How to calculate power provided by a battery

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SUMMARY

The discussion focuses on calculating the energy provided by a 12-volt battery in a circuit with two resistors (1 ohm and 2 ohms) in series. The participant initially calculated power using the formula P = I^2R, resulting in 3 watts over 3 seconds, equating to 9 joules. However, using P = IV yielded 12 watts, leading to 36 joules. The discrepancy arises from the different contexts of power dissipation in resistors versus power supplied by the battery, highlighting the need to consider the entire circuit configuration, including the 9V battery's influence.

PREREQUISITES
  • Understanding of Ohm's Law
  • Familiarity with series circuits
  • Knowledge of power equations (P = IV, P = I^2R)
  • Basic concepts of energy (joules) and power (watts)
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  • Study circuit analysis techniques for series and parallel circuits
  • Learn about the impact of multiple voltage sources in a circuit
  • Explore the concept of energy conservation in electrical circuits
  • Investigate the role of internal resistance in batteries
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Homework Statement



The diagram is of a circuit with 2 resistors in series (1 ohm and 2 ohms) connected to 2 batteries (a 9 V battery and a 12 V battery). The question asks, how much energy is provided by the 12-volt battery in 3 seconds?

Homework Equations



Power = IV = I^2R = V^2/R

I = current, V = voltage, R = resistance

The Attempt at a Solution



I used P = I^2R at first, since I already calculated I (= 1 amp) for a previous part. Thus, P = ((1)^2)*(1+2) = 3 watts. It's over 3 seconds, so it would be 3 watts * 3 seconds = 9 joules. However, when I use P = IV, I get a different answer; P = (1)*(12) = 12 watts. 12 watts * 3 sec = 36 joules.

How is this possible? I thought the equations were equivalent--if anyone could shed some light on this, that would be great! Thank you so much.
 
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How much energy is being supplied by the 9v source?
 
miaou5 said:
I already calculated I (= 1 amp) for a previous part.

That suggests the 12V and 9V batteries are arranged to give 3V across the resistors. Correct?

I used P = I^2R at first, since I already calculated I (= 1 amp) for a previous part. Thus, P = ((1)^2)*(1+2) = 3 watts

That's the equation for the power dissipated in the resistors which might be different to that delivered by the 12V battery. aralbrec has given you a good clue as to why these two are not the same thing. What's happening to the 9V battery?
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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