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Explain comparison of emf of two cell by using potentiometer with necessary diagiam. |
Answer» Solution : As shown in figure a battery having emf E and internal resistance r variable resistance R and switch `K_(1)` is connected between TWO end A and B of POTENTIOMETER. `rArr` To compare emf of two cell `epsilon_(1) and epsilon_(2) `, positive terminal of both cell is connected with A and negative terminal of cell is connected with point 1 and 2 of three way switch with terminal 3 of three way switch galvanometer and jockey key is connected in series. Jockey key can be moved on wire. `rArr` First point 1 and 3 of three way switch are connected hence `epsilon_(1) `cellis in the cell By sliding jockey key on wire obtained point `N_(1)` such that galvanometer shows zero deflection. Let `AN_(1) = l_(1)` By using Krichhoff.s second law for loop, `AN_(1) G31A` `phi l_(1) + 0 - epsilon_(1) = 0` `therefore phi l_(1) = epsilon_(1)"" ` .... (1) `rArr` Now connect point 2 and 3 and obtain point `N_(1)` such that galvanometer show zero deflection, `AN_(2) = l_(2)` For `AN_(2) G 32 A `loop, `phi l_(1) + 0 - epsilon_(2) = 0` `therefore phi l_(2)= epsilon "" `..... (2) By taking RATIO of (1) and (2), ` (epsilon_(1))/(epsilon_(2)) = (l_(1))/(l_(2)) `....(2) when emf of one of the cell is given then from value of`l_(1) and l_(2)` by using EQUATION (3)value of unknown emf can be obtained. `rArr` By using potentiometer potential difference of very small value can be MEASURED. |
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