{"id":6896,"date":"2024-01-24T14:05:34","date_gmt":"2024-01-24T14:05:34","guid":{"rendered":"https:\/\/theexampillar.com\/ncert\/?p=6896"},"modified":"2024-01-24T14:10:19","modified_gmt":"2024-01-24T14:10:19","slug":"ncert-solutions-class-10-science-chapter-12-electricity","status":"publish","type":"post","link":"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-10-science-chapter-12-electricity\/","title":{"rendered":"NCERT Solutions Class 10 (Science) Chapter 12 (Electricity)"},"content":{"rendered":"<h2 style=\"text-align: center;\"><span style=\"font-family: georgia, palatino, serif;\"><strong><span style=\"color: #000000;\">NCERT Solutions Class 10 Science\u00a0<\/span><\/strong><\/span><\/h2>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\">The NCERT Solutions in English Language for Class 10 Science <strong>Chapter &#8211; 12 (Electricity) <\/strong>has been provided here to help the students in solving the questions from this exercise.\u00a0<\/span><\/p>\n<h2 style=\"text-align: center;\"><span style=\"font-family: georgia, palatino, serif;\"><strong><span style=\"color: #000000;\">Chapter &#8211; 12 (Electricity)\u00a0<\/span><\/strong><\/span><\/h2>\n<p style=\"text-align: center;\"><span style=\"color: #ff0000; font-size: 14pt; font-family: georgia, palatino, serif;\"><strong>Questions<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"font-family: georgia, palatino, serif;\"><strong><span style=\"color: #000000;\">1. <\/span><span style=\"color: #000000;\">What does an electric circuit mean ?\u00a0<\/span><\/strong><\/span><br \/>\n<span style=\"color: #000000; font-family: georgia, palatino, serif;\"><strong>Answer &#8211; <\/strong>An electric circuit consists of electric devices, switching devices, source of electricity, etc. that are connected by conducting wires.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"font-family: georgia, palatino, serif;\"><strong><span style=\"color: #000000;\">2. <\/span><span style=\"color: #000000;\">Define the unit of current.\u00a0<\/span><\/strong><\/span><br \/>\n<span style=\"font-family: georgia, palatino, serif;\"><span style=\"color: #000000;\"><strong>Answer &#8211; <\/strong><\/span><span style=\"color: #000000;\">Unit of electric current is ampere. Electric current in a conductor is said to be 1 A if 1 coulomb charge flows through the cross-section of the conductor in 1 second.<\/span><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"font-family: georgia, palatino, serif;\"><strong><span style=\"color: #000000;\">3. <\/span><span style=\"color: #000000;\">Calculate the number of electrons consisting one coulomb of charge.\u00a0<\/span><\/strong><\/span><br \/>\n<span style=\"font-family: georgia, palatino, serif;\"><span style=\"color: #000000;\"><strong>Answer &#8211;\u00a0<\/strong><\/span>Charge on one electron = 1.6 x 10<sup>-19<\/sup>\u00a0coulomb.<\/span><br \/>\n<span style=\"font-family: georgia, palatino, serif;\">No of electron in one coulomb of charge = 1\/1.6 x 10<sup>-19<br \/>\n<\/sup>= 6.25 x 10<sup>18<\/sup><\/span><\/p>\n<p style=\"text-align: center;\"><span style=\"color: #ff0000; font-size: 14pt; font-family: georgia, palatino, serif;\"><strong>Questions<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"font-family: georgia, palatino, serif;\"><strong><span style=\"color: #000000;\">1. <\/span><span style=\"color: #000000;\">Name a device that helps to maintain a potential difference across a conductor. (CBSE 2008)<\/span><\/strong><\/span><br \/>\n<span style=\"font-family: georgia, palatino, serif;\"><span style=\"color: #000000;\"><strong>Answer &#8211; <\/strong><\/span><span style=\"color: #000000;\">A cell or battery.<\/span><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"font-family: georgia, palatino, serif;\"><strong><span style=\"color: #000000;\">2. <\/span><span style=\"color: #000000;\">What is meant by saying that a potential difference between two points is 1 V ?<br \/>\n<\/span><\/strong><span style=\"color: #000000;\"><strong>Answer &#8211; <\/strong><\/span><span style=\"color: #000000;\">Potential difference between two points is 1 V if 1 joule work is done in moving 1 coulomb charge from one point to another point.<\/span><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"font-family: georgia, palatino, serif;\"><strong><span style=\"color: #000000;\">3. <\/span><span style=\"color: #000000;\">How much energy is given to each coulomb of charge passing through a 6 V battery ?<\/span><\/strong><\/span><br \/>\n<span style=\"font-family: georgia, palatino, serif;\"><span style=\"color: #000000;\"><strong>Answer &#8211; <\/strong><\/span><span style=\"color: #000000;\">Energy = Charge \u00d7 Potential difference<br \/>\n= 1 C \u00d7 6 V = 6 J.<\/span><\/span><\/p>\n<p style=\"text-align: center;\"><span style=\"color: #ff0000; font-size: 14pt; font-family: georgia, palatino, serif;\"><strong>Questions<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\"><strong>1. On what factors does the resistance of a conductor depend ?<\/strong><\/span><br \/>\n<span style=\"color: #000000; font-family: georgia, palatino, serif;\"><strong>Answer &#8211; <\/strong>The resistance of a conductor depends upon the following factors:<\/span><\/p>\n<ul style=\"text-align: justify;\">\n<li><span style=\"color: #000000; font-family: georgia, palatino, serif;\">Length of the conductor. <\/span><\/li>\n<li><span style=\"color: #000000; font-family: georgia, palatino, serif;\">Cross-sectional area of the conductor.<\/span><\/li>\n<li><span style=\"color: #000000; font-family: georgia, palatino, serif;\">Material of the conductor<\/span><\/li>\n<li><span style=\"color: #000000; font-family: georgia, palatino, serif;\">Temperature of the conducto\u00a0<\/span><\/li>\n<\/ul>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\"><strong>2. Will current flow more easily through a thick wire or a thin wire of the same material, when connected to the same source Why ?<\/strong><\/span><br \/>\n<span style=\"color: #000000; font-family: georgia, palatino, serif;\"><strong>Answer &#8211;\u00a0 <\/strong>The current flows more easily through a thick wire as compared to thin wire of the same material, when connected to the same source. It is due to the reason that resistance increases with decrease in thickness.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\"><strong>3. <b>Let the resistance of an electric 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?<br \/>\n<\/b><\/strong><strong>Answer &#8211; <\/strong>It is given that resistance R of the electrical component remains constant but the potential difference across the ends of the component decreases to half of its value.