Calculate Power Wasted in Cell: Help with EMF & Resistance

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To calculate the power wasted in a cell with an EMF of 1.5V and an internal resistance of 0.5 ohms connected to a 2.5 ohm resistor, first determine the current, which is 0.5A. The power delivered to the resistor is calculated using the formula P = I^2 * R, resulting in 0.625W for the 2.5 ohm resistor. The total power supplied by the cell can be found using the formula P = I^2 * (R + r), yielding 1.25W. The power wasted in the cell is then the difference between the total power supplied and the power delivered to the resistor, which is 0.625W. Thus, the power wasted in the internal resistance of the battery is indeed the power dissipated in the internal resistor.
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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.
 
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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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