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

Explanation

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

Resistance across the battery arm AC
`R_(AC) = (15 xx 18)/(15+ 18) = 8 .18 Omega`
Current through battery ` I = (10 V)/(8.18 Omega ) = 1.22 A`