<\/span><br \/>\n<span style=\"color: #000000; font-family: georgia, palatino, serif;\">Hence, as per Ohm\u2019s law, new current also decreases to half of its original value.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\"><strong>4. Why are coil of electric toasters and electric irons made of an alloy rather than a pure metal ?<\/strong><\/span><br \/>\n<span style=\"color: #000000; font-family: georgia, palatino, serif;\"><strong>Answer &#8211; <\/strong>The resistivity of an alloy is higher than the pure metal. Moreover, at high temperatures, the alloys do not melt readily. Hence, the coils of heating appliances such as electric toasters and electric irons are made of an alloy rather than a pure metal.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\"><strong>5. <\/strong><b>Use the data in Table 12.2 to answer the following:<br \/>\n<\/b>(a) Which among iron and mercury is a better conductor?<\/span><br \/>\n<span style=\"color: #000000; font-family: georgia, palatino, serif;\">(b) Which material is the best conductor?<\/span><br \/>\n<span style=\"color: #000000; font-family: georgia, palatino, serif;\"><strong>Answer &#8211;\u00a0<\/strong>Table 12.2 Electrical resistivity of some substances at 20\u00b0C<\/span><\/p>\n<table style=\"border-collapse: collapse; width: 52.3412%; height: 390px;\">\n<tbody>\n<tr style=\"height: 24px;\">\n<td style=\"width: 33.3333%; height: 24px;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\"><strong>\u2212<\/strong><\/span><\/td>\n<td style=\"width: 33.3333%; height: 24px;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\"><strong>Material<\/strong><\/span><\/td>\n<td style=\"width: 33.3333%; height: 24px;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\"><strong>Resistivity (\u03a9 m)<\/strong><\/span><\/td>\n<\/tr>\n<tr style=\"height: 27px;\">\n<td style=\"width: 33.3333%; height: 243px;\" rowspan=\"9\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\"><strong>Conductors<\/strong><\/span><br \/>\n<span style=\"color: #000000; font-family: georgia, palatino, serif;\"><strong><br \/>\n<\/strong><strong><br \/>\n<\/strong><strong><br \/>\n<\/strong><strong><br \/>\n<\/strong><strong><br \/>\n<\/strong><strong><br \/>\n<\/strong><strong><br \/>\n<\/strong><strong><br \/>\n<\/strong><strong><br \/>\n<\/strong><\/span><\/td>\n<td style=\"width: 33.3333%; height: 27px;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\">Silver<\/span><\/td>\n<td style=\"width: 33.3333%; height: 27px;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\">1.60 \u00d7 10<sup>\u22128<\/sup><\/span><\/td>\n<\/tr>\n<tr style=\"height: 27px;\">\n<td style=\"width: 33.3333%; height: 27px;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\">Copper<\/span><\/td>\n<td style=\"width: 33.3333%; height: 27px;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\">1.62 \u00d7 10<sup>\u22128<\/sup><\/span><\/td>\n<\/tr>\n<tr style=\"height: 27px;\">\n<td style=\"width: 33.3333%; height: 27px;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\">Aluminium<\/span><\/td>\n<td style=\"width: 33.3333%; height: 27px;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\">2.63 \u00d7 10<sup>\u22128<\/sup><\/span><\/td>\n<\/tr>\n<tr style=\"height: 27px;\">\n<td style=\"width: 33.3333%; height: 27px;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\">Tungsten<\/span><\/td>\n<td style=\"width: 33.3333%; height: 27px;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\">5.20 \u00d7 10<sup>\u22128<\/sup><\/span><\/td>\n<\/tr>\n<tr style=\"height: 27px;\">\n<td style=\"width: 33.3333%; height: 27px;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\">Nickel<\/span><\/td>\n<td style=\"width: 33.3333%; height: 27px;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\">6.84 \u00d7 10<sup>\u22128<\/sup><\/span><\/td>\n<\/tr>\n<tr style=\"height: 27px;\">\n<td style=\"width: 33.3333%; height: 27px;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\">Iron<\/span><\/td>\n<td style=\"width: 33.3333%; height: 27px;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\">10.0 \u00d7 10<sup>\u22128<\/sup><\/span><\/td>\n<\/tr>\n<tr style=\"height: 27px;\">\n<td style=\"width: 33.3333%; height: 27px;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\">Chromium<\/span><\/td>\n<td style=\"width: 33.3333%; height: 27px;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\">12.9 \u00d7 10<sup>\u22128<\/sup><\/span><\/td>\n<\/tr>\n<tr style=\"height: 27px;\">\n<td style=\"width: 33.3333%; height: 27px;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\">Mercury<\/span><\/td>\n<td style=\"width: 33.3333%; height: 27px;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\">94.0 \u00d7 10<sup>\u22128<\/sup><\/span><\/td>\n<\/tr>\n<tr style=\"height: 27px;\">\n<td style=\"width: 33.3333%; height: 27px;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\">Manganese<\/span><\/td>\n<td style=\"width: 33.3333%; height: 27px;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\">1.84 \u00d7 10<sup>\u22126<\/sup><\/span><\/td>\n<\/tr>\n<tr style=\"height: 24px;\">\n<td style=\"width: 33.3333%; height: 96px;\" rowspan=\"3\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\"><strong>Alloys<\/strong><\/span><br \/>\n<span style=\"color: #000000; font-family: georgia, palatino, serif;\"><strong><br \/>\n<\/strong><strong><br \/>\n<\/strong><\/span><\/td>\n<td style=\"width: 33.3333%; height: 24px;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\">Constantan (alloy of Cu and Ni)<\/span><\/td>\n<td style=\"width: 33.3333%; height: 24px;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\">49 \u00d7 10<sup>\u22126<\/sup><\/span><\/td>\n<\/tr>\n<tr style=\"height: 24px;\">\n<td style=\"width: 33.3333%; height: 24px;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\">Manganin\u00a0(alloy of Cu, Mn and Ni)<\/span><\/td>\n<td style=\"width: 33.3333%; height: 24px;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\">44 \u00d7 10<sup>\u22126<\/sup><\/span><\/td>\n<\/tr>\n<tr style=\"height: 48px;\">\n<td style=\"width: 33.3333%; height: 48px;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\">Nichrome\u00a0(alloy of Ni, Cr, Mn and Fe)<\/span><\/td>\n<td style=\"width: 33.3333%; height: 48px;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\">100 \u00d7 10<sup>\u22126<\/sup><\/span><\/td>\n<\/tr>\n<tr style=\"height: 27px;\">\n<td style=\"width: 33.3333%; height: 27px;\" rowspan=\"5\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\"><strong>Insulators<\/strong><\/span><br \/>\n<span style=\"font-family: georgia, palatino, serif;\"><span