 3 \u03a9 and 6 \u03a9 are in series than resultant resistant resistant say (R) can be calculated as
\n
\nAnd if R and 2 \u03a9 in series then resultant resistor (R1<\/sup>) can be calculated as
\nR1<\/sup> = R + 2\u03a9
\n= 2\u03a9 + 2\u03a9
\nR1<\/sup> = 4 \u03a9
\n<\/p>\n(b) If all the resistance are in parallel than resultant resistance (R) can be calculated as follows:
\n<\/p>\n
Question 5.
\nWhat is (a) the highest, (b) the lowest total resistance that can be secured by combinations of four coils of resistance 4 \u03a9, 8 \u03a9, 12 \u03a9, 24 \u03a9 ?
\nAnswer:
\n(a) When the four resistances are in parallel, the resultant resistance will be lowest.
\nLet the resultant resistance be ‘R’ so
\n<\/p>\n
(b) When the four resistance are in series, the resultant resistance will be highest.
\nLet the resultant resistant be ‘R’ so.
\nR1<\/sub> = 4\u03a9 + 8\u03a9 + 12\u03a9 + 24\u03a9
\nR1<\/sub> = 48 \u03a9<\/p>\nIn-text Questions (Page 2018)<\/span><\/p>\nQuestion 1.
\nWhy does the cord of an electric heater not glow while the heating element does?
\nAnswer:
\nThe cord of an electric heater not glow because it is less resistive while the heating element glow because it is purely resistive.<\/p>\n
Question 2.
\nCompute the heat generated while transferring 96000 coulomb of charge in one hour through a potential difference of 50 V.
\nAnswer:
\nWe know that the heat generated in time is
\nH = V I t
\nNow I = \\(\\frac{\\mathrm{Q}}{\\mathrm{t}}=\\frac{96000 \\mathrm{C}}{60 \\times 60 \\mathrm{Sec}}\\)
\n= 26.67 Amp.
\nnow, H = V I t
\n= 50 volt \u00d7 26.67 Amp. \u00d7 3600 Sec
\n= 4800600 J<\/p>\n
Question 3.
\nAn electric iron of resistance 20 \u03a9 takes a current of 5 A. Calculate the heat developed in 30 sec.
\nAnswer:
\nGiven I = 5 A
\nR = 20 W
\nt = 30 S
\nH = ?
\nH = I2<\/sup> R t
\n= (5 A)2<\/sup> \u00d7 20 \u03a9 \u00d7 30 Sec
\n= 25 A2<\/sup> \u00d7 20 \u03a9 \u00d7 30Sec
\n= 15000 J<\/p>\n<\/p>\n
In-text Questions (Page 220)<\/span><\/p>\nQuestion 1.
\nWhat determines the rate at energy is delivered by a current ?
\nAnswer:
\nIt is electric power (p) given by the formula,
\nP = V I = I2<\/sup>R = \\(\\frac{\\mathrm{V}^{2}}{\\mathrm{R}}\\)<\/p>\nQuestion 2.
\nAn electric motor takes 5 A from a 220 V line. Determine the power of the motor and the energy consumed in 2 h.
\nAnswer:
\nGiven I = 5 A
\nV = 220 V
\nsince P = V \u00d7 I
\n= 220 V \u00d7 5 A
\n= 1100 V – A
\nP = 1100 W
\nEnergy consumed in ‘2’ hours
\n= V \u00d7 I \u00d7 t
\n= 220 volt \u00d7 5 A \u00d7 2 \u00d7 60 \u00d7 60 sec
\n= 7920000 J<\/p>\n
Class 10 Science Chapter 12 Electricity Textbook Questions and Answers<\/h3>\n
Page no. 221<\/span><\/p>\nQuestion 1.
\nA 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’ then the ration R\/R’ is
\n(a) \\(\\frac { 1 }{ 25 }\\)
\n(b) \\(\\frac { 1 }{ 5 }\\)
\n(c) 5
\n(d) 25
\nAnswer:
\n(d) 25<\/p>\n
Question 2.
