This method may be more intuitive for some people.

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The discussion focuses on calculating the energy transfer rate to a 1.5V cell during charging, with a potential difference of 1.9V across the terminals. The user calculates the current through a resistor and finds the power delivered to the cell using the formula P = VI, resulting in 0.6W. Questions arise about the validity of using other power equations like P = I²R, which are deemed inappropriate for the cell since its voltage remains constant regardless of current. The conversation emphasizes that the power dissipated in the resistor should not be confused with the power supplied to the battery. Ultimately, the conservation of energy approach is suggested as a valid method for understanding the power dynamics in the circuit.
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Homework Statement


upload_2019-2-11_19-20-51.png

Question - Calculate the rate of energy transferred to the 1.5V cell during charging if the potential between X and Y is 1.9V

Homework Equations


P = VI

The Attempt at a Solution


I think I understand how to get the answer. If XY has 1.9V then the resistor will get 0.4V. This will give a current of 0.44A (0.4V/0.6Ω). Using P=VI for the cell, I get 0.44A x 1.5V = 0.6W. However, thinking about it more I have a few questions. I was thinking about why I couldn't use other circuit power equations - P = I2R and P = V2/R but then I realized this would give the power delivered to the internal resistor and not the cell. However, if we work backwards and use P = V2/R for the cell to find R then what does this mean? i.e. we know it has a power of 0.6W and a voltage of 1.5V so R = V2/P = 3.75Ω. Is this right or is it not valid to use it in this context?

Thanks for any help offered.
 

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Jimmy87 said:
P = I2R
is for resistors where V = I##\times##R. In a (ideal) chemical cell the power is used for something, well, chemical :smile: and the voltage does not depend on the current. So you have to revert to P = V##\times##I

In your exercise the non-ideal chemical cell is modeled as an ideal cell with an internal series resistance.
To answer your question you could check that with e.g. a 0.1 ##\Omega## you would get a different answer.
 
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BvU said:
is for resistors where V = I##\times##R. In a (ideal) chemical cell the power is used for something, well, chemical :smile: and the voltage does not depend on the current. So you have to revert to P = V##\times##I

In your exercise the non-ideal chemical cell is modeled as an ideal cell with an internal series resistance.
To answer your question you could check that with e.g. a 0.1 ##\Omega## you would get a different answer.

Thanks. So if I have understood you correctly you're saying that we can't use P = I2R because this equation comes from V = IR where the voltage across something depends on its current whereas for a cell its voltage is fixed regardless of the current through it?
 
Yep.
 
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P=I^2R would be the power dissipated in the resistor rather than put into the battery.

One other way to do it would be to apply conservation of energy...

Power into battery = power supplied by the battery charger - power dissipated in resistor
= (1.9*0.44) - (0.4*0.44)
= 1.5*0.44
= 0.6W
 
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