style=\"color: #000000;\"><strong><br \/>\n<\/strong><\/span><span style=\"color: #000000;\"><strong><br \/>\n<\/strong><\/span><span style=\"color: #000000;\"><strong><br \/>\n<\/strong><\/span><span style=\"color: #000000;\"><strong><br \/>\n<\/strong><\/span><\/span><\/td>\n<td style=\"width: 33.3333%; height: 27px;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\">Glass<\/span><\/td>\n<td style=\"width: 33.3333%; height: 27px;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\">10<sup>10<\/sup>\u00a0\u2212 10<sup>14<\/sup><\/span><\/td>\n<\/tr>\n<tr>\n<td style=\"width: 33.3333%;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\">Hard rubber<\/span><\/td>\n<td style=\"width: 33.3333%;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\">10<sup>13<\/sup>\u00a0\u2212 10<sup>16<\/sup><\/span><\/td>\n<\/tr>\n<tr>\n<td style=\"width: 33.3333%;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\">Ebonite<\/span><\/td>\n<td style=\"width: 33.3333%;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\">10<sup>15<\/sup>\u00a0\u2212 10<sup>17<\/sup><\/span><\/td>\n<\/tr>\n<tr>\n<td style=\"width: 33.3333%;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\">Diamond<\/span><\/td>\n<td style=\"width: 33.3333%;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\">10<sup>12<\/sup>\u00a0\u2212 10<sup>13<\/sup><\/span><\/td>\n<\/tr>\n<tr>\n<td style=\"width: 33.3333%;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\">Paper (dry)<\/span><\/td>\n<td style=\"width: 33.3333%;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\">10<sup>12<\/sup><\/span><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\">(a) Resistivity of iron = 10.0\u00d710<sup>\u22128<\/sup>\u03a9m<br \/>\nResistivity of mercury =94.0\u00d710<sup>\u22128 <\/sup>\u03a9m.<br \/>\nResistivity of mercury is more than that of iron. This implies that iron is a better conductor than mercury.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\">(b) It can be observed from Table 12.2 that the resistivity of silver is the lowest among the listed materials. Hence, it is the best conductor.<\/span><\/p>\n<p style=\"text-align: center;\"><span style=\"color: #ff0000; font-size: 14pt; font-family: georgia, palatino, serif;\"><strong>Questions<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\"><strong>1. Draw a schematic diagram of a circuit consisting of a batteries of three of 2 V each, a 5 \u03a9 resistor, 8 \u03a9 resistor and a 12 \u03a9 resistor and a plug key, all connected in series.\u00a0<\/strong><\/span><br \/>\n<span style=\"color: #000000; font-family: georgia, palatino, serif;\"><strong>Answer &#8211; <\/strong>The schematic diagram of circuit is as follows:<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-medium wp-image-7390\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2024\/01\/NCERT-Class-10-Science-Solution-Ch-12-Ans.1-300x123.png\" alt=\"NCERT Class 10 Science Solution\" width=\"300\" height=\"123\" srcset=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2024\/01\/NCERT-Class-10-Science-Solution-Ch-12-Ans.1-300x123.png 300w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2024\/01\/NCERT-Class-10-Science-Solution-Ch-12-Ans.1.png 355w\" sizes=\"auto, (max-width: 300px) 100vw, 300px\" \/><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\"><strong>2. Redraw the circuit of question 12, putting an ammeter to measure the current through the resistor and a voltmeter to measure the potential difference across 12 \u03a9 resistor. What would be the reading in the ammeter ?<\/strong><\/span><br \/>\n<span style=\"color: #000000; font-family: georgia, palatino, serif;\"><strong>Answer &#8211;\u00a0<\/strong>Here ammeter A has been joined in series of circuit and voltmeter V is joined in parallel to 12 ohms\u2019 resistor. <\/span><br \/>\n<span style=\"color: #000000; font-family: georgia, palatino, serif;\">Total voltage of battery V = 3\u00d72 = 6 V. <\/span><br \/>\n<span style=\"color: #000000; font-family: georgia, palatino, serif;\">Total resistance R = R<sub>1<\/sub>+ R<sub>2<\/sub>+ R<sub>3<\/sub> = 5 \u03a9 + 8 \u03a9 + 12 \u03a9 = 25 \u03a9<\/span><br \/>\n<span style=\"color: #000000; font-family: georgia, palatino, serif;\">Ammeter reading (current) = I = V\/R = 6\/25 = 0.24 A. <\/span><br \/>\n<span style=\"color: #000000; font-family: georgia, palatino, serif;\">Voltmeter reading = IR = 0.24 \u00d7 12 = 2.88 V.<\/span><\/p>\n<p style=\"text-align: center;\"><span style=\"color: #ff0000; font-size: 14pt; font-family: georgia, palatino, serif;\"><strong>Questions<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\"><strong>1.\u00a0Judge the equivalent resistance when the following are connected in parallel :<br \/>\n<\/strong>(a) 1 \u03a9 and 10<sup>6<\/sup>\u00a0\u03a9<\/span><br \/>\n<span style=\"color: #000000; font-family: georgia, palatino, serif;\">(b) 1 \u03a9 and 10<sup>3<\/sup>\u00a0\u03a9 and 10<sup>6<\/sup>\u00a0\u03a9.<\/span><br \/>\n<span style=\"color: #000000; font-family: georgia, palatino, serif;\"><strong>Answer &#8211;\u00a0 <\/strong>When the resistances are joined in parallel, the resultant resistance in parallel arrangement is given by:<\/span><br \/>\n<span style=\"color: #000000; font-family: georgia, palatino, serif;\">1\/R = 1\/R<sub>1<\/sub>\u00a0+ 1\/R<sub>2<\/sub>\u00a0+ 1\/R<sub>3<br \/>\n<\/sub><strong>(a)\u00a0<\/strong>1\/R = 1\/1+ 1\/10<sup>6<\/sup>\u00a0= 1+ 10<sup>-6<br \/>\n<\/sup>R = 1 \u03a9<\/span><br \/>\n<span style=\"color: #000000; font-family: georgia, palatino, serif;\"><strong>(b)\u00a0<\/strong>1\/R = 1\/1 + 1\/10<sup>3<\/sup>\u00a0+ 1\/10<sup>6<\/sup>\u00a0= 1 + 10<sup>-3\u00a0<\/sup>+ 10<sup>-6<br \/>\n<\/sup>R = 1 \u03a9<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\"><strong>2. An electric lamp of 100 W, a toaster of resistance 50 \u03a9, and a water filter of resistance 500 \u03a9 are connected in parallel to 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 ?<\/strong><\/span><br \/>\n<span style=\"color: #000000; font-family: georgia, palatino, serif;\"><strong>Answer &#8211;\u00a0<\/strong>Here, voltage (V) = 220 V<\/span><br \/>\n<span style=\"color: #000000; font-family: georgia, palatino, serif;\">R<sub>1\u00a0<\/sub>= 100 \u03a9, <\/span><br \/>\n<span style=\"color: #000000; font-family: georgia, palatino, serif;\">R<sub>2<\/sub> = 50 \u03a9 <\/span><br \/>\n<span style=\"color: #000000; font-family: georgia, palatino, serif;\">R<sub>3\u00a0<\/sub>= 500 \u03a9 <\/span><br \/>\n<span style=\"color: #000000; font-family: georgia, palatino, serif;\">1\/R = 1\/R<sub>1<\/sub>\u00a0+ 1\/R<sub>2<\/sub>\u00a0+ 1\/R<sub>3<br \/>\n<\/sub>1\/R = 1\/100+1\/50 +1\/500 = 16\/500 <\/span><br \/>\n<span style=\"color: #000000; font-family: georgia, palatino, serif;\">R = 500\/16 = 31.25 \u03a9 <\/span><br \/>\n<span style=\"color: #000000; font-family: georgia, palatino, serif;\">The resistance of electric iron, which draws as much current as all three appliances take together = R = 31.25 \u03a9. <\/span><br \/>\n<span style=\"color: #000000; font-family: georgia, palatino, serif;\">Current passing through electric iron (I) = V\/R = 220\/31.25 = 7.04 A.