A potentiometer wire of length 4 m and resistance 5 is connected in series with a resistance of 992 and a cell of e.m.f. 4 V with internal resistance 3 . The length of 0.75 m on potentiometer wire balances the e.m.f. of

Step-by-step solution

Total resistance of the primary circuit is given by R_ total = R_w + R_s + r Substituting the given values, R_ total = 5 + 992 + 3 = 1000 Current in the primary circuit is I = E R_ total = 4 1000 = 4 10⁻³ A Potential difference across the potentiometer wire is V_w = I R_w = 4 10⁻³ 5 = 20 10⁻³ V Potential gradient of the potentiometer wire is k = V_w L = 20 10⁻³ 4 = 5 10⁻³ V/m The e.m.f. balanced by a length l = 0.75 m is E' = k l = 5 10⁻³ 0.75 = 3.75 10⁻³ V = 3.75 mV