A wire of uniform cross section and resistance 4 ohms is bent in the shape of square ABCD. Point A is connected to a point P on DC by a wire AP of resistance 1 ohm . When a potential difference is applied between A and C , the points B and P are seen to be in same potential. What is the resistance of part DP ?

Step-by-step solution

Each side of the square has resistance 4 4 = 1 . Let R_ DP = x , so R_ PC = 1-x . Since B and P are at the same potential, the A-P section resistance equals the P-C section resistance (by symmetry of the voltage divider). From A to P: direct wire AP (1 ) is in parallel with path A D P (1+x) : R_ AP(eq) = 1 (1+x) 1+(1+x) = 1+x 2+x Setting R_ AP(eq) = R_ PC : 1+x 2+x = 1-x 1+x = (1-x)(2+x) = 2+x-2x-x^2 x^2 + 2x - 1 = 0 x = -2+ 8 2 = -1+ 2 R_ DP = 2 -1