
An ammeter having resistance ${R_0}$ shows full deflection when the current $I$ flow from it. What resistance should be connected to it in series so that it shows deflection $\dfrac{I}{n}$ when the same current flows.
A) $\dfrac{{{R_0}}}{n}$
B) ${R_0}\left( {n - 1} \right)$
C) $\dfrac{{{R_0}}}{{n + 1}}$
D) None of these
Answer
214.5k+ views
Hint: To solve this question, we have to understand the arrangement in which a galvanometer can be converted to an ammeter. To convert a galvanometer to ammeter, a resistor of very low resistance called shunt resistance should be connected in parallel to the galvanometer.
Complete step by step solution:
An ammeter is a device used to measure the magnitude of current flowing through the wire. It is connected in the series with the branch in which current has to be measured.
An ammeter can be constructed from a galvanometer by connecting a low resistance in parallel to the galvanometer as shown:

If G is the resistance of the galvanometer, $I$ is the current to be measured and ${I_g}$ is the current flowing through the galvanometer, the value of shunt resistance connected to the galvanometer is given by –
$S = \dfrac{{G{I_g}}}{{I - {I_g}}}$
Thus, the complete setup can be summarised as the schematic figure of an ammeter with a resistance which is given here, as ${R_0}$ with the current I flowing through the ammeter.

Let R be another resistance connected in series with ${R_0}$ such that the current flowing through the ammeter shall be $\dfrac{I}{n}$ as shown:

In both the cases, the voltage drop across the ammeter remains constant since the current flowing is the same in both the cases.
Voltage drop in the first case: $V = I{R_0}$
Voltage drop in the second case: $V = \dfrac{I}{n}\left( {R + {R_0}} \right)$
Equating them, we get –
$I{R_0} = \dfrac{I}{n}\left( {R + {R_0}} \right)$
$ \Rightarrow {R_0} = \dfrac{{R + {R_0}}}{n}$
$ \Rightarrow n{R_0} = R + {R_0}$
$ \Rightarrow R = n{R_0} - {R_0}$
$ \Rightarrow R = {R_0}\left( {n - 1} \right)$
Hence, the correct option is Option B.
Note: The students must note the fact that in an ammeter, the shunt resistor used, has a very low value, usually in micro- or milli- ohms. If you encounter any problem on calculating the shunt resistance in an ammeter and you do not get a low resistance, it is sure that the solution is headed in the wrong direction.
Complete step by step solution:
An ammeter is a device used to measure the magnitude of current flowing through the wire. It is connected in the series with the branch in which current has to be measured.
An ammeter can be constructed from a galvanometer by connecting a low resistance in parallel to the galvanometer as shown:

If G is the resistance of the galvanometer, $I$ is the current to be measured and ${I_g}$ is the current flowing through the galvanometer, the value of shunt resistance connected to the galvanometer is given by –
$S = \dfrac{{G{I_g}}}{{I - {I_g}}}$
Thus, the complete setup can be summarised as the schematic figure of an ammeter with a resistance which is given here, as ${R_0}$ with the current I flowing through the ammeter.

Let R be another resistance connected in series with ${R_0}$ such that the current flowing through the ammeter shall be $\dfrac{I}{n}$ as shown:

In both the cases, the voltage drop across the ammeter remains constant since the current flowing is the same in both the cases.
Voltage drop in the first case: $V = I{R_0}$
Voltage drop in the second case: $V = \dfrac{I}{n}\left( {R + {R_0}} \right)$
Equating them, we get –
$I{R_0} = \dfrac{I}{n}\left( {R + {R_0}} \right)$
$ \Rightarrow {R_0} = \dfrac{{R + {R_0}}}{n}$
$ \Rightarrow n{R_0} = R + {R_0}$
$ \Rightarrow R = n{R_0} - {R_0}$
$ \Rightarrow R = {R_0}\left( {n - 1} \right)$
Hence, the correct option is Option B.
Note: The students must note the fact that in an ammeter, the shunt resistor used, has a very low value, usually in micro- or milli- ohms. If you encounter any problem on calculating the shunt resistance in an ammeter and you do not get a low resistance, it is sure that the solution is headed in the wrong direction.
Recently Updated Pages
Wheatstone Bridge Explained: Working, Formula & Uses

Young’s Double Slit Experiment Derivation Explained

Chemical Equation - Important Concepts and Tips for JEE

JEE Main 2022 (July 29th Shift 1) Chemistry Question Paper with Answer Key

Conduction, Transfer of Energy Important Concepts and Tips for JEE

JEE Analytical Method of Vector Addition Important Concepts and Tips

Trending doubts
JEE Main 2026: Application Form Open, Exam Dates, Syllabus, Eligibility & Question Papers

JEE Main Correction Window 2026 Session 1 Dates Announced - Edit Form Details, Dates and Link

Equation of Trajectory in Projectile Motion: Derivation & Proof

JEE Main 2026 Application Login: Direct Link, Registration, Form Fill, and Steps

Hybridisation in Chemistry – Concept, Types & Applications

Angle of Deviation in a Prism – Formula, Diagram & Applications

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

Collision: Meaning, Types & Examples in Physics

How to Convert a Galvanometer into an Ammeter or Voltmeter

Atomic Structure: Definition, Models, and Examples

Degree of Dissociation: Meaning, Formula, Calculation & Uses

Average and RMS Value in Physics: Formula, Comparison & Application

