# Using Potential Dividers to find EMFs

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## Using Potential Dividers to find EMFs

Potential dividers can be used to find the emf of a cell.

*Remember:*

This ratio can be used to measure all sorts of resistances, potential differences and emfs.

In this circuit, both cells are trying to push current through the wire at P, but in opposite directions. Which cell wins depends on the position of the sliding contact.

*Look:*

The pd between A and sliding contact = 2V (from top cell).

So the 1V cell is overwhelmed and loses. Current is pushed the **wrong** way through it by the 2V cell.

**Now look at this alternative:**

Here the sliding contact is at A so there's **no potential difference between the sliding contact and A from the top cell.** Hence the 1V cell wins and pushes current in the other direction past P.

But at the mid-point of the wires - notice what happens!

The pd between A and the sliding contact = 1V (due to top cell) and 1V due to lower cell.

Neither wins. They cancel out. **No current flows past P!**

We have found the **balance point (or null point).** This allows us to calculate the emf of the bottom cell. No current is flowing through it, so no pd is lost in the internal resistance. The emf of the bottom cell is exactly the same as the pd across the wire (from A to the sliding contact). And we can calculate the pd across the wire by measuring the length of the wire - because that's the beauty of potential dividers!

*Example:*

**What's the emf of cell X if cell Y has emf of 3V and the balance point is 40cm (see diagram) along the wire?**

(Total wire length = 1m).

*Answer:*

**Use the fact that:**

ratio of lengths of the wire = ratio of pds across the wire.

X = 1.2V