\nWhich of the following terms does not represent electric power in a circuit ?
\n(a) I2<\/sup>R
\n(b) IR2<\/sup>
\n(c) V I
\n(d) \\(\\frac{\\mathrm{V}^{2}}{\\mathrm{R}}\\)
\nAnswer:
\n(b) IR2<\/sup><\/p>\nQuestion 3.
\nAn electric bulb is rated 230 V and 100 W. When it is operated on 110 V, the power consumed will be –
\n(a) 100 W
\n(b) 75 W
\n(c) 50 W
\n(d) 25 W
\nAnswer:
\n(d) 25 W<\/p>\n
Question 4.
\nTwo conducting wires of the same material and of equal lengths and equal diameters are first connected in series and then parallel in an electric circuit. The ratio of heat produced in series and parallel combinations would be:
\n(a) 1 : 2
\n(b) 2 : 1
\n(c) 1 : 4
\n(d) 4 : 1
\nAnswer:
\n(c) 1 : 4<\/p>\n
<\/p>\n
Question 5.
\nHow is a voltmeter connected in the circuit to measure the potential difference between two points ?
\nAnswer:
\nA voltmeter is connected in the circuit in parallel to measure the potential difference between two points.<\/p>\n
Question 6.
\nA copper wire has diameter 0.5 mm and resistivity of 1.6 \u00d7 10-8<\/sup> \u2126 m. What will be the length of this wire to make its resistance 10 W ? How much does the resistance change if the diameter is doubled ?
\nAnswer:
\nGiven
\nDiameter = 0.5 mm
\nso radius (r) = \\(\\frac {0.5}{2}\\) mm
\n= \\(\\frac{0.5}{2 \\times 10^{3}}\\) metre
\nResistivity (\u03c1) = 1.6 \u00d7 10 10-8<\/sup> \u2126m
\nR = 10 W
\nA = \u03c0r\u00b2
\nl = ?
\nWe know that
\n
\n
\nSecond Case
\n<\/p>\nQuestion 7.
\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:
\nResistance (R) = tan Q = \\(\\frac{\\mathrm{BC}}{\\mathrm{AB}}\\)
\n<\/p>\n
<\/p>\n
Question 8.
\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:
\nBy Ohm’s Law
\n<\/p>\n
Question 9.
\nA battery of 9 V is connected in series with resistors of 0.2 \u2126, 0.3 \u2126, 0.5 \u2126, and 12 \u2126 respectively. How much current would flow through the 12 \u2126 resistor ?
\nAnswer:
\nLet the resultant resistor be R.
\nSo, R = 0.2 \u2126 + 0.3 \u2126 + 0.4 \u2126 + 0.5 \u2126 + 12 \u2126
\nR = 13.4 \u2126
\nBy Ohm’s Law
\nV = I \u00d7 R
\nI = \\(\\frac{V}{R}=\\frac{9 V}{13.4} \\mathrm{Ohm}\\)
\n= 0.671 A.
\nIn series current remain same in all the resistors. So current in 12 \u2126 resistor will be 0.671 Amp.<\/p>\n
Question 10.
\nHow many 176 \u2126 resistors (in parallel) are required to carry 5 A on a 220 V line ?
\nAnswer:
\nBy Ohm’s Law
\nV= I \u00d7 R
\nR = \\(\\frac{\\mathrm{V}}{\\mathrm{I}}=\\frac{220 \\mathrm{Volt}}{5 \\text { Ampere }}\\)
\n= 44 Ohm
\nLet ‘n’ resistors are connected in parallel.
\n
\nSo the no. of resistors will be ‘4’.<\/p>\n
Question 11.
\nShow how you would connect three resistors each of resistance 6 \u2126, so that the combination has a resistance of (i) 9 \u2126, (b) 4 \u2126.
\nAnswer:
\n(i)
\n
\nTwo resistance in parallel and their resultant should be in series with the third resistance.
