Exam

JEE Main

Year

2025

Date

3 April

Shift

Morning Shift

Subject

Physics

Chapter

Current Electricity

JEE Main 2025 · 3 April · Morning Shift · Physics · Current Electricity

In the figure shown below, a resistance of 150.4 is connected in...

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 _____ mA.

Step-by-step solution

$$\begin{aligned} & \mathrm{R}_{\mathrm{eq}}=\mathrm{R}_1+\mathrm{R}_2 \\ & \mathrm{R}_{\mathrm{eq}}=150.4+\frac{240 \times 10}{250} \\ & =150.4+9.6=160 \Omega \\ & \mathrm{I}_1=\frac{\mathrm{I} \mathrm{R}_2}{240} \\ & \mathrm{I}_1=\frac{\mathrm{I} \times 9.6}{240} \\ & =\frac{20}{160} \times \frac{9.6}{2400}=\frac{1}{200}=5 \times 10^{-3} \mathrm{~A}=5 \mathrm{~mA}\end{aligned}$$