In the given circuit, the current flowing through the resistance 20 Ω is 0.3 A , while the ammeter reads 0.9 A . The value of R 1 is \ \ \ \ \ \ \ Ω .

Physics · Current Electricity

JEE Main 2024 — 29 January, Shift 2 — Question 56

In the given circuit, the current flowing through the resistance $20 \Omega$ is 0.3 A , while the ammeter reads 0.9 A . The value of $R_{1}$ is $\_\_\_\_\_\_\_$ $\Omega$ .

Answer: 30

Numerical answer — enter this value.

Step-by-step solution

Given,

$i_{1}=0.3 \mathrm{~A}, \mathrm{i}_{1}+\mathrm{i}_{2}+\mathrm{i}_{3}=0.9 \mathrm{~A}$ So, $\mathrm{V}_{\mathrm{AB}}=\mathrm{i}_{1} \times 20 \Omega=20 \times 0.3 \mathrm{~V}=6 \mathrm{~V}$ $\mathrm{i}_{2}=\frac{6 \mathrm{~V}}{15 \Omega}=\frac{2}{5} \mathrm{~A}$ $\mathrm{i}_{1}+\mathrm{i}_{2}+\mathrm{i}_{3}=\frac{9}{10} \mathrm{~A}$

$\frac{3}{10}+\frac{2}{5}+\mathrm{i}_{3}=\frac{9}{10}$ $\frac{7}{10}+\mathrm{i}_{3}=\frac{9}{10}$

$\mathrm{i}_{3}=0.2 \mathrm{~A}$ 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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Exam: JEE Main 2024
Paper: 29 January, Shift 2
Subject: Physics
Chapter: Current Electricity
Topic: Circuit Analysis, Kirchhoff's Law and Nodal Analysis

In the given circuit, the current in resistance R 3 is :

Answer: 30

Step-by-step solution

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