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Describe the use of a potentiometer to compare the emf's of two cells by the sum difference method. |
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Answer» SOLUTION :A BETTERY of stable emf E is used to set up a potential gradient `V//L` along the potentiometer WIRE, where `V equiv p.d` across total length L of the wire. The positive terminal of cell 1 of emf `E_(1)` is connected to the higher potential terminal A of the potentiometer, the negative terminal is connected to the galvanometer through the reversing key. the other terminal of the galvanometer is connected to a pencil jockey. Cell 2 of emf `E_(2)` is connected across the remaining TWO opposite terminals of the reversing key. `E_(1)` should be greater than `E_(2)`, and `E` should be greater than `E_(1)+E_(2)`. Inserting two plugs in the reversing key in positions `1-1`, the two cells assist each other so that the net emf is `E_(1)+E_(2)`. the jockey is tapped along the wire to locate the null point D. If the null point is a distance `l_(1)` from A, `E_(1)+E_(2)=l_(1) (V//L)`  For the same potential gradient, the plugs are now inserted into position `2-2`. The emf `E_(2)` then opposes `E_(1)` and the net emf is `E_(1)-E_(2)`. The new null point D' is, say, a distance `l_(2)` from A and `E_(1)-E_(2)=l_(2) (V//L)` `:. (E_(1)+E_(2))/(E_(1)-E_(2))=l_(1)/l_(2)` or `E_(1)/E_(2)=(l_(1)+l_(2))/(l_(1)-l_(2))` The experiment is repeated for different potential gradients. |
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