Battery Problem, Power delivered ,energy dissipated

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SUMMARY

The discussion focuses on calculating the power delivered by an automotive battery with an emf of 11.1 V and a terminal voltage of 10.6 V while supplying a current of 20.0 A to a starter motor. The power delivered by the battery is calculated using the formula P = E * I, resulting in 222 W. The power delivered to the starter motor is determined using the terminal voltage, yielding 212 W. The energy dissipated in the battery over 6.9 seconds is calculated using the difference in power and the time, leading to a decrease in chemical energy of the battery.

PREREQUISITES
  • Understanding of electromotive force (emf) and terminal voltage
  • Familiarity with electrical power calculations (P = E * I)
  • Knowledge of energy dissipation concepts in electrical systems
  • Basic grasp of internal resistance in batteries
NEXT STEPS
  • Study internal resistance calculations in batteries
  • Learn about energy conservation in electrical circuits
  • Explore the impact of battery chemistry on performance
  • Investigate the relationship between current, voltage, and power in automotive applications
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Automotive engineers, electrical engineering students, and anyone involved in battery technology and energy management in vehicles will benefit from this discussion.

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Homework Statement


An automotive battery has an emf of 11.1 V. When supplying power to the starter motor, the current in the battery is 20.0 A, and the terminal voltage of the battery is 10.6 V.

a.)How much power is delivered by the battery due to the chemical reactions within the battery when the current in the battery is 20 A?

b.)How much of this power is delivered to the starter when the current in the battery is 20 A?

c.)By how much does the chemical energy of the battery decrease if the current in the starter is 20 A for 6.9 s?

d.)How much energy is dissipated in the battery during those 6.9 seconds?

Homework Equations



P=(E)(I);where E is the emf
Ps=VI; power to the starter V=10.6 in this case.
E=Pt; t is time P is (E)(I) power
Q=(P-Ps)t; the energy would be dissipated in the form of Heat (Q)
r=\DeltaV/I; internal resistance inside the battery, delta V would be 11.1-10.6
Not sure if internal resistance is needed I didn't use it. but it might be needed

The Attempt at a Solution


P=11.1*20=222W? or kW? got it wrong so I tried 222^-6 kw wrong
Then I am just lost, any help would be greatly appreciated
 
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I really need so type of help.
 
Try drawing a picture, with the physical battery being a box that has two things inside of it: the emf source and the internal resistance. Then connect the box's external terminals to "starter motor." Draw an arrow to show where the given value of current flows. When you use your formula P=EI, you will have to make a choice which voltage is the right one to substitute into it, either the 11.1 V or the 10.6 V.
 

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