Answer to Question #97557 in Electrical Engineering for Ojugbele Daniel

Question

A D.C wheatstone bridge has a 5.0V supply connected between A and C and a galvanometer of resistance 50ohms between point B and D what will be the current through the galvanometer when the resistance in the bridge arms are AB=120ohms, BC=120ohms, CD=120ohms, DA=120ohms.

Answer 1

Since all resistances are equal, the potential difference between points B and D is 0 and the current thus will be 0 as well. The potential difference between B and D:

$V=\bigg(\frac{R_2}{R_1+R_2}-\frac{R_4}{R_3+R_4}\bigg)V_{P.S.}=\\ \space\\ =\bigg(\frac{120}{120+120}-\frac{120}{120+120}\bigg)\cdot5=0\text{ V}.$