A potentiometer wire of length 1.0 m has a resistance of 15 ohm. It is connected to a 5 V source in series with a resistance of `5 Omega`. Determine the emf of the primary cell which has a balance point at 60 cm.

Explanation

Here, `L = 1 m = 100 cm, R_(0) = 10 Omega, V = 6V`,
`R = 5 Omega, l = 40 cm, epsilon = ?`
Current in potentiometer wire, `I = V/(R_(0) + R)`
Potential difference across the potentiometer wire `= IR_(0) = V/((R_(0) +R)) xx R_(0)`
Potential gradient, `K = (VR_(0))/((R_(0) +R)) 1/L`
EMF of the primary cell, `epsilon = (V R_(0) l)/((R_(0)+ R)L)`
`= ( 6 xx 10 xx 40)/((10 +5 )xx 100) = 1.6V`