Describe how potentiometer is used to compare the e.m.f.s of two cells by combination method.
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=) For this , the necessary circuit is shown in figure . Across the ends A and B of the potentiometer wire are joined a storage cell B1 , rheostat Rh and key K1 . The positive pole of B1 is connected to the end A of the wire . Now , the positive poles of the two cells E1 and E2 whose emf's are to be compared are connected to A and the negative poles to the jockey J through a two-way key K2 and a shunted galvanometer G .
The key K1 and is closed so that a potential difference is established between the ends of the wire AB . Now , by the means of the key K2 , the cell E1 is included in the circuit and the null-point is determined by the jockey . Suppose the null-point on the wire is at a length l1 from the point A . Then , we have
E1 = K l1 ,
where K is the potential-gradient along the wire , Similarly , the other cells E2 is included in the circuit and again the null-point is determined . Let the length to this null-point from the point A be l2 . Then ,
E2 = K l2 .
E1 / E2 = l1 / l2 .
From this , the ratio E1 and E2 can be calculated . Since , the measurements l1 and l2 are taken in the condition of no current , the internal resistances of the sources of emf do not enter the picture . If one of the two cells is a standard cell , whose emf is known , then the emf of the other cell can be determined .
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__________________________
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=) For this , the necessary circuit is shown in figure . Across the ends A and B of the potentiometer wire are joined a storage cell B1 , rheostat Rh and key K1 . The positive pole of B1 is connected to the end A of the wire . Now , the positive poles of the two cells E1 and E2 whose emf's are to be compared are connected to A and the negative poles to the jockey J through a two-way key K2 and a shunted galvanometer G .
The key K1 and is closed so that a potential difference is established between the ends of the wire AB . Now , by the means of the key K2 , the cell E1 is included in the circuit and the null-point is determined by the jockey . Suppose the null-point on the wire is at a length l1 from the point A . Then , we have
E1 = K l1 ,
where K is the potential-gradient along the wire , Similarly , the other cells E2 is included in the circuit and again the null-point is determined . Let the length to this null-point from the point A be l2 . Then ,
E2 = K l2 .
E1 / E2 = l1 / l2 .
From this , the ratio E1 and E2 can be calculated . Since , the measurements l1 and l2 are taken in the condition of no current , the internal resistances of the sources of emf do not enter the picture . If one of the two cells is a standard cell , whose emf is known , then the emf of the other cell can be determined .
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HOPE , IT HELPS ... ✌️
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