Four resistors of resistances `2Omega, 1Omega, 3Omega, 4Omega` are arranged in a cyclic order to form a Wheatstone's network ABCD. The junctions B and D are connected to a galvanometer of resistance `2Omega`. A current of 0.1 A enters the junction 'A'. Calculate the current in the galvanometer.

Explanation

Applying KCL to the node A, `I_(1) + I_(2) = 0.1`
`thereforeI_(2)=0.1-I_(1)` …(1)
Applying KCL to the node B,
`I_(1)=I_(3)+I_(g)`
So that, `I_(3)=I_(1)-I_(g)` …(2)
Applying KCL to the node `D,I_(2)+I_(g)=I_(i)` …(2)
substituting for `I_(3) I_(4)=0.1-I_(1)+I_(g)` ...(3)
Applying KVL to the mesh ABDA,
`2I_(1)+2I_(g)-4(0.1-I_(1))=0`
`6I_(1)+2I_(g)=0.4` ....(4)
Applying KVL to the mesh BCDB we write
`1I_(g)-3(0.1-I_(1)+I_(g))-2I_(g)=0`
`+3I_(1)-5I_(g)+(I_(1)-I_(g))=0.3`
`4I_(1)-6I_(g)=0.3` ...(5)
by using determinant method, equations ( 4) and (5) may be solved.
i.e., `Delta=|{:(6,2),(4,-6):}|=-36-8=-44`,
`DeltaI_(1)=|{:(0.4,2),(0.3,-6):}|=-2.4-0.6=-3.0`,
`DeltaI_(g)=|{:(6,0.4),(4,0.3):}|=-1.8-1.6=0.2`
`thereforeI_(g)=(DeltaI_(g))/(Delta)=(0.2)/(-44)=-4.545` mA
The current in the galvanomelcr is 4.545 mA from D to B.