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A potentiometer wire of length 1 m is connected to a driver cellof emf 3 V as shown in the Fig. When a cell of 1.5 V emf is used in the secondary circuit, the balance point is found to be 60 cm. On replacing this cell and using a cell of unknown emf, the balance point shifts to 80 cm.(i)Calculate unknown emf of the cell.(ii) Explain with reason, whether the circuit works, if the driver cell is replaced with a cell of emf 1V.(iii) Does the high resistance R, used in the secondary circuit affect the balance point? Justify your answer. |
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Answer» Solution :(i) Here EMF of first cell `epsi_1 = 1.5 V, l_1 = 60 cm " and " l_2 = 80 cm` ` therefore ` emf of 2nd cell `epsi_2 = epsi_1 l_2/l_1 = 1.5 xx 80/60 = 2.0 V` (ii) If the driver cell of 3 V emf is replaced with a cell of emf 1 V, the circuit will not work. It is because, in order to obtain a NULL point on the potentiometer wire the fall in potential due to driver cell must be greater than the emf of the cell, whose emf is to be determined. (iii) The high resistance R, used in the SECONDARY circuit does not affect the balance point because atthe TIME of null point no current flows in the secondary circuit. |
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