Calculate Power Wasted in Cell: Help with EMF & Resistance

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SUMMARY

The discussion focuses on calculating the power wasted in a cell with an electromotive force (EMF) of 1.5V and an internal resistance of 0.5 ohms when connected to a 2.5 ohm resistor. The user correctly calculated the current as 0.5A and the terminal potential difference (pd) as 1.25V. The power wasted in the cell can be determined using the formula P = I²r, where I is the current and r is the internal resistance. The power wasted in the internal resistance of the battery is equal to the power dissipated across it.

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a cell of emf 1.5V and internal resistance 0.5 ohms is connected to a 2.5 ohm resistor.

ive calculated the current as 0.5A
terminal pd as 1.25V
and calculated the power delivered to the 2.5 ohm resistor.

now i have to calculate the power wasted in the cell?

ive got the equation Ie = I^2 + I^2r
which is the power supplied by the cell = the power delivered to R + the power wasted in the cell.

but I am so confused. on rearranging it and subsituting the numbers in?

can anyone shed any light on how i do this? and if so can you give me the equation and talk me through it please..

thanks :)
 
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Hint:
How much power is dissipated in the internal resistance of the battery?

A battery can only generate power but a resistor can "waste" it.
 
Last edited:
ahh.. so i jus find out the amount of power delivered to the internal resistor and then that's my wasted power?
 
Yes, that's right.

Asking it like that just turned a simple problem into a difficult one.
 

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