In the above network , find the current through the `20 Omega` resistance and current through the battery .

Explanation

The circuit can be redrawn as
Here `(R_(1))/(R_(2))= (R_(3))/( R_(4))= (6)/(5)`, so wheatstone bridge principle is satisfied . Hence, current through he `20 Omega ` resistance `=0` Now,`20 Omega ` can be eliminated from the galvanometer arm BD.

Resistance across the battery arm AC
`R_(AC) = ( 15 xx 18)/( 15+18) = 8.18 Omega `
Current through battery `l= ( 10V)/( 8.18Omega) = 1.22 A`