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Explain method to measure internal resistance of cell by using potentiometer. |
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Answer» Solution :`rArr`As shown in FIGURE between point A and C, positive terminal of BATTERY (B) variable resistance R and key (K) are connected. `rArr` CELL whose internal resistance (r) is to be measured is connected in parallel with key `(K_(2))` and resistance box (R). `rArr` Positive of cell `(epsilon)` is connected with A and negative terminal of cell is connected with galvanometer Another end of galvanometer is connected with jockey key and moved on wire AC. When key `K_(2)` is open the null point obtained for zero DEFLECTION of galvanometer be `N_(1)`. Assume `AN_(1) = l_(1)` `thereforeepsilon = phi l_(1) ""` ...(1) `""` (For open circuit condition ) `rArr` When switch `K_(2)` is closed then current will pass through resistance box R and cell `(epsilon)`. Let null point obtained in this condition be `N_(2)`. Let `AN_(2) = l_(2)` `therefore ` If V is terminal voltage of cell `.epsilon.` then, `V = phi l_(2) ` .... (2) By taldng ratio of (1) and (2), `(epsilon)/(V) = (l_(1))/(l_(2))` but `epsilon ` = I (R + r) and V = IR `therefore (I (R + r))/(IR ) = (l_(1))/(l_(2))` `therefore (R + r)/(R) = (l_(1))/(l_(2))` `therefore 1 + (r)/(R) = (l_(1))/(l_(2))` `therefore (r)/(R) = (l_(1))/(l_(2)) - 1 ` `therefore r = R [ (l_(1) - l_(2))/(l_(2)) ] " or " r = R [ (l_(1))/(l_(2)) - 2 ] ` Advantage of potentiometer is suoh that cµrrent do not flow throughcell whose internal ressistance is to be measured. Hence, it do not affect internal resistance of the cell. |
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