InterviewSolution
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InterviewSolution
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Potentiometer wire `PQ` of `1 m` length is connected to a standard cell `E_(1)`. Another cell `E_(2)` of emf `1.02 V` is connected with a resistance `r` and a switch `S` as shown in the circuit diagram. With switch `S` open, the null position is obtained at a distance of `51 cm` from `P`. (i) Calculate the potential gradient of the potentiometer wire. (ii) Find the emf of cell `E_(1)`. (iii) When switch `S` is closed, will the null point move toward `P` or toward `Q`? Give reason for your answer. |
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Answer» (i) Potential gradient is `k = (V)/(l) = (1.02)/(0.51) = 2 Vm^(-1)` (ii) The emf of the cell is `E_(1) = k_(l) = 2 xx 1= 2 V` (iii) When the swich `S` is closed, there is no shift in the position of the null point because it depends on the potential gradient along the poteniometer wire (which depends on the emf of battery `E_(1)` and the resistance of the potentiometer wire circul and length of potentiometer) and the emf of cell `E_(2)`, which does not change when the swich `S` is closed. |
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