ABCD forms the four arms of a Wheatstone's network. The arms AB, BC, CD, DA are of `10Omega, 15Omega, 25Omega` and `20Omega` respectively. The junctions B and D are connected to the ends of galvanometer of resistance `25Omega`. If the source voltage is 5V, r = 0, then calculate the current in the galvanometer. Assume the branch currents across AB, BC, CD, DA and BD as `i_(1),i_(3),-i_(4),-i_(2),` and `i_(g)` respectively.

Explanation

Applying KCL to the nodes A, B, C and D
we get `i_(2)=i-i_(1) , i_(3)=i_(1)-i_(g)`,
`i_(3)+i_(4)=i,thereforei_(4)=i-i_(1)+i_(g)`
Applying KVL to the mesh ABDA we get,
`10i_(1)+25i_(g)-20i_(2)=0`
or `2i_(1)+5i_(g)-4i_(2)=0`
Applying KVL to the mesh BCDB we get,
`15(i_(1)-i_(g))-25(i-i_(1)+i_(g))-25i_(g)=0`
`15i_(1)-15i_(g)-25i+25i_(1)-25i_(g)-25i_(g)=0`
`40i_(1)-65i_(g)-25(i_(1)+i_(2))=0`
`15i_(1)-65i_(g)-25i_(2)=0`
or `3i_(1)-13i_(g)-5i_(2)=0`
Applying KVL to the mesh ADCA we get,
`20i_(2)+25(i_(2)+i_(g))=5`
`45i_(2)+25i_(g)=5`
or `9i_(2)+5i_(g)=1` .....(3)
Using determinant method, `Delta=|{:(2,5,-4),(3,-13,-5),(0,5,9):}|= 2(- 117 + 25)-3( 45 + 20) + 0`
i.e., `Delta=-184-195 = - 379`, `Deltai_(g)=|{:(2,0,-4),(3,0,-5),(0,1,9):}|= 0+0+1 (- 10+12)` , `Deltai_(g)=2`
`i_(g)=(Deltai_(g))/(Delta)=2/(-379)=-5.28` mA , Current in the galvanometer= 5.28 mA