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A cell in secondary circuit gives null deflection for 2.5 m length of potentiometer having 10 m length of wire . If the length of the potentiometer the cell in the primary , the position of the nul point now is |
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Answer» 3.5m `R_(eq) = (Rxx50)/(R+50)` From ohm 's law , `V= iR_eq1` `(100)/(3) = i((50R)/(R+50))` `i= (100)/(3) ((R+50)/(50R)` and `50kOmega` resistance is connected in series as shown in the figure , `R(eq) =((50R)/(R+50))+50` `100 = [(100)/(3)((R+50)/(50R)][ (50R)/(R+50) +50]` `Rightarrow 100 = (100)/(3) [R+50)/(50R)][(50R)/(R+50)+50]` `Rightarrow (3xx50R)/(R+50) = (50R)/(R+50)+50` `Rightarrow (2xx50R)/(R+50)= 50` `Rightarrow 2R= R+50` `R = 50kOmega` 
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