In a potentiometer circuit, the emf of driver cell is 2V and internal resistance is `0.5 Omega `. The potentiometer wire is 1m long . It is found that a cell of emf 1V and internal resistance `0.5 Omega ` is balanced against 60 cm length of the wire. Calculate the resistance of potentiometer wire.

Explanation

Here, `i_(0) = ( E_(0))/(R_(0)+r)= ( 2)/(R_(0)+0.5)`, when `R_(0)` is the resistance of potentiometer wire
Given `V_9CO) = 1V [ :' ` There is no deflection ]
`implies V_(AB) = (1)/(6) xx 100 = ( 10 )/( 6) V ( :' (V)/(l)=` Same through out the wire )

Now, `V_(AB) =i_(0)R_(0)`
`=.- (10)/( 6) = (2)/( (R_(0)+0.5))R_(0)`
`implies R_(0)+ 0.5= 1.2 R_(0)`
or `R_(0)= 2.5 Omega `