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Example 25.2 provides an analysis of a real battery with internal resistance. The device attached across the terminals (see the diagram) is called the "load." The voltage across the terminals is also the voltage across out device (RL, the load.) Use the same values for the ideal battery voltage and internal resistance as in the example (12 V and 0.02 respectively.) Find voltage across the terminals, the current, and the power dissipated in the load when the load resistance is RL = 1k, much larger than the internal resistance. Report all your results to 2 significant figures. Find voltage across the terminals, the current, and the power dissipated in the load when the load resistance is 1m, much smaller than the internal resistance. Report all your results to 2 significant figures. For what ratio of RL/Rint (load over internal resistance) is the power to the load maximum? (It might help you to plot the power as a function of x = RL/Rint.)

Example 25.2 provides an analysis of a real battery with internal resistance. The device attached across the terminals (see the diagram) is called the "load." The voltage across the terminals is also the voltage across out device (RL, the load.) Use the same values for the ideal battery voltage and internal resistance as in the example (12 V and 0.02 respectively.) Find voltage across the terminals, the current, and the power dissipated in the load when the load resistance is RL = 1k, much larger than the internal resistance. Report all your results to 2 significant figures. Find voltage across the terminals, the current, and the power dissipated in the load when the load resistance is 1m, much smaller than the internal resistance. Report all your results to 2 significant figures. For what ratio of RL/Rint (load over internal resistance) is the power to the load maximum? (It might help you to plot the power as a function of x = RL/Rint.)

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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Problem 1 Example 25.2 provides an analysis of a real battery with internal resistance. The device attached across the terminals (see the diagram) is called the "load." The voltage across the terminals is also the voltage across out device (RL, the load.) Use the same values for the ideal battery voltage and internal resistance as in the example (12 V and 0.02 2 respectively.) Find voltage across the terminals, the current, and the power dissipated in the load when the load resistance is R₁ = 1kn, much larger than the internal resistance. Report all your results to 2 significant figures. Find voltage across the terminals, the current, and the power dissipated in the load when the load resistance is 1m2, much smaller than the internal resistance. Report all your results to 2 significant figures. For what ratio of RL/Rint (load over internal resistance) is the power to the load maximum? (It might help you to plot the power as a function of x = RL/Rint.)
Transcribed Image Text:Problem 1 Example 25.2 provides an analysis of a real battery with internal resistance. The device attached across the terminals (see the diagram) is called the "load." The voltage across the terminals is also the voltage across out device (RL, the load.) Use the same values for the ideal battery voltage and internal resistance as in the example (12 V and 0.02 2 respectively.) Find voltage across the terminals, the current, and the power dissipated in the load when the load resistance is R₁ = 1kn, much larger than the internal resistance. Report all your results to 2 significant figures. Find voltage across the terminals, the current, and the power dissipated in the load when the load resistance is 1m2, much smaller than the internal resistance. Report all your results to 2 significant figures. For what ratio of RL/Rint (load over internal resistance) is the power to the load maximum? (It might help you to plot the power as a function of x = RL/Rint.)
EXAMPLE 25.2 Internal Resistance: Starting a Car Your car has a 12-V battery with internal resistance 0.020 . When the starter motor is cranking, it draws 125 A. What's the voltage across the battery terminals while starting? INTERPRET This problem is about a real battery connected to a load, as in Fig. 25.8. We identify one resistor as the internal resistance and the load resistance as the starter motor. DEVELOP Figure 25.9 is our sketch, showing the internal resistance in series with the load. The current is the same everywhere in a series circuit, so we can use Ohm's law to find the voltage across Rint-Sub- tracting that voltage from the battery's emf will then tell what's left across the load. EVALUATE For the internal resistance, Ohm's law gives Vint= IRint = (125 A) (0.020 f2) = 2.5 V That leaves 12 V - 2.5 V or 9.5 V across the battery terminals. ASSESS Make sense? That 9.5 V is substantially less than the bat- tery's 12-V rating, so we're hardly treating it ideally. But the starter These points are the battery terminals. R₁0020 ns "int" ε = 12v! I=125 A _MW FIGURE 25.9 Sketch for Example 25.2. motor runs only briefly; most of the time the load on the battery- headlights, ignition system, electronics, and so on-draws far less current and so the battery behaves essentially like an ideal 12-V emf. A battery voltage of 9-11 V is typical during starting; much less than 9 V indicates a weak battery, a defective starter, or very cold weather.
Transcribed Image Text:EXAMPLE 25.2 Internal Resistance: Starting a Car Your car has a 12-V battery with internal resistance 0.020 . When the starter motor is cranking, it draws 125 A. What's the voltage across the battery terminals while starting? INTERPRET This problem is about a real battery connected to a load, as in Fig. 25.8. We identify one resistor as the internal resistance and the load resistance as the starter motor. DEVELOP Figure 25.9 is our sketch, showing the internal resistance in series with the load. The current is the same everywhere in a series circuit, so we can use Ohm's law to find the voltage across Rint-Sub- tracting that voltage from the battery's emf will then tell what's left across the load. EVALUATE For the internal resistance, Ohm's law gives Vint= IRint = (125 A) (0.020 f2) = 2.5 V That leaves 12 V - 2.5 V or 9.5 V across the battery terminals. ASSESS Make sense? That 9.5 V is substantially less than the bat- tery's 12-V rating, so we're hardly treating it ideally. But the starter These points are the battery terminals. R₁0020 ns "int" ε = 12v! I=125 A _MW FIGURE 25.9 Sketch for Example 25.2. motor runs only briefly; most of the time the load on the battery- headlights, ignition system, electronics, and so on-draws far less current and so the battery behaves essentially like an ideal 12-V emf. A battery voltage of 9-11 V is typical during starting; much less than 9 V indicates a weak battery, a defective starter, or very cold weather.
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