
In an ammeter \[0.2\%\] of main current passes through the galvanometer. If resistance of galvanometer is G, the resistance of the ammeter will be:
Answer
164.4k+ views
Hint: Given that some percentage of main current passes through the galvanometer, this is a case of shunt resistance. As a result, the circuit will have a low resistance value.
Formula Used: The given circuit arrangement is in parallel. Therefore, the concept of parallel connection considering, same potential difference and different current will be used. Hence according to Ohm’s law,
\[V = IR\]
Where V is the potential difference which is the same for parallel combination of resistors, I is the current and R is the resistance.
Also, to find out the equivalent resistance across a parallel circuit, the formula used will be
\[\dfrac{1}{{{R_{eq}}}} = \dfrac{1}{{{R_1}}} + \dfrac{1}{{{R_2}}} + \dfrac{1}{{{R_3}}} + ......\]
Complete step by step solution:
Given that the resistance of the galvanometer is G. Also current through the galvanometer is \[0.2\% \] of main current. That is,
\[{I_G} = \dfrac{{0.2}}{{100}}I = \dfrac{1}{{500}}I\]
The current through the shunt will be
\[{I_S} = I - \dfrac{1}{{500}}I\]
\[\Rightarrow {I_S} = \dfrac{{500I - I}}{{500}}\]
\[\Rightarrow {I_S} = \dfrac{{499}}{{500}}I\]
Since, shunt and galvanometer are connected in parallel, therefore, potential difference will be same for both. It can be written that
\[{V_S} = {V_G}\]
Using Ohm’s Law,
\[{I_S}{R_S} = {I_V}{R_V}\]
Substituting the values, we get
\[\dfrac{{499}}{{500}}I \times S = \dfrac{1}{{500}}I \times G\]
\[\Rightarrow S = \dfrac{G}{{499}}\]
Since, the circuit is in parallel connection, therefore, the equivalent resistance will be the reciprocal sum of individual resistances. The resistance across ammeter (shunt) is given by
\[\dfrac{1}{{{R_A}}} = \dfrac{1}{{{R_G}}} + \dfrac{1}{{{R_S}}}\]
\[\Rightarrow \dfrac{1}{{{R_A}}} = \dfrac{1}{G} + \dfrac{1}{S}\]
\[\Rightarrow \dfrac{1}{{{R_A}}} = \dfrac{1}{G} + \dfrac{{\dfrac{1}{G}}}{{499}}\]
\[\Rightarrow \dfrac{1}{{{R_A}}} = \dfrac{{500}}{G}\]
\[\therefore {R_A} = \dfrac{G}{{500}}\]
The resistance of the ammeter will be \[\dfrac{G}{{500}}\].
Note: It is important to remember that shunt is a device which, when there is a connection between two points, provides a low resistance path to current to flow through. It is used to measure current through the circuit. Since it creates a low resistance path for current in the circuit, therefore, it is also used to protect the galvanometer from high current.
Formula Used: The given circuit arrangement is in parallel. Therefore, the concept of parallel connection considering, same potential difference and different current will be used. Hence according to Ohm’s law,
\[V = IR\]
Where V is the potential difference which is the same for parallel combination of resistors, I is the current and R is the resistance.
Also, to find out the equivalent resistance across a parallel circuit, the formula used will be
\[\dfrac{1}{{{R_{eq}}}} = \dfrac{1}{{{R_1}}} + \dfrac{1}{{{R_2}}} + \dfrac{1}{{{R_3}}} + ......\]
Complete step by step solution:
Given that the resistance of the galvanometer is G. Also current through the galvanometer is \[0.2\% \] of main current. That is,
\[{I_G} = \dfrac{{0.2}}{{100}}I = \dfrac{1}{{500}}I\]
The current through the shunt will be
\[{I_S} = I - \dfrac{1}{{500}}I\]
\[\Rightarrow {I_S} = \dfrac{{500I - I}}{{500}}\]
\[\Rightarrow {I_S} = \dfrac{{499}}{{500}}I\]
Since, shunt and galvanometer are connected in parallel, therefore, potential difference will be same for both. It can be written that
\[{V_S} = {V_G}\]
Using Ohm’s Law,
\[{I_S}{R_S} = {I_V}{R_V}\]
Substituting the values, we get
\[\dfrac{{499}}{{500}}I \times S = \dfrac{1}{{500}}I \times G\]
\[\Rightarrow S = \dfrac{G}{{499}}\]
Since, the circuit is in parallel connection, therefore, the equivalent resistance will be the reciprocal sum of individual resistances. The resistance across ammeter (shunt) is given by
\[\dfrac{1}{{{R_A}}} = \dfrac{1}{{{R_G}}} + \dfrac{1}{{{R_S}}}\]
\[\Rightarrow \dfrac{1}{{{R_A}}} = \dfrac{1}{G} + \dfrac{1}{S}\]
\[\Rightarrow \dfrac{1}{{{R_A}}} = \dfrac{1}{G} + \dfrac{{\dfrac{1}{G}}}{{499}}\]
\[\Rightarrow \dfrac{1}{{{R_A}}} = \dfrac{{500}}{G}\]
\[\therefore {R_A} = \dfrac{G}{{500}}\]
The resistance of the ammeter will be \[\dfrac{G}{{500}}\].
Note: It is important to remember that shunt is a device which, when there is a connection between two points, provides a low resistance path to current to flow through. It is used to measure current through the circuit. Since it creates a low resistance path for current in the circuit, therefore, it is also used to protect the galvanometer from high current.
Recently Updated Pages
Uniform Acceleration - Definition, Equation, Examples, and FAQs

JEE Main 2021 July 25 Shift 1 Question Paper with Answer Key

JEE Main 2021 July 22 Shift 2 Question Paper with Answer Key

JEE Atomic Structure and Chemical Bonding important Concepts and Tips

JEE Amino Acids and Peptides Important Concepts and Tips for Exam Preparation

JEE Electricity and Magnetism Important Concepts and Tips for Exam Preparation

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

Atomic Structure - Electrons, Protons, Neutrons and Atomic Models

Displacement-Time Graph and Velocity-Time Graph for JEE

JEE Main 2025: Derivation of Equation of Trajectory in Physics

Electric field due to uniformly charged sphere class 12 physics JEE_Main

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

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

JEE Advanced Weightage 2025 Chapter-Wise for Physics, Maths and Chemistry

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

Degree of Dissociation and Its Formula With Solved Example for JEE

Wheatstone Bridge for JEE Main Physics 2025

Instantaneous Velocity - Formula based Examples for JEE
