ABCD is a Wheatstone bridge in which the resistanvely `2Omega , 4 Omega , 6Omega and 8Omega` . The points A and C are connected to the termicals of a cell of emf 2 v and negligible internal resistance . The points B and D are connected to a galvanometer of resistance `50Omega` . Using Kirchhoff's laws find the current flowing through the galvanometer .

Explanation

In the circuit of let , `V_(C)=0, "then" V_(A)=2V, " again , if " V_(B)=V_(1)andV_(B)-V_(D)=V, "then" V_(D)=V_(1)-V`
Now , applying Kirchhoff's 1 st law at the junction B , we get
`(V_(1)-2)/(2)+(V_(1))/(4)+(V)/(50)=0`
`" or, " (50V_(1)-100+25V_(1)+2V)/(100)=0`
`or, " 75V_(1)+2V=100 " " ...(1)`
Again applying Kirchhoff's 1 st law at the junction D , we get
`(V_(1)-V-2)/(8)+(V_(1)-V)/(6)+(-V)/(50)=0`
`" or, " (75V_(1)-75V-150+100V_(1)+100V-12V)/(600)=0`
`or, " 175V_(1)-187V=150 " " ...(2)`
Doing `(1)xx7-(2)xx3` we get ,
`575V=250 " or, " V=(250)/(575)V`
`thereforeI_(G)=(V)/(50)=(250)/(575xx50)=(1)/(115)=0.0087A=8.7mA`