In an experiment with a potentiometer, VB = 10V. R is adjusted to be 50Ω (Fig. 3.9). A student wanting to measure voltage E1 of a battery (approx. 8V) finds no null point possible. He then diminishes R to 10Ω and is able to locate the null point on the last (4th) segment of the potentiometer. Find the resistance of the potentiometer wire and potential drop per unit length across the wire in the second case.



Consider R to be the resistance of the potentiometer wire.


When R=50Ω, no null point is found,


Effective value of resistance = R + 50Ω


I =


IR = < 8V


10RV < 400VΩ + 8RV


R < 200Ω


Similarly, when R=10Ω, null point is found,


Hence,





Also,






Now, In the second case,


Potential drop across 400cm > 8V


Potential drop across 300cm < 8V


Let k be potential drop per unit length




And,





Hence resistance of the potentiometer wire lies between 160Ω and 200Ω and the potential drop per unit length lies between 8/3Vm-1 and 2Vm-1.


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