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\"><strong>3. <\/strong><b>What are the advantages of connecting electrical devices in parallel with the battery instead of connecting them in series?<\/b><br \/>\n<strong>Answer &#8211;\u00a0<\/strong>Advantage of connecting electrical devices in parallel with the battery are as follows:<\/span><br \/>\n<span style=\"color: #000000; font-family: georgia, palatino, serif;\">(i) Voltage across each connecting electrical device is same and device take current as per its resistance.<\/span><br \/>\n<span style=\"color: #000000; font-family: georgia, palatino, serif;\">(ii) Separate on\/off switches can be applied across each device.<\/span><br \/>\n<span style=\"color: #000000; font-family: georgia, palatino, serif;\">(iii) Total resistance in parallel circuit decreases, hence, a great current may be drawn from cell.<\/span><br \/>\n<span style=\"color: #000000; font-family: georgia, palatino, serif;\">(iv) If one electrical device is damaged; then other devices continue to work properly.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\"><strong>4. <b>How can three resistors of resistance 2 \u03a9, 3 \u03a9 and 6 \u03a9 be connected to give a total resistance of<br \/>\n(a) 4 \u03a9<br \/>\n(b) 1 \u03a9?<br \/>\n<\/b><\/strong><strong>Answer &#8211; <\/strong>There are three resistors of resistances 2 \u03a9, 3 \u03a9, and 6 \u03a9 respectively.<br \/>\n<strong>(a)\u00a0<\/strong>The following circuit diagram shows the connection of the three resistors.<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-7392\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2024\/01\/NCERT-Class-10-Science-Solution-Ch-12-Ans.3.jpg\" alt=\"NCERT Class 10 Science Solution\" width=\"177\" height=\"162\" \/><br \/>\nHere, 6 \u03a9 and 3 \u03a9 resistors are connected in parallel. <\/span><br \/>\n<span style=\"color: #000000; font-family: georgia, palatino, serif;\">Therefore, their equivalent resistance will be given by<\/span><br \/>\n<span style=\"color: #000000; font-family: georgia, palatino, serif;\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-7393\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2024\/01\/NCERT-Class-10-Science-Solution-Ch-12-Ans.3i.png\" alt=\"NCERT Class 10 Science Solution\" width=\"131\" height=\"61\" \/><\/span><br \/>\n<span style=\"color: #000000; font-family: georgia, palatino, serif;\">This equivalent resistor of resistance 2 \u03a9 is connected to a 2 \u03a9 resistor in series. <\/span><br \/>\n<span style=\"color: #000000; font-family: georgia, palatino, serif;\">Therefore, equivalent resistance of the circuit = 2 \u03a9 + 2 \u03a9 = 4 \u03a9<\/span><br \/>\n<span style=\"color: #000000; font-family: georgia, palatino, serif;\">Hence, the total resistance of the circuit is 4 \u03a9.\u00a0<\/span><\/p>\n<p><span style=\"color: #000000; font-family: georgia, palatino, serif;\"><strong>(b)<\/strong> The following\u00a0circuit diagram shows the connection of the three resistors.<\/span><br \/>\n<span style=\"color: #000000; font-family: georgia, palatino, serif;\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-7394\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2024\/01\/NCERT-Class-10-Science-Solution-Ch-12-Ans.3a.jpg\" alt=\"\" width=\"203\" height=\"80\" \/><\/span><br \/>\n<span style=\"color: #000000; font-family: georgia, palatino, serif;\">All the resistors are connected in series. Therefore, their equivalent resistance will be given as.\u00a0<\/span><br \/>\n<span style=\"color: #000000; font-family: georgia, palatino, serif;\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-7395\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2024\/01\/NCERT-Class-10-Science-Solution-Ch-12-Ans.3bi.png\" alt=\"NCERT Class 10 Science Solution\" width=\"201\" height=\"61\" \/><\/span><br \/>\n<span style=\"color: #000000; font-family: georgia, palatino, serif;\">Therefore, the total resistance of the circuit is 1 \u03a9.\u00a0<\/span><\/p>\n<p><span style=\"font-family: georgia, palatino, serif; color: #000000;\"><strong style=\"text-align: justify;\">5. What 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?<br \/>\n<\/strong><strong>Answer &#8211;\u00a0<\/strong>There are four coils of resistances 4 \u03a9, 8 \u03a9, 12 \u03a9 and 24 \u03a9 respectively.<br \/>\n<strong>(a)<\/strong> If these coils are connected in series, then the equivalent resistance will be the highest, given by the sum 4 \u03a9, 8 \u03a9, 12 \u03a9 + 24 \u03a9 = 48 \u03a9<br \/>\n<strong>(b)<\/strong> If these coils are connected in parallel, then the equivalent resistance will be the lowest, given by<\/span><br \/>\n<span style=\"font-family: georgia, palatino, serif; color: #000000;\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-7396\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2024\/01\/NCERT-Class-10-Science-Solution-Ch-12-Ans.5.png\" alt=\"NCERT Class 10 Science Solution\" width=\"275\" height=\"61\" \/><\/span><br \/>\n<span style=\"font-family: georgia, palatino, serif; color: #000000;\">Therefore, 2\u00a0\u03a9 is the lowest total resistance.<\/span><\/p>\n<p style=\"text-align: center;\"><span style=\"color: #ff0000; font-size: 14pt; font-family: georgia, palatino, serif;\"><strong>Questions<\/strong><\/span><\/p>\n<p><span style=\"color: #000000; font-family: georgia, palatino, serif;\"><strong style=\"text-align: justify;\">1. Why does the cord of an electric heater not glow while the heating element does?<br \/>\n<\/strong><strong>Answer &#8211;\u00a0<\/strong>The heating element of an electric heater is a resistor. The amount of heat produced by it is proportional to its resistance. The resistance of the element of an electric heater is very high. As current flows through the heating element, it becomes too hot and glows red. On the other hand, the resistance of the cord is low. It does not become red when current flows through it.<\/span><\/p>\n<p><span style=\"color: #000000; font-family: georgia, palatino, serif;\"><strong>2. Compute the heat generated while transferring 96000 coulomb of charge in one hour through a potential difference of 50 V.