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2₂ 83Figure 25.1a V = 8.3A-S V.41.5 (41.572 5 The emf and the internal resistance of a battery are as shown. In Figure 25.1a, a current of 8.3 A is drawn the battery when a resistor is connected across the terminals. The power dissipated by the resistor A) 530 W B) 790 W C) 700 W D) 440 W (+² 95.0 V 5Ω 8.3 A = from is closest to: E) 620 W

2₂ 83Figure 25.1a V = 8.3A-S V.41.5 (41.572 5 The emf and the internal resistance of a battery are as shown. In Figure 25.1a, a current of 8.3 A is drawn the battery when a resistor is connected across the terminals. The power dissipated by the resistor A) 530 W B) 790 W C) 700 W D) 440 W (+² 95.0 V 5Ω 8.3 A = from is closest to: E) 620 W

College Physics
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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**Educational Content Transcription:** --- **Problem 3:** The emf and the internal resistance of a battery are shown in Figure 251a. In Figure 251a, a current of 8.3 A is drawn from the battery when a resistor is connected across the terminals. The power dissipated by the resistor is closest to: Options: - A) 530 W - B) 790 W - C) 700 W - D) 440 W - E) 620 W **Diagram Explanation (Figure 251a):** The diagram depicts a simple electrical circuit with a battery that has an emf of 950 V and an internal resistance of 5 Ω. A current of 8.3 A flows through the circuit. **Handwritten Calculations:** - The voltage drop across the internal resistance (V = I * r): - V = 8.3 A * 5 Ω = 41.5 V - The voltage across the terminals (V_terminal = emf - V): - V_terminal = 950 V - 41.5 V = 908.5 V - The power dissipated by the resistor (P = I * V_terminal): - P = 8.3 A * 908.5 V = 7540.55 W (closest option is B) 790 W) Additional information is written on an adjacent page with various circuit diagrams, which might not directly relate to this problem. **Note:** The diagrams and calculations are vital for visualizing the problem and understanding the application of circuit laws to solve for power dissipation. --- This transcription is intended to aid students in solving circuit problems and understanding the principles of emf, internal resistance, and power calculation.
Transcribed Image Text:**Educational Content Transcription:** --- **Problem 3:** The emf and the internal resistance of a battery are shown in Figure 251a. In Figure 251a, a current of 8.3 A is drawn from the battery when a resistor is connected across the terminals. The power dissipated by the resistor is closest to: Options: - A) 530 W - B) 790 W - C) 700 W - D) 440 W - E) 620 W **Diagram Explanation (Figure 251a):** The diagram depicts a simple electrical circuit with a battery that has an emf of 950 V and an internal resistance of 5 Ω. A current of 8.3 A flows through the circuit. **Handwritten Calculations:** - The voltage drop across the internal resistance (V = I * r): - V = 8.3 A * 5 Ω = 41.5 V - The voltage across the terminals (V_terminal = emf - V): - V_terminal = 950 V - 41.5 V = 908.5 V - The power dissipated by the resistor (P = I * V_terminal): - P = 8.3 A * 908.5 V = 7540.55 W (closest option is B) 790 W) Additional information is written on an adjacent page with various circuit diagrams, which might not directly relate to this problem. **Note:** The diagrams and calculations are vital for visualizing the problem and understanding the application of circuit laws to solve for power dissipation. --- This transcription is intended to aid students in solving circuit problems and understanding the principles of emf, internal resistance, and power calculation.
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