# A potentiometer wire of length 1 m has a resistance of 10 Ω. It is connected to a 6 V battery in series with a resistance of 5 Ω. Determine the emf of the primary cell which gives a balance point at 40 cm.

2
Log in to add a comment

Log in to add a comment

m =100cm,

R =10 ohm,

E0=6v,

R'=5 ohm.

I =E0/(R+R') = 6/(10+5) =6/15 A.

VAB = IR

=(6/15)×10

=60/15 Volt.

VAB/I =4/100 =0.04,

E0 =,(VAB /I )X

(x=balance point =40cm)

=0.04 ×40 = 1.6 Volt.