<br \/>\nAnswer &#8211;<br \/>\n<\/strong>Charge transferred (Q) = 96000 C, time = 1 hour = 60 x 60 = 3600 s and potential difference (V) = 50 V. <\/span><br \/>\n<span style=\"color: #000000; font-family: georgia, palatino, serif;\">Heat generated (H) = VIt = V.Q = 50 x 96000 = 4800000 j = 4.8 x 10<sup>6<\/sup>\u00a0j.<\/span><\/p>\n<p><span style=\"color: #000000; font-family: georgia, palatino, serif;\"><strong>3.<\/strong> <b>An electric iron of resistance 20 takes a current of 5 A. Calculate the heat developed in 30 s.<br \/>\n<\/b><strong>Answer &#8211;\u00a0<\/strong>Resistance of electric iron (R) = 20 \u03a9, current (I) = 5 A and time = 30 s.<\/span><br \/>\n<span style=\"color: #000000; font-family: georgia, palatino, serif;\">Heat generated (H) = I2Rt = 5<sup>2<\/sup>\u00a0x 20 x 30 = 15000 j.<\/span><\/p>\n<p style=\"text-align: center;\"><span style=\"color: #ff0000; font-size: 14pt; font-family: georgia, palatino, serif;\"><strong>Questions<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\"><b>1. What determines the rate at which energy is delivered by a current?<br \/>\n<strong>Answer &#8211;\u00a0<\/strong><\/b>Electric power determines the rate at which energy is delivered by a current.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\"><b>2. An electric motor takes 5 A from a 220 V line. Determine the power of the motor and energy consumed in 2 h.<br \/>\n<\/b><strong>Answer &#8211;<br \/>\n<\/strong>It is given that current drawn by electric motor (I) = 5 A. the line voltage V = 220 V time (t) = 2 h. <\/span><br \/>\n<span style=\"color: #000000; font-family: georgia, palatino, serif;\">Power of motor (P) = P = VI = 220 \u00d7 5 = 1100 W and the energy consumed (E) = Pt <\/span><br \/>\n<span style=\"color: #000000; font-family: georgia, palatino, serif;\">1100 W \u00d7 2 h = 2200 Wh or 2.2 kWh.<\/span><\/p>\n<p style=\"text-align: center;\"><span style=\"color: #ff0000; font-size: 14pt; font-family: georgia, palatino, serif;\"><strong>Exercise<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\"><strong>1. 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\u2019, then the ratio R\/R\u2019 is<\/strong><\/span><br \/>\n<span style=\"color: #000000; font-family: georgia, palatino, serif;\">(a)\u00a01\/25<\/span><br \/>\n<span style=\"color: #000000; font-family: georgia, palatino, serif;\">(b)\u00a01\/5<\/span><br \/>\n<span style=\"color: #000000; font-family: georgia, palatino, serif;\">(c)\u00a05<\/span><br \/>\n<span style=\"color: #000000; font-family: georgia, palatino, serif;\">(d)\u00a025<\/span><br \/>\n<span style=\"color: #000000; font-family: georgia, palatino, serif;\"><strong>Answer &#8211;\u00a0<\/strong>(d)\u00a025<br \/>\nResistance of each part = R\/5<br \/>\nAll the five parts are connected in parallel. Hence, equivalent resistance (R\u2019) is given as<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-7401\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2024\/01\/NCERT-Class-10-Science-Solution-Ch-12-Ans.1d.png\" alt=\"NCERT Class 10 Science Solution\" width=\"273\" height=\"127\" \/><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\"><strong>2. Which of the following terms does not represent electrical power in a circuit ?<\/strong><\/span><br \/>\n<span style=\"color: #000000; font-family: georgia, palatino, serif;\">(a)\u00a0I<sup>2<\/sup>R<\/span><br \/>\n<span style=\"color: #000000; font-family: georgia, palatino, serif;\">(b)\u00a0IR<sup>2<\/sup><\/span><br \/>\n<span style=\"color: #000000; font-family: georgia, palatino, serif;\">(c)\u00a0VI<\/span><br \/>\n<span style=\"color: #000000; font-family: georgia, palatino, serif;\">(d)\u00a0V<sup>2<\/sup>\/R<\/span><br \/>\n<span style=\"color: #000000; font-family: georgia, palatino, serif;\"><strong>Answer &#8211;\u00a0<\/strong>(b)\u00a0IR<sup>2<br \/>\n<\/sup>Electrical power is given by the expression P = VI &#8212;&#8212;&#8212;- (i) <\/span><br \/>\n<span style=\"color: #000000; font-family: georgia, palatino, serif;\">According to Ohm&#8217;S law, V = IR &#8212;&#8212;&#8212;&#8212;&#8212;- (ii) <\/span><br \/>\n<span style=\"color: #000000; font-family: georgia, palatino, serif;\">Where, V = Potential difference, I = Current, <i>R<\/i>\u00a0= Resistance<\/span><br \/>\n<span style=\"color: #000000; font-family: georgia, palatino, serif;\"><em>P = VI<br \/>\n<\/em>From equation (i), it can be written <\/span><br \/>\n<span style=\"color: #000000; font-family: georgia, palatino, serif;\"><i>P<\/i>\u00a0= (<i>IR<\/i>) \u00d7<i>\u00a0I<br \/>\nP = I<sup>2<\/sup>R<br \/>\n<\/i>From equation (ii), it can be written<\/span><br \/>\n<span style=\"color: #000000; font-family: georgia, palatino, serif;\"><i>I = V\/R<br \/>\nP = V \u00d7 V\/R<br \/>\nP = V<sup>2<\/sup>\/R<br \/>\nP = VI = I<sup>2<\/sup>R = V<sup>2<\/sup>\/R<br \/>\n<\/i>Power\u00a0<i>P<\/i>\u00a0cannot be expressed as<i>\u00a0IR<\/i><sup>2<\/sup>.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\"><strong>3. <\/strong><strong>An electric bulb is rated as 220 V and 100 W. When it is operated on 110 V the power consumed will be<\/strong><\/span><br \/>\n<span style=\"color: #000000; font-family: georgia, palatino, serif;\">(a)\u00a0100 W<\/span><br \/>\n<span style=\"color: #000000; font-family: georgia, palatino, serif;\">(b)\u00a075 W<\/span><br \/>\n<span style=\"color: #000000; font-family: georgia, palatino, serif;\">(c)\u00a050 W<\/span><br \/>\n<span style=\"color: #000000; font-family: georgia, palatino, serif;\">(d)\u00a025 W<\/span><br \/>\n<span style=\"color: #000000; font-family: georgia, palatino, serif;\"><strong>Answer &#8211;\u00a0<\/strong>(d)\u00a025 W<br \/>\n<i>P = VI = V<sup>2<\/sup>\/R<br \/>\n<\/i><em>R = <i>V<sup>2<\/sup>\/P<br \/>\n<\/i><\/em>Power rating,\u00a0<i>P\u00a0<\/i>= 100 W <\/span><br \/>\n<span style=\"color: #000000; font-family: georgia, palatino, serif;\">Voltage,\u00a0<i>V\u00a0<\/i>= 220 V<\/span><br \/>\n<span style=\"color: #000000; font-family: georgia, palatino, serif;\">Resistance,\u00a0<i>R\u00a0<\/i>= 220<sup>2<\/sup>\/100 = 484 \u03a9<\/span><br \/>\n<span style=\"color: #000000; font-family: georgia, palatino, serif;\">The resistance of the bulb remains constant if the supply voltage is reduced to 110 V. If the bulb is operated on 110 V, then the energy consumed by it is given by the expression for power as<\/span><br \/>\n<span style=\"color: #000000; font-family: georgia, palatino, serif;\">\u2234 <em>P&#8217;<\/em> = <em>V&#8217;<sup>2<\/sup>\/R<br \/>\nP&#8217; = <\/em>110<sup>2<\/sup>\/484 = 25 W<\/span><br \/>\n<span style=\"color: #000000; font-family: georgia, palatino, serif;\">Therefore, the power consumed will be 25 W.