\n
\nTwo resistance in series and their resultant should be parallel with the third resistance.<\/p>\n
Question 12.
\nSeveral electric bulbs designed to be used on a 220 V electric supply line, are rated 10 \u2126. How many lamps can be connected in parallel with each other across the two wires of 220 V line if the maximum allowable current is 5 ?
\nAnswer:
\nThe resistance of an electric bulb can be calculated as follows :
\nP = \\(\\frac{\\mathrm{V}_{2}}{\\mathrm{R}}\\)
\nR = \\(\\frac{220 \\times 220}{10}\\) = 22 \u00d7 220
\n= 4840 Ohm.
\nIf the allowable current is 5 A and voltage is 220 V than resistance
\nR1<\/sup> = \\(\\frac{V}{I}=\\frac{220 \\mathrm{~V}}{5 \\mathrm{Amp}}\\)
\n= 44 Ohm.
\nLet the number of lamps are ‘n’, so
\n
\nSo the number of lamps will be 110.<\/p>\nQuestion 13.
\nA hot plate of an electric oven connected to a 220 V line has two resistance coils A and B, each of 24 \u2126 resistance, which may be used separately, in series or in parallel. What are the currents in the three cases ?
\nAnswer:
\nCase I when the coils are in series,
\nLet resultant resistance be R, so
\nR = 24 \u2126 + 24 \u2126 = 48 \u2126
\nNow, I = \\(\\frac{\\mathrm{V}}{\\mathrm{R}}=\\frac{220 \\mathrm{~V}}{48 \\Omega}\\) = 4.58 A
\nCase when the coil ‘A’ and ‘B’ are in parallel, let the resultant resistance be ‘R’
\n<\/p>\n
Question 14.
\nCompare the power used in the 2 \u2126 resistor in each of the following circuit (i) a 6 V battery in series with 1 \u2126 and 2 \u2126 resistors, and (ii) a 4 V battery in parallel with 12 \u2126 and 2 \u2126 resistors.
\nAnswer:
\n(i) The resultant resistance (R)
\n= 1 \u2126 + 2 \u2126 = 3 \u2126
\n= (R1<\/sub>)(R2<\/sub>)
\n
\n(ii) When a battery in parallel with 12 \u2126 and 2 \u2126 resistors, then voltage remain same in both the resistance. Current will be different.
\nSo power (P) = \\(\\frac{V^{2}}{R}=\\frac{(4)^{2}}{2}\\)
\n= \\(\\frac {16}{2}\\) = 8 J
\n\u2234 \\(\\frac{P}{P^{1}}=\\frac{8 J}{8 J}\\)
\nP : P1<\/sup> = 1 : 1<\/p>\n<\/p>\n
Question 15.
\nTwo lamps, one rated 100 W at 220 V, and the other 60 W at 220 V, are connected in parallel to electric mains supply. What current is drawn from the line if the supply voltage is 220 V ?
\nAnswer:
\nResistance of bulb rated 100 W at 220 say R1
\n
\nfor the bulb rated 60 W at 220 V, say resistance is R2<\/sub>
\nSo, R2<\/sub> = \\(\\frac{\\mathrm{V}^{2}}{\\mathrm{P}}=\\frac{220 \\times 220}{60}\\)
\n= 806.6 Ohm
\nWhen R1<\/sub> and R2<\/sub> are in series, the resultant resistance say R, so
\n<\/p>\nQuestion 16.
\nWhich uses more energy, a 250 W.T.V. set in 1 hr, or a 1200 W toaster in 10 minutes ?
\nAnswer:
\nEnergy used by T.V. in one second = 250 J
\nSo energy used by T.V. in 1 hr i.e., 3600 sec
\n= 250 \u00d7 3600 J = 900000 J
\nEnergy used by a toaster in one second = 1200 J
\nSo energy used by a toaster in 10 minutes i.e., 600 sec
\n= 1200 \u00d7 600 J = 720000 J
\nHence T.V. will used more energy than a toaster.<\/p>\n
Question 17.
