What it shows
On a uniform wire carrying a steady current, the voltage falls evenly along its length. The balance lengths for two cells are in the same ratio as their emfs, and no current is drawn from the cells at balance.
The science
When a steady current I flows through a uniform wire of resistance per unit length ρ, the potential drop along a length L is k L with a constant k = I ρ. A cell of emf E is connected so that it opposes the wire’s voltage through a sensitive detector. At the balance length ℓ the detector reads zero, so E = k ℓ, and no current flows from the cell, which means its internal resistance does not matter. For two cells E1 / E2 = ℓ1 / ℓ2.
Materials
Cost: under Rs 500. Many items come from home or from waste; spare parts such as LEDs, motors and wire are in the store.
- About 1 m of constantan or nichrome wire stretched on a wooden board with a metre scale
- A driver supply of two or three D cells and a rheostat or fixed resistor
- Two test cells (for example a 1.5 V AA and a used cell of a different make)
- A multimeter on its lowest mA or microamp range as a null detector
- A jockey made from a nail or a paper clip tip
- A two-way key from clips and a pin
- Screws or drawing pins to hold the wire ends
Safety
For this project: An adult must check the wiring; keep the key open between readings to save the driver cells, and use only low voltage cells.
Never use mains electricity: power models with cells, batteries or a low-voltage adapter. Sharp and hot tools, flames and lab chemicals need an adult or your teacher beside you. Keep magnets, button cells and small parts away from young children, and wash your hands after making.
All the safety rulesStep by step
7 steps. Do them in order, and test each part before you go on to the next. Write what you notice in a notebook as you go.
Stretch the wire tight along the metre scale and fix both ends under screws, keeping it straight and not touching anything.
Connect the driver cells, a switch and the rheostat across the two ends of the wire so the current flows along the whole wire.
Connect the positive of test cell 1 to the end of the wire where the driver’s positive is connected, and its negative through the detector to the jockey.
Touch the jockey lightly at several places on the wire (never drag it) and find the point where the detector reading is zero or changes direction; record the length ℓ1.
Replace cell 1 by cell 2 with the same polarity and find ℓ2 in the same way.
Compute E1 / E2 = ℓ1 / ℓ2 and compare with the ratio of the voltmeter readings of the two cells.
Repeat with different rheostat settings to show the ratio of lengths stays the same although each length changes.
What you should see
The ratio ℓ1 / ℓ2 matches the voltmeter ratio of the emfs to within a few percent, and it stays constant even when the driver current is changed.
Viva questions
A teacher or a judge will ask how it works. Try to answer each question before you open it, then compare.
1Why is the potentiometer better than a voltmeter for emf?
At balance it draws no current from the cell, so it measures the true emf, not the terminal voltage.
2Why must the driver’s emf be larger than the test cell’s?
Otherwise the potential drop along the whole wire can never match the cell’s emf and no balance point exists.
3Why must the wire be uniform?
So that the potential drop is exactly proportional to length.
4What does a longer wire do to the accuracy?
A longer wire gives a longer balance length and a smaller percentage error in measuring it.
Take it further
Find the internal resistance of a cell by balancing it with and without a known resistor across it.
Change one thing, measure what happens, and add a table or chart of your readings: that is what turns a model into your own investigation.
Free ebooks for this project
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