Problem 1Example 25.2 provides an analysis of a real battery with internal resistance. The device attachedacross the terminals (see the diagram) is called the "load." The voltage across the terminals is alsothe voltage across out device (RL, the load.) Use the same values for the ideal battery voltage andinternal 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 loadresistance is R₁ = 1kn, much larger than the internal resistance. Report all your results to 2significant figures.Find voltage across the terminals, the current, and the power dissipated in the load when the loadresistance is 1m2, much smaller than the internal resistance. Report all your results to 2 significantfigures.For what ratio of RL/Rint (load over internal resistance) is the power to the load maximum? (Itmight help you to plot the power as a function of x = RL/Rint.) EXAMPLE 25.2Internal Resistance: Starting a CarYour car has a 12-V battery with internal resistance 0.020 . Whenthe starter motor is cranking, it draws 125 A. What's the voltageacross 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 andthe load resistance as the starter motor.DEVELOP Figure 25.9 is our sketch, showing the internal resistance inseries with the load. The current is the same everywhere in a seriescircuit, 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 leftacross the load.EVALUATE For the internal resistance, Ohm's law givesVint= IRint = (125 A) (0.020 f2) = 2.5 VThat 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 starterThese points arethe battery terminals.R₁0020 ns"int"ε = 12v!I=125 A_MWFIGURE 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 lesscurrent and so the battery behaves essentially like an ideal 12-V emf.A battery voltage of 9-11 V is typical during starting; much lessthan 9 V indicates a weak battery, a defective starter, or very coldweather.

Answered: Image /qna-images/answer/0cfce327-7278-496e-a66a-13d660be82e2.jpg

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.)

University Physics Volume 2

18th Edition

ISBN:9781938168161

Author:OpenStax

Publisher:OpenStax

Chapter10: Direct-current Circuits

Section: Chapter Questions

Problem 74AP: Consider a circuit that consists of a real battery with an emf and an internal resistance of r...

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