Consider a potentiometer circuit where the primary loop contains a battery of emf E and internal resistance r connected across a potentiometer wire of resistance 15r and length 600 cm . A secondary branch, containing a battery of emf E /2 , a resistor r , and a galvanometer, is connected between one end (point A ) of the wire and a jockey. The positive terminals of both batteries are joined at point A . If the jockey

Step-by-step solution

Let the total length of the potentiometer wire be L = 600 cm and its total resistance be R = 15r . The jockey divides the wire into two segments at a distance l = 560 cm from point A. Let the point where the jockey touches be J, and the other end of the wire be B. The resistance of the segment AJ is: R_ AJ = l L R = 560 600 15r = 14r The resistance of the remaining segment JB is: R_ JB = 15r - 14r = r We can simplify the primary circuit using Thevenin's theorem across the points A and J. First, we find the open-cir