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A battery does 1.922 J of work to transfer 0.089 C of charge from the negative to the positive terminal. What is the emf of this battery?

Respuesta :

We can find the EMF as follows:

[tex]\begin{gathered} EMF=\frac{W}{q} \\ where: \\ W=1.922J \\ q=0.089C \\ so: \\ EMF=\frac{1.922}{0.089} \\ EMF=21.5955V \end{gathered}[/tex]

Answer:

21.5955V

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