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\"><strong>4. Two 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<\/strong><\/span><br \/>\n<span style=\"color: #000000; font-family: georgia, palatino, serif;\">(a) 1 : 2<\/span><br \/>\n<span style=\"color: #000000; font-family: georgia, palatino, serif;\">(b)\u00a02 : 1<\/span><br \/>\n<span style=\"color: #000000; font-family: georgia, palatino, serif;\">(c)\u00a01 : 4<\/span><br \/>\n<span style=\"color: #000000; font-family: georgia, palatino, serif;\">(d) 4 : 1<\/span><br \/>\n<span style=\"color: #000000; font-family: georgia, palatino, serif;\"><strong>Answer &#8211; <\/strong>(c) 1 : 4<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\"><strong>5. How is voltmeter connected in circuit to measure the potential difference between two points ?<br \/>\n<\/strong><strong>Answer &#8211; \u00a0<\/strong>To measure the potential difference between two points, a voltmeter should be connected in parallel to the points.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\"><strong>6. A copper wire has diameter 0.5 mm and resistivity of 1.6 x 10<sup>-8<\/sup>\u00a0\u00a0\u03a9 m. What will be the length of this wire to make its resistance 10\u00a0\u03a9 ? How much does the resistance change if diameter is doubled ?<\/strong><\/span><br \/>\n<span style=\"color: #000000; font-family: georgia, palatino, serif;\"><strong>Answer &#8211; <\/strong>Diameter of wire (d) = 0.5 mm, <\/span><br \/>\n<span style=\"color: #000000; font-family: georgia, palatino, serif;\">Resistivity (\u03c1) 1.6 x 10<sup>-8<\/sup> \u03a9m, <\/span><br \/>\n<span style=\"color: #000000; font-family: georgia, palatino, serif;\">Resistance (R) = 10 \u03a9.<\/span><br \/>\n<span style=\"color: #000000; font-family: georgia, palatino, serif;\">R = \u03c1L\/A <\/span><br \/>\n<span style=\"color: #000000; font-family: georgia, palatino, serif;\">L = \u03c0D<sup>2<\/sup>R\/4\u03c1 <\/span><br \/>\n<span style=\"color: #000000; font-family: georgia, palatino, serif;\">= 22 \u00d7 (5 \u00d7 10<sup>-4<\/sup>) 2 \/ 7 \u00d7 4 \u00d7 1.6 \u00d710<sup>-8<\/sup>\u00a0<\/span><br \/>\n<span style=\"color: #000000; font-family: georgia, palatino, serif;\">= 122.5 m <\/span><br \/>\n<span style=\"color: #000000; font-family: georgia, palatino, serif;\">If the diameter is doubled for given length of given material resistance is inversely proportional to the cross-section area of wire.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\"><strong>7. The values of current I flowing in a given resistor for the corresponding values of potential difference V across the resistor are given below :<br \/>\n<\/strong><\/span><\/p>\n<table style=\"border-collapse: collapse;\" border=\"1\">\n<tbody>\n<tr>\n<td style=\"width: 16.6667%;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\"><i>I\u00a0<\/i>(amperes )<\/span><\/td>\n<td style=\"width: 16.6667%;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\">0.5<\/span><\/td>\n<td style=\"width: 16.6667%;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\">1.0<\/span><\/td>\n<td style=\"width: 16.6667%;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\">2.0<\/span><\/td>\n<td style=\"width: 16.6667%;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\">3.0<\/span><\/td>\n<td style=\"width: 16.6667%;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\">4.0<\/span><\/td>\n<\/tr>\n<tr>\n<td style=\"width: 16.6667%;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\"><i>V\u00a0(volts)<\/i><\/span><\/td>\n<td style=\"width: 16.6667%;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\">1.6<\/span><\/td>\n<td style=\"width: 16.6667%;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\">3.4<\/span><\/td>\n<td style=\"width: 16.6667%;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\">6.7<\/span><\/td>\n<td style=\"width: 16.6667%;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\">10.2<\/span><\/td>\n<td style=\"width: 16.6667%;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\">13.2<\/span><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p><span style=\"color: #000000; font-family: georgia, palatino, serif;\">Plot a graph between <i>V<\/i>\u00a0and\u00a0<i>I<\/i>\u00a0and calculate the resistance of that resistor.<\/span><br \/>\n<span style=\"color: #000000; font-family: georgia, palatino, serif;\"><strong>Answer &#8211;<br \/>\n<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\">The plot between voltage and current is called\u00a0<i>IV<\/i>\u00a0characteristic. The voltage is plotted on\u00a0<i>x<\/i>-axis and current is plotted on\u00a0<i>y<\/i>-axis. The values of the current for different values of the voltage are shown in the given table.<\/span><\/p>\n<table style=\"border-collapse: collapse; width: 46.1825%;\" border=\"1\">\n<tbody>\n<tr>\n<td style=\"width: 24.4681%;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\"><i>V\u00a0(volts)<\/i><\/span><\/td>\n<td style=\"width: 11.8477%;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\">1.6<\/span><\/td>\n<td style=\"width: 15.5263%;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\">3.4<\/span><\/td>\n<td style=\"width: 15.5263%;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\">6.7<\/span><\/td>\n<td style=\"width: 16.0526%;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\">10.2<\/span><\/td>\n<td style=\"width: 15.7895%;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\">13.2<\/span><\/td>\n<\/tr>\n<tr>\n<td style=\"width: 24.4681%;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\"><i>I\u00a0<\/i>(amperes)<\/span><\/td>\n<td style=\"width: 11.8477%;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\">0.5<\/span><\/td>\n<td style=\"width: 15.5263%;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\">1.0<\/span><\/td>\n<td style=\"width: 15.5263%;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\">2.0<\/span><\/td>\n<td style=\"width: 16.0526%;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\">3.0<\/span><\/td>\n<td style=\"width: 15.7895%;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\">4.0<\/span><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p><span style=\"color: #000000; font-family: georgia, palatino, serif;\">The\u00a0<i>IV<\/i>\u00a0characteristic of the given resistor is plotted in the following figure.