Electricity PDF Past Paper 2024-25
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This document provides a table of electrical resistivities and example problems related to calculating electricity parameters. It covers topics such as conductors, insulators, alloys, and how their properties relate to electrical applications.
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have very low resistivity in the range of 10–8 Ω m to 10–6 Ω m. They are good conductors of electricity. Insulators like rubber and glass have resistivity of the order of 1012 to 1017 Ω m. Both the resistance and resistivity of a material vary with temperature. Table 11.2 reveals that the resist...
have very low resistivity in the range of 10–8 Ω m to 10–6 Ω m. They are good conductors of electricity. Insulators like rubber and glass have resistivity of the order of 1012 to 1017 Ω m. Both the resistance and resistivity of a material vary with temperature. Table 11.2 reveals that the resistivity of an alloy is generally higher than that of its constituent metals. Alloys do not oxidise (burn) readily at high temperatures. For this reason, they are commonly used in electrical heating devices, like electric iron, toasters etc. Tungsten is used almost exclusively for filaments of electric bulbs, whereas copper and aluminium are generally used for electrical transmission lines. Table 11.2 Electrical resistivity* of some substances at 20°C Material Ω m) Resistivity (Ω Conductors Silver 1.60 × 10–8 Copper 1.62 × 10–8 Aluminium 2.63 × 10–8 Tungsten 5.20 × 10–8 Nickel 6.84 × 10–8 Iron 10.0 × 10–8 Chromium 12.9 × 10–8 Mercury 94.0 × 10–8 Manganese 1.84 × 10–6 Alloys Constantan 49 × 10–6 (alloy of Cu and Ni) Manganin 44 × 10–6 (alloy of Cu, Mn and Ni) Nichrome 100 × 10–6 (alloy of Ni, Cr, Mn and Fe) Insulators Glass 1010 – 1014 Hard rubber 1013 – 1016 Ebonite 1015 – 1017 Diamond 1012 - 1013 Paper (dry) 1012 * You need not memorise these values. You can use these values for solving numerical problems. Example 11.3 (a) How much current will an electric bulb draw from a 220 V source, if the resistance of the bulb filament is 1200 Ω? (b) How much current will an electric heater coil draw from a 220 V source, if the resistance of the heater coil is 100 Ω? Solution (a) We are given V = 220 V; R = 1200 Ω. From Eq. (12.6), we have the current I = 220 V/1200 Ω = 0.18 A. (b) We are given, V = 220 V, R = 100 Ω. From Eq. (11.6), we have the current I = 220 V/100 Ω = 2.2 A. Note the difference of current drawn by an electric bulb and electric heater from the same 220 V source! Electricity 179 2024-25
