In a potentiometer, a wire of length 10 ~m having resistance 50 is used. A battery of 5 ~V and a resistor of 450 are connected in series to the wire. If an unknown battery of emf E balances the potentiometer at 450 ~cm , then the value of E is

Step-by-step solution

Given length of wire, l=10 ~m , resistance of wire, R=50 , emf of battery, E₁=5 ~V , balancing length, x=450 ~cm =4.5 ~m series resistor, R₁=450 Current, i= E₁ R+R₁ = 5 50+450 = 5 500 =0.01 ~A So, V=i R=.01 50=0.5 ~V emf of primary cell, E= V_x l = 0.5 4.50 10 =0.225 ~V