<\/span><br \/>\n<span style=\"color: #000000; font-family: georgia, palatino, serif;\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-medium wp-image-7402\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2024\/01\/NCERT-Class-10-Science-Solution-Ch-12-Ans.7-300x221.jpg\" alt=\"NCERT Class 10 Science Solution\" width=\"300\" height=\"221\" srcset=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2024\/01\/NCERT-Class-10-Science-Solution-Ch-12-Ans.7-300x221.jpg 300w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2024\/01\/NCERT-Class-10-Science-Solution-Ch-12-Ans.7.jpg 304w\" sizes=\"auto, (max-width: 300px) 100vw, 300px\" \/><\/span><br \/>\n<span style=\"color: #000000; font-family: georgia, palatino, serif;\">The slope of the line gives the value of Resistance(R) as,<\/span><br \/>\n<span style=\"color: #000000; font-family: georgia, palatino, serif;\">Slope = 1\/R = BC\/AC = 2\/6.8<\/span><br \/>\n<span style=\"color: #000000; font-family: georgia, palatino, serif;\">R = 6.8\/2 = 3.4 \u03a9<\/span><br \/>\n<span style=\"color: #000000; font-family: georgia, palatino, serif;\">Therefore,the resistance of register is 3.4\u03a9.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\"><strong>8. When a 12 V battery is connected across an unknown resistance, there is a current of 2.5 mA in the circuit. Find the value of resistance of the resistor.<\/strong><\/span><br \/>\n<span style=\"color: #000000; font-family: georgia, palatino, serif;\"><strong>Answer &#8211; <\/strong>Voltage of battery = V = 12 V, <\/span><br \/>\n<span style=\"color: #000000; font-family: georgia, palatino, serif;\">Current (I) = 2.5 mA = 2.5 \u00d7 10<sup>-3<\/sup>\u00a0A<\/span><br \/>\n<span style=\"color: #000000; font-family: georgia, palatino, serif;\">Resistance (R) = V\/I = 12V\/ 2.5 \u00d7 10<sup>-3<\/sup>\u00a0A <\/span><br \/>\n<span style=\"color: #000000; font-family: georgia, palatino, serif;\">= 4800 \u03a9.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\"><strong>9. A battery of 9 V is connected in series with resistors of 0.2 \u00a0\u03a9, 0.3 \u00a0\u03a9, 0.4 \u00a0\u03a9, 0.5\u00a0\u00a0\u03a9 and 12 \u00a0\u03a9 respectively. How much current would flow through 12 \u03a9 resistor ?<br \/>\n<\/strong><strong>Answer &#8211; <\/strong>Potential difference (V) = 9 V.<\/span><br \/>\n<span style=\"color: #000000; font-family: georgia, palatino, serif;\">Total resistance (R) = R<sub>1<\/sub>+ R<sub>2<\/sub>+ R<sub>3<\/sub>+R<sub>4<\/sub>\u00a0+R<sub>5<br \/>\n<\/sub>= 0.2 +0.3 + 0.5 + 0.5 + 12 = 13.4 \u03a9 <\/span><br \/>\n<span style=\"color: #000000; font-family: georgia, palatino, serif;\">Current in the circuit (I) = V\/R = 9 V \/ 13. 4 \u03a9 = 0.67 A. <\/span><br \/>\n<span style=\"color: #000000; font-family: georgia, palatino, serif;\">In series circuit same current flows through all the resistance, hence current of 0.67 A will flow through 12 \u03a9 resistor.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\"><strong>10. How many 176 \u00a0\u03a9 resistors (in parallel) are required to carry 5 A on 220 V line ?<\/strong><\/span><br \/>\n<span style=\"color: #000000; font-family: georgia, palatino, serif;\"><strong>Answer &#8211;\u00a0<\/strong>Let a resistor of 176 \u03a9 are joined in parallel. Then their combined resistance (R)<\/span><br \/>\n<span style=\"color: #000000; font-family: georgia, palatino, serif;\">1\/R = 1\/176 + 1\/176 \u2026\u2026 times = n\/176 or R = 176\/n \u03a9 <\/span><br \/>\n<span style=\"color: #000000; font-family: georgia, palatino, serif;\">It is given that V= 220 V and I = 5 A <\/span><br \/>\n<span style=\"color: #000000; font-family: georgia, palatino, serif;\">R = V\/I or 176\/n = 220\/5 = 44 \u03a9 <\/span><br \/>\n<span style=\"color: #000000; font-family: georgia, palatino, serif;\">n = 176\/44 = 4<\/span><br \/>\n<span style=\"color: #000000; font-family: georgia, palatino, serif;\">4 resistors should be joined in parallel.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\"><b>11. Show how you would connect three resistors, each of resistance 6 \u03a9 so that the combination has resistance of (i) 9 \u03a9 (ii) 4 \u03a9.<br \/>\n<\/b><strong>Answer &#8211;\u00a0 <\/strong>It is given here that R<sub>1<\/sub>\u00a0= R<sub>2<\/sub>\u00a0= 6 \u03a9.<\/span><br \/>\n<span style=\"color: #000000; font-family: georgia, palatino, serif;\">(i) To get net resistance of 9 \u03a9 we should join three resistors as below:<\/span><\/p>\n<p><span style=\"color: #000000; font-family: georgia, palatino, serif;\"><img decoding=\"async\" class=\"img-responsive\" src=\"https:\/\/www.pw.live\/files\/che34_1a.png\" alt=\"NCERT Solutions Class 10 Science chapter 12-Electricity\/image007.png\" \/><\/span><\/p>\n<p><span style=\"color: #000000; font-family: georgia, palatino, serif;\">(ii) To get 4 \u03a9 net resistance we should join three resistors as below:<\/span><\/p>\n<p><span style=\"color: #000000; font-family: georgia, palatino, serif;\"><img decoding=\"async\" class=\"img-responsive\" src=\"https:\/\/www.pw.live\/files\/che34_1b.png\" alt=\"chapter 12-Electricity NCERT Solutions Class 10 Science\/image008.jpg\" \/><\/span><\/p>\n<p><span style=\"color: #000000; font-family: georgia, palatino, serif;\"><b>12. Several electric bulbs designed to be used on a 220 V electric supply line, are rated 10 W. 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 A?<br \/>\n<\/b><strong>Answer &#8211;\u00a0 <\/strong>Each bulb is rated as 10 W, 220 V, <\/span><br \/>\n<span style=\"color: #000000; font-family: georgia, palatino, serif;\">It draws a current (I) = P\/V = 10 W\/220 <\/span><br \/>\n<span style=\"color: #000000; font-family: georgia, palatino, serif;\">V = 1\/22 A. <\/span><br \/>\n<span style=\"color: #000000; font-family: georgia, palatino, serif;\">As the maximum allowable current is 5 A and all lamps are connected in parallel, hence maximum number of bulbs joined in parallel with each other = 5 \u00d7 22 = 110.<\/span><\/p>\n<p><span style=\"color: #000000; font-family: georgia, palatino, serif;\"><b>13. A hot plate of an electric oven connected to a 220 V line has two resistance coils A and B. Each of 24 \u03a9 resistances, which may be used separately, in series or in parallel. What are the currents in the three cases?<br \/>\n<\/b><strong>Answer &#8211;\u00a0 <\/strong>It is given that potential difference (V) = 220 V.