Question
A potentiometer wire of length 1 m has a resistance of 5 Ω. It is connected to a 6 V battery in series with a resistance of 10 Ω. Determine the emf of the primary cell which gives a balance point at 80
cm.
Solution
Length (l) = 1 m Resistance of potentiometer wire = 10Ω Current (I) in potentiometer wire = V/(R + R') = 6/(10 + 5) = 6/15 = 0.4 A Potential gradient (k) = IRp/L = (0.4 × 10)/1 = 4 Vm-1 e = kl = 4 × 0.8 = 3.2 V
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