Physics · Current Electricity

JEE Main 2024 — 5 April, Shift 1 — Question 47

In the given figure $\mathrm{R}_{1}=10 \Omega, \mathrm{R}_{2}=8 \Omega, \mathrm{R}_{3}=4 \Omega$ and $\mathrm{R}_{4}=8 \Omega$ . Battery is ideal with emf 12 V . Equivalent

resistant of the circuit and current supplied by battery are respectively.

Option A:
(1) $12 \Omega$ and 11.4 A
Option B:
$10.5 \Omega$ and 1.14 A
Option C:
$10.5 \Omega$ and 1 A
Option D:
$12 \Omega$ and 1 A
Correct

Answer: D

Step-by-step solution

Here $\mathrm{R}_{2}, \mathrm{R}_{3}, \mathrm{R}_{4}$ are in parallel

$\frac{1}{\mathrm{R}_{234}}=\frac{1}{\mathrm{R}_{2}}+\frac{1}{\mathrm{R}_{3}}+\frac{1}{\mathrm{R}_{4}}$

$\mathrm{R}_{234}=2 \Omega$

$\mathrm{R}_{234}$ is in series with $\mathrm{R}_{1}$ so $\mathrm{R}_{\mathrm{eq}}=\mathrm{R}_{234}+\mathrm{R}_{1}=2+10=12 \Omega$ $\mathrm{i}=\frac{12}{12}=1 \mathrm{Amp}$

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Exam: JEE Main 2024
Paper: 5 April, Shift 1
Subject: Physics
Chapter: Current Electricity
Topic: Circuit Analysis, Kirchhoff's Law and Nodal Analysis

In the experiment to determine the galvanometer resistance by half-deflection method, the plot of 1/theta vs the resistance ( R) of the…

Answer: D

Step-by-step solution

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