1. The problem statement, all variables and given/known data A battery if emf E0=12v connected across a 4m long uniform wire having resistance 4Ω. The cells of small emf E1=2v and E2=4v having internal resistance 2Ω and 6Ω respectively, are connected as shown in figure below. If galvanometer shows no deflection at the point N, then distance of N from point A is equal to a. 1/6 cm b. 25 cm c. 75cm d. 50 cm 2. Relevant equations V=IR Kirchoff's laws V=kl (v=voltage, k=voltage per unit length, l=length) => for potentiometer 3. The attempt at a solution Since the potentiometer is in balanced condition no current passes through galvanometer. So applying V=IR for potentiometer 12v=(8+4)I 12=12I I=1A BY applying V=IR to potentiometer wire V=4I=4.1 V=4v Potential drop per unit length k=4v/4m K=1 vm-1 I have no idea to find potential of the points of circuit which is connected to the potentiometer. How to find that. When Kirchoff;s laws applied that potential get 0 and balanced length become 0. Applying Kirchoff's Laws 4v=6I1 I1=2/3 A -2v=2I2 I2=-1A Direction of current is opposite Now by V=IR, the potential of circuit become 0. How to complete this.