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There is a potentiometer wire of length 1200 cmand a 60 mA curreut is flowing in it. A battery of emf 5 V and internal resistance of 20 Omegais balanced on this potentiometer wfre with a balancing length 1000 cm. The resistance of the potentiometer wire is |
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Answer» `60 Omega`  LET terminal voltage of PRIMARY BATTERY is`V_(p)`. For any neutral point on potentiometer wire, `V_(p)` remains constant . `therefore ` potential GRADIENT, `(V)/(l) = (V_(p))/(L)` `therefore (5)/(1000) = (V_(p))/(1200)` `therefore= (1200 xx 5)/(1000) = 6V` `therefore ` Resistance of potentiometer wire `R_(p) = (V_(p))/(I)` `thereforeR_(p) = (6)/(60 xx 10^(-3)) ` `therefore R_(p) =100 Omega` |
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