JEE MAIN - Physics (2024 - 29th January Evening Shift - No. 21)
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 _________ $$\Omega$$.
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Explanation
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Given, $$\mathrm{i}_1=0.3 \mathrm{~A}, \mathrm{i}_1+\mathrm{i}_2+\mathrm{i}_3=0.9 \mathrm{~A}$$
So, $$\mathrm{V}_{\mathrm{AB}}=\mathrm{i}_1 \times 20 \Omega=20 \times 0.3 \mathrm{~V}=6 \mathrm{~V}$$
$$\begin{aligned} & \mathrm{i}_2=\frac{6 \mathrm{~V}}{15 \Omega}=\frac{2}{5} \mathrm{~A} \\ & \mathrm{i}_1+\mathrm{i}_2+\mathrm{i}_3=\frac{9}{10} \mathrm{~A} \\ & \frac{3}{10}+\frac{2}{5}+\mathrm{i}_3=\frac{9}{10} \\ & \frac{7}{10}+\mathrm{i}_3=\frac{9}{10} \\ & \mathrm{i}_3=0.2 \mathrm{~A} \\ & \mathrm{So}, \mathrm{i}_3 \times \mathrm{R}_1=6 \mathrm{~V} \\ & (0.2) \mathrm{R}_1=6 \\ & \mathrm{R}_1=\frac{6}{0.2}=30 \Omega \end{aligned}$$
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