Physics · Current Electricity

JEE Main 2024 — 29 January, Shift 2 — Question 45

In the given circuit, the current in resistance $R_{3}$ is :

Question figure

Option A:
1 A
Correct
Option B:
1.5 A
Option C:
3 A
Option D:
2.5 A

Answer: A

Step-by-step solution

$\mathrm{R}_{\mathrm{eq}}=2 \Omega+2 \Omega+1 \Omega=5 \Omega$

\mathrm{i}=\frac{\mathrm{V}}{\mathrm{R}_{\mathrm{eq}}}=\frac{10}{5}=2 \mathrm{~A}

Current in resistance $\mathrm{R}_{3}=2 \times\left(\frac{4}{4+4}\right)$

\begin{gathered}=2 \times \frac{4}{8} =1 \mathrm{~A}\end{gathered}

Answer key and solution verified before publishing.

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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 flowing through the resistance 20 Omega is 0.3 A , while the ammeter reads 0.9 A . The value of R 1 is \…

In the given circuit, the current in resistance R 3 is : | JEE Main 2024 PYQ with Solution · DhiX AI