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A potentiometer wire of length 1 m has a resistance of 5Omega. It is connected to a 8 V battery in series with a resistance of 15Omega. Determine the emf of the primary cell which gives a balance point at 60 cm. |
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Answer» Solution : CURRENT flowing in the potentiometer `I=V/(R_1+R_2)` `=8/(5+15)A=8/20A`=0.4 A Potential drop across the potentiometer wire V=IR =0.4 x 5 =2 V Potential gradient `K=V/t` `=2/1=2Vm^(-1)` `THEREFORE`Unknown EMF of the CELL = kl’ = 2 × 0.6 V = 1.2 V |
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