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Their relation is V=E-Ir. But while charging of secondary cell It become V=E+It.Why is there a change in sign of the equation?

Williamson Rk
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1 Answers1

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$V$ is the potential difference across the terminals of the cell and $E$ is the emf of the cell.
The emf is set up by an electrochemical process within the cell.

The potential difference across the internal resistance of the cell is always $Ir$ with the current flowing from the node at the higher potential to the node at the lower potential.

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$V=E-Ir$

In the top diagram the current $I$ is flowing from left to right so the potential of node $B$ is greater than that at node $C$ by $Ir$.
Starting at node $A$ and going through the cell to node $B$ results in an increase of potential of $E$ and then going through the resistor from node $B$ to node $C$ there is a decrease in potential of $Ir$.
So node $C$ is higher in potential than node $A$ and the potential difference is $E-Ir =V$.
In this instance the cell is a source of electrical energy.

$V=E+Ir$

In the lower diagram the current is flowing from right to left with the potential of node $C'$ higher than that of node $B'$ by $Ir$
Starting at node $A'$ and going through the cell to node $B'$ results in an increase of potential of $E$ and then going through the resistor from node $B'$ to node $C'$ there is an increase in potential of $Ir$.
So node $C'$ is higher in potential than node $A'$ and the potential difference is $E+Ir =V$.
In this instance the cell is a sink of electrical energy.

Farcher
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