Physics - Current Electricity Question with Solution | TestHub

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

So, $V_{AB}=i_1\times 20\Omega =0.3\times 20\mathrm{V}=6\mathrm{V}$

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

Now,

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

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

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

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

So, $i_3\times R_1=6\mathrm{V}$

$\Rightarrow (0.2)R_1=6$

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