In the given circuit, the current flowing through the resistance $20\ \Omega$ is $0.3 \mathrm{~A}$, while the ammeter reads $0.9 \mathrm{~A}$. The value of $\mathrm{R}_1$ is_____________ $\Omega$.
JEE MAIN 2024, Diffcult
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Given, $i_1=0.3 \mathrm{~A}, \mathrm{i}_1+\mathrm{i}_2+\mathrm{i}_3=0.9 \mathrm{~A}$
So, $\mathrm{V}_{\mathrm{AB}}=\mathrm{i}_1 \times 20 \Omega=20 \times 0.3 \mathrm{~V}=6 \mathrm{~V}$
$ \mathrm{i}_2=\frac{6 \mathrm{~V}}{15 \Omega}=\frac{2}{5} \mathrm{~A} $
$ \mathrm{i}_1+\mathrm{i}_2+\mathrm{i}_3=\frac{9}{10} \mathrm{~A} $
$ \frac{3}{10}+\frac{2}{5}+\mathrm{i}_3=\frac{9}{10} $
$ \frac{7}{10}+\mathrm{i}_3=\frac{9}{10} $
$ \mathrm{i}_3=0.2 \mathrm{~A} $
$ \text { So, } \mathrm{i}_3 \times \mathrm{R}_1=6 \mathrm{~V} $
$ (0.2) \mathrm{R}_1=6 $
$ \mathrm{R}_1=\frac{6}{0.2}=30 \Omega$
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