Quantrex Quantrex AcademyJEE · NEET · NDA PYQs with solutions Open app
JEE Main 20246 AprilEvening Shift
Exam
JEE Main
Year
2024
Date
6 April
Shift
Evening Shift
Subject
Physics
Chapter
Current Electricity

JEE Main 2024 · 6 April · Evening Shift · Physics · Current Electricity

In the given figure an ammeter A consists of a 240 coil connected...

In the given figure an ammeter A consists of a 240Ω240 \Omega coil connected in parallel to a 10Ω10 \Omega shunt. The reading of the ammeter is \qquad mA\mathrm{mA} In the given figure an ammeter A consists of a 240 coil connected in parallel to a 10 shunt. The reading of the ammeter

Correct answer

160

Step-by-step solution

Req =140.4+240×10240+10 Req =140.4+2400250 Req. =150Ω\begin{aligned} & =140.4+\frac{240 \times 10}{240+10} \\ & \text{ Req }=140.4+\frac{2400}{250} \\ & \text{ Req. }=150 \Omega \end{aligned}∴ Current in ammeter =24150=160mA\begin{aligned} \therefore \text{ Current in ammeter } & =\frac{24}{150} \\ & =160 \mathrm{ mA}\end{aligned}$
Practice Current Electricity on Quantrex Academy →

More from Current Electricity

A metal wire of cross-sectional area 0.5 mm ^2 and length 100 m is connected across a… JEE Advanced 2026As shown in the figure, the resistance of a galvanometer G can be found by the… JEE Advanced 2026The stored charge in the capacitor in steady state of the following circuit is ________ μ… JEE Main 2026A moving coil of galvanometer when shunted with 2\,Ω resistance gives a full scale… JEE Main 2026Two cells of emfs 1 V and 2 V and internal resistance 2\,Ω and 1\,Ω , respectively… JEE Main 2026Refer to the circuit diagram given below. The heat generated across the 6\,Ω resistance… JEE Main 2026In an experiment to determine the resistance of a given wire using Ohm's law, the… JEE Main 2026Under steady state condition the potential difference across the capacitor in the circuit… JEE Main 2026 Full Current Electricity list All JEE Main PYQs