Explanation Given data: Case (a): A battery is connected to a car radio provides 12.5 V to the radio and assume the radio is modeled as a 6.25 ohms resistor. Case (b): A battery is connected to a set of headlights provides 11.7 V to the headlights and assume the headlights is modeled as a 0.65 ohms resistor. Calculation: The circuit is drawn as per the statement defined in case (a) as shown in Figure 1. In Figure 1, the current i flow in the circuit is calculated by using Ohm’s law. i = 12.5 V 6.25 Ω = 2 A { ∵ 1 A = 1 V 1 Ω } Apply Kirchhoff’s voltage law in Figure 1. V Th = 2 R Th + 12.5 (1) The circuit is drawn as per the statement defined in case (b) as shown in Figure 2. In Figure 2, the current I flow in the circuit is calculated by using Ohm’s law. I = 11.7 V 0.65 Ω = 18 A { ∵ 1 A = 1 V 1 Ω } Apply Kirchhoff’s voltage law in Figure 1. V Th = 18 R Th + 11.7 (2) On equating the equation (1) and (2). 18 R Th + 11.7 = 2 R Th + 12.5 18 R Th − 2 R Th = 12.5 − 11.7 16 R Th = 0.8 Rearrange the equation as follows, R Th = 0.8 16 = 0.05 × 10 3 × 10 − 3 Ω = 50 × 10 − 3 Ω = 50 m Ω { ∵ 1 m = 10 − 3 } Since, the Thevenin resistance is equal to Norton Resistance

10th Edition

ISBN: 9780133760033

Author: James W. Nilsson, Susan Riedel

Publisher: PEARSON

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Thevenin Theorem

Any linear circuit, no matter how complex, maybe simplified to an equivalent circuit utilizing only a single voltage source and series resistance coupled to a load, according to Thevenin's Theorem. The superposition principle says that all foundational e…

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Chapter 4, Problem 70P

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Find the Thevenin and Norton resistance, Thevenin voltage, and Norton current for the battery.

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Chapter 4, Problem 69P

Chapter 4, Problem 71P

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