This method may be more intuitive for some people.

In summary, the power transferred to the 1.5V cell during charging is 0.6W. This can be calculated using the equation P=VI, where V is the potential difference between X and Y (1.9V) and I is the current through the resistor (0.44A). Other circuit power equations such as P=I^2R and P=V^2/R are not applicable in this context as the voltage of a chemical cell does not depend on the current. Alternatively, conservation of energy can also be used to calculate the power transferred to the cell.
  • #1
Jimmy87
686
17

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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  • #2
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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  • #3
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?
 
  • #4
Yep.
 
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  • #5
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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