Question

In the circuit shown in figure, the internal resistance of the cell is negligible. For what value of R, no current flows through the galvanometer?

Answer 1

the two terminals of the galvanometer should be at the same potential
therefore voltage drop across 10 ohm resistance is 6 V

Therefore current in the branch containing the resistances is $\frac{6}{10}=0.6A$

therefore current in the branch containing the source $=1.1-0.6=0.5A$

drop across 20 ohm resistance = drop across $R$

$20\times 0.6=R\times 0.5\Rightarrow R=24\Omega$