JEE Main Class 11 Syllabus 2025 PDF for Paper 1 and 2

Ammeter - Practice Questions & MCQ

Edited By admin | Updated on Sep 18, 2023 18:34 AM | #JEE Main

Quick Facts

  • Ammeter is considered one the most difficult concept.

  • 31 Questions around this concept.

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A galvanometer of resistance 95 \Omega, shunted by a resistance of 50 \, \mathrm{ohm} gives a deflection of 50 divisions when joined in series with a resistance of 20 \, \mathrm{k} \Omega and a 2 volt accumulator, The current sensitivity of galvanometer (in \operatorname{div}/\mu \mathrm{A}) is 

Two resistors 400 ohm and 800 ohm are connected in series with a 6V battery. It is desired to measure the current in the circuit. An ammeter of 10 ohm resistance is used for this purpose. The reading in the ammeter is:

The sensitivity of a galvanometer of resistance 406 ohm is increased by 30 times. The shunt used is :

The sensitivity of a galvanometer of resistance 406 ohm is increased by 30 times.
The shunt used is :
 

Find the reading of the ammeters A_{1}  (in ampere) connected as shown in the network. 

In the circuit as shown in figure, the current in ammeter is:

 

A moving coil galvanometer is converted into an ammeter reading up to 0.03 A by connecting a shunt of resistance 4r across it and into an ammeter reading up to 0.06 A when a shunt of resistance r is connected across it. What is the maximum current which can be sent through this galvanometer, if no shunt is used?

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A galvanometer having 50 divisions provided with a variable shunt s which is used to measure the current when connected in series with a resistance of 90 \Omega and a battery of internal resistance 10 \Omega.It is observed that when the shunt resistances are 10 \Omega and 50 \Omega,the deflection are respectively 9 and 30 division. The resistance of the Galvanometer is :

The reading of ammeter in the  figure shown below is

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Concepts Covered - 1

Ammeter

Ammeter

It is a device used to measure current and always connected in series with the circuit

  • In the above circuit, A represents the ammeter.

  • Conversion of galvanometer into ammeter: Connect a low resistance (shunt) in parallel to the galvanometer.

  • The equivalent resistance of the combination is \frac{G s}{G+s} where G is the galvanometer resistance and s is the shunt resistance.

  • Required shunt s=\frac{i_{g} G}{\left(i-i_{g}\right)}. Where i is the total current, i_{g is the galvanometer current

  • To pass n^{th} part of main current through galvanometer that is If i_g = i/n

Then the required shunt resistance s=\frac{G}{(n-1)}

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Ammeter

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