
In an experiment to determine the resistance of a galvanometer by half deflection method, the circuit shown is used. In one set of readings, if \[R = 10\Omega \] and \[S = 4\Omega \], then the resistance of the galvanometer is:

(A) \[\dfrac{{20}}{3}\Omega \]
(B) \[\dfrac{{40}}{3}\Omega \]
(C) \[\dfrac{{50}}{3}\Omega \]
(D) \[\dfrac{{70}}{3}\Omega \]
Answer
139.8k+ views
Hint: In half deflection method, the value of the resistance across the resistor is such that the deflection made will be half that of the current when the resistance was disconnected. The current is directly proportional to the deflection of a galvanometer pointer.
Formula used: In this solution we will be using the following formulae;
\[G = \dfrac{{RS}}{{R - S}}\] where \[G\] is the resistance of the galvanometer, \[R\] is the resistance in the main circuit and \[S\] is the resistance across the galvanometer
Complete step by step solution:
Generally, the formula using half deflection method the resistance of a galvanometer is given as
\[G = \dfrac{{RS}}{{R - S}}\] where \[G\] is the resistance of the galvanometer, \[R\] is the resistance in the main circuit and \[S\] is the resistance across the galvanometer
Hence, by inserting known values, we have
\[G = \dfrac{{10\left( 4 \right)}}{{10 - 4}} = \dfrac{{40}}{6}\Omega \]
\[ \Rightarrow G = \dfrac{{20}}{3}\Omega \]
Hence, the correct option is A
Note: For clarity, the formula used can be proven as follows:
First, let’s assume the key \[{K_1}\] is the only closed key. In this case the current flowing through the circuit, and hence through \[G\] will be
\[{I_G} = \dfrac{E}{{R + G}}\]
Generally, the deflection shown in the galvanometer is proportional to the current flowing through it. Hence,
\[{I_G} = k\theta \]
Then
\[\dfrac{S}{{S + G}}I = \dfrac{{k\theta }}{2}\]
Now, assuming we close the key \[{K_2}\], it can be proven that the current flowing through the galvanometer becomes
\[I{'_G} = \dfrac{S}{{S + G}}I\]where\[I\] is the new current flowing through the main circuit, hence the current flowing through \[R\].
In the half deflection method, the value of S is adjusted such that the deflection is half of the original value when the key \[{K_2}\] was not closed.
Thus,
\[I{'_G} = \dfrac{{k\theta }}{2}\].
\[ \Rightarrow \dfrac{S}{{S + G}}I = \dfrac{{k\theta }}{2}\]
Dividing equation above by \[\dfrac{S}{{S + G}}I = \dfrac{{k\theta }}{2}\], and simplifying, we have
\[\dfrac{{SI(R + G)}}{{E(S + G)}} = \dfrac{1}{2}\]
Now, the equivalent resistance of the circuit is
\[{R_{eq}} = R + \dfrac{{SG}}{{S + G}}\]. Then the current would be
\[I = \dfrac{E}{{{R_{eq}}}} = \dfrac{E}{{R + \dfrac{{SG}}{{S + G}}}}\]
Then substituting into \[\dfrac{{SI(R + G)}}{{E(S + G)}} = \dfrac{1}{2}\] and then simplifying to make \[G\] subject, we have
\[G = \dfrac{{RS}}{{R - S}}\]
Formula used: In this solution we will be using the following formulae;
\[G = \dfrac{{RS}}{{R - S}}\] where \[G\] is the resistance of the galvanometer, \[R\] is the resistance in the main circuit and \[S\] is the resistance across the galvanometer
Complete step by step solution:
Generally, the formula using half deflection method the resistance of a galvanometer is given as
\[G = \dfrac{{RS}}{{R - S}}\] where \[G\] is the resistance of the galvanometer, \[R\] is the resistance in the main circuit and \[S\] is the resistance across the galvanometer
Hence, by inserting known values, we have
\[G = \dfrac{{10\left( 4 \right)}}{{10 - 4}} = \dfrac{{40}}{6}\Omega \]
\[ \Rightarrow G = \dfrac{{20}}{3}\Omega \]
Hence, the correct option is A
Note: For clarity, the formula used can be proven as follows:
First, let’s assume the key \[{K_1}\] is the only closed key. In this case the current flowing through the circuit, and hence through \[G\] will be
\[{I_G} = \dfrac{E}{{R + G}}\]
Generally, the deflection shown in the galvanometer is proportional to the current flowing through it. Hence,
\[{I_G} = k\theta \]
Then
\[\dfrac{S}{{S + G}}I = \dfrac{{k\theta }}{2}\]
Now, assuming we close the key \[{K_2}\], it can be proven that the current flowing through the galvanometer becomes
\[I{'_G} = \dfrac{S}{{S + G}}I\]where\[I\] is the new current flowing through the main circuit, hence the current flowing through \[R\].
In the half deflection method, the value of S is adjusted such that the deflection is half of the original value when the key \[{K_2}\] was not closed.
Thus,
\[I{'_G} = \dfrac{{k\theta }}{2}\].
\[ \Rightarrow \dfrac{S}{{S + G}}I = \dfrac{{k\theta }}{2}\]
Dividing equation above by \[\dfrac{S}{{S + G}}I = \dfrac{{k\theta }}{2}\], and simplifying, we have
\[\dfrac{{SI(R + G)}}{{E(S + G)}} = \dfrac{1}{2}\]
Now, the equivalent resistance of the circuit is
\[{R_{eq}} = R + \dfrac{{SG}}{{S + G}}\]. Then the current would be
\[I = \dfrac{E}{{{R_{eq}}}} = \dfrac{E}{{R + \dfrac{{SG}}{{S + G}}}}\]
Then substituting into \[\dfrac{{SI(R + G)}}{{E(S + G)}} = \dfrac{1}{2}\] and then simplifying to make \[G\] subject, we have
\[G = \dfrac{{RS}}{{R - S}}\]
Recently Updated Pages
Average fee range for JEE coaching in India- Complete Details

Difference Between Rows and Columns: JEE Main 2024

Difference Between Length and Height: JEE Main 2024

Difference Between Natural and Whole Numbers: JEE Main 2024

Algebraic Formula

Difference Between Constants and Variables: JEE Main 2024

Trending doubts
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility, & More

JEE Main 2025: Derivation of Equation of Trajectory in Physics

A point charge + 20mu C is at a distance 6cm directly class 12 physics JEE_Main

JEE Main Exam Marking Scheme: Detailed Breakdown of Marks and Negative Marking

Learn About Angle Of Deviation In Prism: JEE Main Physics 2025

Electric Field Due to Uniformly Charged Ring for JEE Main 2025 - Formula and Derivation

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

JEE Main 2025: Conversion of Galvanometer Into Ammeter And Voltmeter in Physics

Degree of Dissociation and Its Formula With Solved Example for JEE

Electric field due to uniformly charged sphere class 12 physics JEE_Main

Dual Nature of Radiation and Matter Class 12 Notes: CBSE Physics Chapter 11

Formula for number of images formed by two plane mirrors class 12 physics JEE_Main
