The four arms ABCD of a Wheatstones network have the following resistances:
AB = 2 Ω, BC=4 Ω, CD= 4 Ω and DA= 8 Ω. A galvanometer of resistance 10 Ω is
connected between B and D. Find the current through the galvanometer, when
the potential difference between A and C is 5 V.
Applying Kirchoff's current law at node B we get;
(V1-5)/2 + (V2-V1)/10 -V1/4 = 0
Applying Kirchoff's current law at node D we get;
(V2-5)/8 + (V1-V2)/10 -V2/4 = 0
from here we get the values of V2 and V1 as 14.28 volt , 3.57 volt
Current comes out to be: (14.28 volt - 3.57 volt)/10 =1.071 ampere
Sorry, for late reply!
Though there are other students question being posted we too have subscribed to meritnation!...you simply cant be neglecting us based on the question that we post!..Everyone cant be Einstien!...always the reply is late!...then what is the use of the answer that youexperts give after our scheduled time!..no use of it!...I have my exam tomorrow I just came up with this question as a doubt hoping that you wouldanswer itbut till now there's no reply of the experts!!....I cant bear it at any cost!....i'm fed up with this service!...