5 represents the correct relation between $\varepsilon$ and V?
In an electric circuit, a cell provides energy to push charge. The electromotive force, or emf (represented by $\varepsilon$), is the total energy per unit charge supplied by the cell. It's essentially the potential difference across the cell's terminals when no current is flowing.
However, when a cell is connected in a closed circuit and current ($I$) flows through it, the cell itself has an internal resistance ($r$). This internal resistance causes a drop in potential *within* the cell.
The terminal potential difference ($V$) is the actual potential difference measured across the terminals of the cell when current is flowing. This is the voltage available to the external circuit.
When current ($I$) flows out of the positive terminal of the cell, it passes through the internal resistance ($r$). According to Ohm's Law, the potential drop across this internal resistance is given by:
Potential Drop across internal resistance = $I \times r$
The terminal potential difference ($V$) is the emf ($\varepsilon$) minus the potential drop that occurs within the cell due to its internal resistance. Therefore, the correct relation is:
$V = \varepsilon - Ir$
This equation shows that the terminal voltage ($V$) is always less than the emf ($\varepsilon$) when the cell is discharging (i.e., when current $I$ is flowing out of the positive terminal).
The relationship between the emf ($\varepsilon$), internal resistance ($r$), current ($I$), and terminal potential difference ($V$) for a cell in a closed circuit (discharging) is $V = \varepsilon - Ir$.
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