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Internal Resistance: Starting a Car Your car has a 12-V battery with internal resistance 0.020 2. When the starter motor is cranking, it draws 125 A. What's the voltage across the battery terminals while starting? EXAMPLE 25.2 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. Subtracting 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 2) = 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 battery's 12-V rating, so the battery is hardly behaving ideally. But the starter Rint=0.020 ε =12 Vi These points are the battery terminals. I=125 A 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.

Internal Resistance: Starting a Car Your car has a 12-V battery with internal resistance 0.020 2. When the starter motor is cranking, it draws 125 A. What's the voltage across the battery terminals while starting? EXAMPLE 25.2 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. Subtracting 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 2) = 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 battery's 12-V rating, so the battery is hardly behaving ideally. But the starter Rint=0.020 ε =12 Vi These points are the battery terminals. I=125 A 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.

Physics for Scientists and Engineers, Technology Update (No access codes included)
Physics for Scientists and Engineers, Technology Update (No access codes included)
9th Edition
ISBN:9781305116399
Author:Raymond A. Serway, John W. Jewett
Publisher:Raymond A. Serway, John W. Jewett
Chapter28: Direct-current Circuits
Section: Chapter Questions
Problem 28.59AP: A rechargeable battery has an emf of 13.2 V and an internal resistance of 0.850 . It is charged by a...
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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 2 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.)
Transcribed Image Text: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 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.)
Internal Resistance: Starting a Car Your car has a 12-V battery with internal resistance 0.020 2. When the starter motor is cranking, it draws 125 A. What's the voltage across the battery terminals while starting? EXAMPLE 25.2 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. Subtracting 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 2) = 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 battery's 12-V rating, so the battery is hardly behaving ideally. But the starter Rint=0.020 ε =12 Vi These points are the battery terminals. I=125 A 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:Internal Resistance: Starting a Car Your car has a 12-V battery with internal resistance 0.020 2. When the starter motor is cranking, it draws 125 A. What's the voltage across the battery terminals while starting? EXAMPLE 25.2 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. Subtracting 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 2) = 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 battery's 12-V rating, so the battery is hardly behaving ideally. But the starter Rint=0.020 ε =12 Vi These points are the battery terminals. I=125 A 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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