Q.
A potentiometer wire of length 10m and resistance 20Ω is connected in series with a 15V battery and an external resistance 40Ω . A secondary cell of emf in the secondary circuit is balanced by 240cm long potentiometer wire. The emf E of the cell is
Total resistance or R=20+40 R=60Ω.
Given: V=15 volt current I=RV=6015 I=0.25A
Potential gradient =lV 1020×0.25 =0.5Vm−1
Potential difference across 240 cm E=0.5×2.4E=1.2V