In the figure shown below, a resistance of 150.4 Ω is connected in series to an ammeter A of resistance 240 Ω . A shunt resistance of 10 Ω is connected in parallel with the ammeter. The reading of the ammeter is mA .

Physics · Current Electricity

JEE Main 2025 — 3 April, Morning Shift — Question 64

In the figure shown below, a resistance of $150.4 \Omega$ is connected in series to an ammeter A of resistance $240 \Omega$ . A shunt resistance of $10 \Omega$ is connected in parallel with the ammeter. The reading of the ammeter is $\qquad$ mA .

Answer: 5

Numerical answer — enter this value.

Step-by-step solution

$i_{0}=\frac{20}{150.4+\frac{240 \times 10}{250}}$

$=\frac{20}{150.4+9.6}=\frac{20}{160}=\frac{1}{8} \mathrm{~A}$

$i_{A}=\frac{1}{8} \times \frac{10}{250} \times 1000 \mathrm{~mA}$

=5 \mathrm{~mA}

Answer key and solution verified before publishing.

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Exam: JEE Main 2025
Paper: 3 April, Morning Shift
Subject: Physics
Chapter: Current Electricity
Topic: Electrical Measuring Instruments

A wire of length 25 m and cross-sectional area 5 mm^2 having resistivity 2 × 10^-6 Omega m is bent into a complete circle. The resistance…

Answer: 5

Step-by-step solution

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