In the given circuit, the current flowing through the resistance 20 Ω is 0 . 3 A , while the ammeter reads 0…

Given, $i_1=0.3 \mathrm{A}, i_1+i_2+i_3=0.9 \mathrm{A}$

So, $V_{\mathrm{AB}}={\mathrm{i}}_1\times 20 \Omega =0.3\times 20 \mathrm{V}=6 \mathrm{V}$

$i_2=\frac{6 \mathrm{V}}{15\Omega }=\frac{2}{5} \mathrm{A}$

Now,

$i_1+i_2+i_3=\frac{9}{10} \mathrm{A}$

$\Rightarrow \frac{3}{10}+\frac{2}{5}+i_3=\frac{9}{10}$

$\Rightarrow \frac{7}{10}+i_3=\frac{9}{10}$

$\Rightarrow i_3=0.2 \mathrm{A}$

So, $i_3\times R_1=6 \mathrm{V}$

$\Rightarrow (0.2)R_1=6$

$\Rightarrow R_1=\frac{6}{0.2}=30 \Omega$