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Question: A milli ammeter of 10 \[\Omega \]internal resistance has a full scale deflection current of 10 mA . ...

A milli ammeter of 10 Ω\Omega internal resistance has a full scale deflection current of 10 mA . To read up to 10 A, the resistance to be connected is
A. 100999Ω\dfrac{{100}}{{999}}\Omega
B. 10999Ω\dfrac{{10}}{{999}}\Omega
C. 1099Ω\dfrac{{10}}{{99}}\Omega
D. 10090Ω\dfrac{{100}}{{90}}\Omega

Explanation

Solution

The electrical resistance of an object is a measure of its resistance to the flow of electric current in electronics and electromagnetism. Electrical conductance is the ease with which an electric current travels in the reciprocal quantity. Electrical resistance and mechanical friction have some conceptual similarities. The ohm (Ω\Omega ) is the SI unit for electrical resistance.

Complete step by step answer:
An object's resistance is mostly determined by the material it is constructed of. Electrical insulators, such as rubber, have a very high resistance and low conductivity, whereas electrical conductors, such as metals, have a very low resistance and great conductivity.

Resistivity or conductivity are used to measure this connection. Resistance and conductance are not just determined by the nature of a material; they are also influenced by the size and shape of an item, as these qualities are widespread rather than intensive.

The resistance should be connected in parallel to raise the full scale deflection (FSD) of the milliammeter from 10 mA to 10 A is given by,
R2=R1n1{R_2} = \dfrac{{{R_1}}}{{n - 1}}
Where R2{R_2} is the connecting resistance,R1{R_1} is the milliampere's internal resistance, and nn is the scale multiplication factor.
n = FSD of ammeterFSD of multimeter{\text{n = }}\dfrac{{{\text{FSD of ammeter}}}}{{{\text{FSD of multimeter}}}}
We have
R1=10Ω{R_1} = 10\Omega
n=10  A10  mA n=10000  mA10  mA n=1000\Rightarrow {\text{n}} = \dfrac{{10\;{\text{A}}}}{{10\;{\text{mA}}}} \\\ \Rightarrow {\text{n}}= \dfrac{{10000\;{\text{mA}}}}{{10\;{\text{mA}}}} \\\ \Rightarrow {\text{n}}= 1000
Using these values we get
R2=R1(n1) R2=10(10001) R2=10999Ω{{\text{R}}_2} = \dfrac{{{{\mathbf{R}}_1}}}{{({\text{n}} - 1)}} \\\ \Rightarrow {{\text{R}}_2}= \dfrac{{10}}{{(1000 - 1)}} \\\ \Rightarrow {{\text{R}}_2}= \dfrac{{10}}{{999}}\Omega
R2=10999Ω\therefore {{\text{R}}_2} = \dfrac{{10}}{{999}}\Omega

Hence option D is correct.

Note: VV and II are exactly proportional to each other over a large range of materials and circumstances, hence R and G are constants (although they will depend on the size and shape of the object, the material it is made of, and other factors like temperature or strain). Ohm's law describes this proportionality, and materials that meet it are known as ohmic materials.