Is the emf of a cell equal to the total potential drop in the external circuit of the cell?

Answer 1

The EMF of a cell is the voltage across the terminals at zero current. This is the quoted cell voltage but as soon as a current is dawn from the cell, the voltage will drop. It's due to the internal resistance of the cell.

In a circuit diagram, a cell is often shown as a voltage source (a perfect source) and a resistor in series to represent the internal resistance. Using Ohms Law, it can be seen that as soon as a current flows, a voltage will be developed across the internal resistance, so reducing the voltage that is seen at the terminals of the cell. The higher the current draw, the higher the voltage drop inside the cell.

Normally, the voltage drop is minimal but in most cells, as it loses charge, the internal resistance rises. Eventually it will reach the point where most of the voltage is dropped across the internal resistance, leaving little to drive the intended load. Often, if a battery is removed from a device and measured, the voltage will be measured as equal to or very close to the quoted cell voltage. It is easy to make a judgment that a battery is good when it is almost dead. The only way to confirm the state of the battery is to measure the voltage at the terminals while the load is attached. The results can be very different to the off load voltage.

Alkaline cells have a low internal resistance compared to other dry cells. This makes them well suited for high current drain applications. The internal resistance also rises more slowly than most other cells, so they remain useful far longer than zinc-carbon types.