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A conductivity cell with electrodes of area 4 cm² separated by 2 cm contains pure water of conductivity 8 * 10⁻⁷ S cm⁻¹. Find (a) the resistance of the water and (b) the current that flows when a potential difference of 1 V is applied.

1. R = 6.25 * 10⁵ ohm, I = 1.6 * 10⁻⁶ A
2. R = 6.25 * 10⁴ ohm, I = 1.6 * 10⁻⁵ A
3. R = 1.6 * 10⁶ ohm, I = 6.25 * 10⁻⁷ A
4. R = 3.2 * 10⁵ ohm, I = 3.1 * 10⁻⁶ A

Correct answer: R = 6.25 * 10⁵ ohm, I = 1.6 * 10⁻⁶ A

Solution

The conductivity kappa fixes the resistance through the cell geometry (cell constant l/A). The tiny conductivity of pure water gives a very large resistance and hence a very small current.

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