explain why the terminal voltage is smaller than EMF?
because EMF is eletromotive force,equals the total amount of energy gained per unit charge when there is no current.however if there is current.some energy is lost across the internal resistance.with less energy dissipated across the terminal circuit.there would be less energy per columb charge thus corresponds to a lower terminal voltage.
does it sound professinoa?i didnt learn much,need much improvement.
2006-12-30 20:13:47 · 3 answers · asked by Anonymous in Science & Mathematics ➔ Engineering
because terminal voltage is the value of the difference between 2 poles, while emf is the resultant obtained from the whole battery thus it is greater.
2006-12-30 20:18:30 · answer #1 · answered by john d 1 · 0⤊ 0⤋
Given a hard and fast resistive load, modern-day will improve with voltage in a linear vogue. So if the voltage axis is useful (increasing) the present would be too, ensuing in an outstanding slope.
2016-11-25 02:23:56 · answer #2 · answered by ? 4 · 0⤊ 0⤋
use this relation. V(potential drop)=E(emf)-IR.
I is current; R is internal resistance. clearly, E>V i.e E= V+IR.
V is lesser than EMF due to the potential drop due to internal resistance
2006-12-31 02:25:52 · answer #3 · answered by IN PURSUIT OF WISDOM 2 · 0⤊ 0⤋