\nAn electric heater of resistance 8 W draws 15 A from the service mains 2 hours. Calculate the rate of which heat is developed in the heater.
\nAnswer:
\nThe rate at which heat is developed is called power.
\nSo P = V \u00d7 I
\n= \\(\\frac{\\mathrm{V}^{2}}{\\mathrm{R}}\\)
\nSince V = I \u00d7 R
\nSo, = I2<\/sup> \u00d7 R
\n= (15A)2<\/sup> \u00d7 8 Ohm
\n= 225 A2<\/sup> \u00d7 8 Ohm
\n= 1800 J\/Sec<\/p>\nQuestion 18.
\nExplain the following:
\n(a) Why is the tungsten used almost exclusively for filament of electric lamps ?
\n(b) Why are the conductors of electric heating devices, such as bread-toasters and electric irons, made of an alloy rather than a pure metal ?
\n(c) Why is the series arrangement not used for domestic circuits.
\n(d) How does the resistance of a wire vary with its area on cross-section ?
\n(e) Why are copper and aluminium wires usually employed for electricity transmission ?
\nAnswer:
\n(a) The resistivity of tungsten does not change with high temperature. It has very high melting point. It can operate upto 2700\u00b0 C. So tungsten is almost used exclusively for filament of electricity lamps.
\n(b) Because alloys do not oxidise on heating at high temperature.
\n(c) Because when light are in series if one bulb fuses or one tube light stops working, the circuit breaks and all light go off.
\n(d) The resistance of a given wire is inversely proportional to area of cross-section of the wire.
\n(e) Because copper and aluminium have low resistivity that is why they are good conductors of electricity so they are usually employed for electricity transmission.<\/p>\n
Class 10 Science Chapter 12 Electricity Textbook Activities<\/h3>\n
Activity 12.1 (Page 203)<\/span><\/p>\n\n- Set up a circuit as shown in Fig,, consisting of a nichrome wire XY of length, say 0.5 m, an ammeter, a voltmeter and four cells of 1.5 V each. (Nichrome is an alloy of nickel, chromium, manganese, and iron metals.)<\/li>\n<\/ul>\n
<\/p>\n
\n- First use only one cell as the source in the circuit. Note the reading in the ammeter 1, for the current and reading of the voltmeter V for the potential difference across the nichrome wire XY in the circuit. Tabulate them in the Table given.<\/li>\n
- 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.<\/li>\n
- Repeat the above steps using three cells and then four cells in the circuit separately.<\/li>\n<\/ul>\n
Question 1.
\nCalculate the ratio of V to I for each pair of potential difference V and current J.
\nAnswer:
\n<\/p>\n
Question 2.
\nPlot a graph between V and I, and observe the nature of the graph.
\nAnswer:
\nIn this Activity, you will find that approximately the same value for V\/I is obtained in each case. Thus the V-I graph is a straight line that passes through the origin of the graph, as shown in Fig. Thus, V\/I is a constant ratio.
\n<\/p>\n
<\/p>\n
Activity 12.1 (Page 205)<\/span><\/p>\n\n- Take a nichrome wire, a torch bulb, a 10 W bulb and an ammeter (0 – 5 A range), a plug key and some connecting wires.<\/li>\n
- 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 Fig.<\/li>\n<\/ul>\n
<\/p>\n
\n- 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. ]<\/li>\n
- Replace the nichrome wire with the torch bulb in the circuit and find the current through it by measuring the reading of the ammeter.<\/li>\n
- Now repeat the above step with the 10 W bulb in the gap XY.<\/li>\n
- Are the ammeter readings differ for different components connected in the gap XY? What do the above observations indicate?<\/li>\n
- You may repeat this Activity by keeping any material component in the gap. Observe the ammeter readings in each case. Analyse the observations.<\/li>\n<\/ul>\n
Question 1.
\nIn this Activity we observe that the current is different for different components. Why do they differ?
\nAnswer:
\nCertain 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.<\/p>\n
Activity 12.3 (Page 206)<\/span><\/p>\n