<\/span><br \/>\n<span style=\"color: #000000; font-family: georgia, palatino, serif;\">Resistance of coil (A) = Resistance of coil (B) = 24 \u03a9 <\/span><br \/>\n<span style=\"color: #000000; font-family: georgia, palatino, serif;\">(i) When either coil is used separately, the circuit (I) = V\/R = 220 V\/ 24 \u03a9 = 9.2 A. <\/span><br \/>\n<span style=\"color: #000000; font-family: georgia, palatino, serif;\">(ii) When two coils are used in series total resistance (R) = R<sub>1<\/sub>\u00a0+ R<sub>2<\/sub> = 24 +24 = 48 \u03a9 <\/span><br \/>\n<span style=\"color: #000000; font-family: georgia, palatino, serif;\">Current flowing (I) = V\/ R = 220 V\/ 48 \u03a9 = 4.6 A. <\/span><br \/>\n<span style=\"color: #000000; font-family: georgia, palatino, serif;\">(iii) When two coils are joined in parallel. Total resistance (R) = 1\/24 + 1\/24 <\/span><br \/>\n<span style=\"color: #000000; font-family: georgia, palatino, serif;\">= 2\/24, R = 12 \u03a9. <\/span><br \/>\n<span style=\"color: #000000; font-family: georgia, palatino, serif;\">Current (I) = V\/R = 220 V \/ 12 \u03a9 = 18.3 A.<\/span><\/p>\n<p><span style=\"color: #000000; font-family: georgia, palatino, serif;\"><b>14. Compare the power used in the 2 \u03a9 resistor in each of the following circuits:<br \/>\n<\/b>(i) a 6-volt battery in series with 1 \u03a9 and 2 \u03a9 resistors and,<\/span><br \/>\n<span style=\"color: #000000; font-family: georgia, palatino, serif;\">(ii) a 4 V battery in parallel with 12 \u03a9 and \u03a9 resistors.<\/span><br \/>\n<span style=\"color: #000000; font-family: georgia, palatino, serif;\"><strong>Answer &#8211;<br \/>\n<\/strong>(i) When a 2 \u03a9 resistor is joined t a 6 V battery in series with 1 \u03a9 and 2 \u03a9 resistors. Total resistance (R) = 2 + 1 + 2 = 5 \u03a9. <\/span><br \/>\n<span style=\"color: #000000; font-family: georgia, palatino, serif;\">Current (I) = 6 V\/5 \u03a9 = 1.2 A <\/span><br \/>\n<span style=\"color: #000000; font-family: georgia, palatino, serif;\">Power used in 2 A resistor = I<sup>2<\/sup>R = 2.88 W<\/span><\/p>\n<p><span style=\"color: #000000; font-family: georgia, palatino, serif;\"><strong>(ii)<\/strong> When 2 \u03a9 resistor is joined to a 4 V battery in parallel with 12 \u03a9 resistor and 2 \u03a9 resistors, the current flowing in 2 \u03a9 = 4 V\/ 2 \u03a9 = 2 A\/. <\/span><br \/>\n<span style=\"color: #000000; font-family: georgia, palatino, serif;\">Power used in 2 \u03a9 resistor = I<sup>2<\/sup>R = 8 W <\/span><br \/>\n<span style=\"color: #000000; font-family: georgia, palatino, serif;\">Ratio = 2.88\/8 = 0.36 : 1<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><b>15. Two 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?<br \/>\n<\/b><strong>Answer &#8211;<br \/>\n<\/strong>Current drawn by 1<sup>st<\/sup> lamp rated 100 W at 220 V = P\/V = 100\/ 220 = 5\/11 A.<br \/>\nCurrent drawn by 2<sup>nd<\/sup> lamp rated 60 W at 220 V = 60\/220 = 3\/11 A.<br \/>\nIn parallel arrangement the total current = I1 +I2 = 3\/11+ 5\/11 = 8\/11 = 0.73 A.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><b>16. Which uses more energy, a 250 W TV set in 1 hour, or a 1200 W toaster in 10 minutes?<br \/>\n<\/b><strong>Answer &#8211;<br \/>\n<\/strong>Energy used by a TV set of power 250 W in 1 hour = P x t = 250 Wh.<br \/>\nEnergy used by toaster of power 1200 W in 10 minute (10\/60 h)<br \/>\n= P x t = 1200 W x 10\/60 h = 200 Wh.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><b>17. An electric heater of resistance 8 draws 15 A from the service mains for 2 hours. Calculate the rate at which heat is developed in the heater.<br \/>\n<\/b><strong>Answer &#8211;<br \/>\n<\/strong>Resistance of electric heater (R) = 8 \u03a9, current (I) = 15 A.<br \/>\nRate at which heat developed in the heater = I2Rt\/t = 15 x 15 x 8 = 1800 W.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><b>18. Explain the following:<br \/>\n<\/b><strong>(a) Why is the tungsten used almost exclusively for filament of electric lamps?<br \/>\n<\/strong><strong>Answer &#8211;\u00a0<\/strong>For filament of electric lamp we require a strong metal with high melting point. Tungsten is used exclusively for filament of electric lamps because its melting point is extremely high.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(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?<br \/>\n<\/strong><strong>Answer &#8211;\u00a0<\/strong>Conductors of electric heating devices are made of an alloy rather than a pure metal due to high resistivity than pure metal and high melting point to avoid getting oxidized at high temperature.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(c) Why is the series arrangement not used for domestic circuits?<br \/>\n<\/strong><strong>Answer &#8211;\u00a0<\/strong>Series arrangement is not used for domestic circuits as current to all appliances remain same in spite of different resistance and every appliance cannot be switched on\/ off independently.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(d) How does the resistance of wire vary with its area of cross-section?<br \/>\n<\/strong><strong>Answer &#8211;\u00a0<\/strong>Resistance of a wire is inversely proportional to its cross-section area.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(e) Why are copper and aluminium wires usually employed for electric transmission?<br \/>\n<\/strong><strong>Answer &#8211; <\/strong>Copper and aluminium wires are usually employed for electricity transmission because they are good conductor with low resistivity. They are ductile also to be drawn into thin wires.<\/span><\/p>\n<p><span style=\"font-family: georgia, palatino, serif;\"><a href=\"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-10-science\" target=\"_blank\" rel=\"noopener\"><b><i>Go Back To Chapters<\/i><\/b><\/a><\/span><\/p>\n","protected":false},"excerpt":{"rendered":"<p>NCERT Solutions Class 10 Science\u00a0 The NCERT Solutions in English Language for Class 10 Science Chapter &#8211; 12 (Electricity) has been provided here to help the students in solving the questions from this exercise.\u00a0 Chapter &#8211; 12 (Electricity)\u00a0 Questions 1. What does an electric circuit mean ?\u00a0 Answer &#8211; An electric circuit consists of electric [&hellip;]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1125],"tags":[1385,1133,1410,1409,1388,5],"class_list":["post-6896","post","type-post","status-publish","format-standard","hentry","category-class-10-science","tag-class-10-ncert-science-solutions","tag-class-10-ncert-solutions","tag-ncert-class-10-science-chapter-12-electricity","tag-ncert-class-10-science-chapter-12-solutions","tag-ncert-class-10-science-solutions","tag-ncert-solutions-in-english"],"yoast_head":"<!-- This site is optimized with the Yoast SEO Premium plugin v22.9 (Yoast SEO v27.3) - https:\/\/yoast.com\/product\/yoast-seo-premium-wordpress\/ -->\n<title>NCERT Solutions Class 10 (Science) Chapter 12 (Electricity) | TheExamPillar NCERT<\/title>\n<meta name=\"description\" content=\"NCERT Solutions Class 10 Science\u00a0The NCERT Solutions in English Language for Class 10 Science Chapter - 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