Electricity 14 PDF
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This document covers electricity, specifically series circuits. It provides formulas and examples of calculations relating to Ohm's law and resistances. It's likely instructional or part of a textbook.
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On applying Ohm’s law to the three resistors separately, we further have V1 = I R1 [11.13(a)] V2 = I R2 [11.13(b)] and V3 = I R3...
On applying Ohm’s law to the three resistors separately, we further have V1 = I R1 [11.13(a)] V2 = I R2 [11.13(b)] and V3 = I R3 [11.13(c)] From Eq. (11.11), I R = I R1 + I R2 + I R3 or Rs = R1 +R2 + R3 (11.14) We can conclude that when several resistors are joined in series, the resistance of the combination Rs equals the sum of their individual resistances, R1, R2, R3, and is thus greater than any individual resistance. Example 11.7 An electric lamp, whose resistance is 20 Ω, and a conductor of 4 Ω resistance are connected to a 6 V battery (Fig. 11.9). Calculate (a) the total resistance of the circuit, (b) the current through the circuit, and (c) the potential difference across the electric lamp and conductor. Figure 11.9 An electric lamp connected in series with a resistor of 4 Ω to a 6 V battery Solution The resistance of electric lamp, R 1 = 20 Ω, The resistance of the conductor connected in series, R 2 = 4 Ω. Then the total resistance in the circuit R = R1 + R2 Rs = 20 Ω + 4 Ω = 24 Ω. The total potential difference across the two terminals of the battery V = 6 V. Now by Ohm’s law, the current through the circuit is given by I = V/Rs = 6 V/24 Ω = 0.25 A. 184 Science 2024-25
