Answer to Question #97557 in Electrical Engineering for Ojugbele Daniel

Question #97557
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.
1
Expert's answer
2019-11-01T06:02:07-0400

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=(R2R1+R2R4R3+R4)VP.S.= =(120120+120120120+120)5=0 V.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}.



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