question_answer

A milliammeter of range 10 mA has a coil of resistance\[1\Omega \]. To use it as an ammeter of range 1 A, the required shunt must have a resistance of

A)  \[\frac{1}{101}\Omega \]                           

B)  \[\frac{1}{100}\Omega \]

C)  \[\frac{1}{99}\Omega \]                             

D)  \[\frac{1}{9}\Omega \]

Correct Answer: C

Solution :

 Idea An ammeter have low resistance in parallel and its value is given by \[S=\frac{{{i}_{g}}r}{i-{{i}_{g}}}\] Given, \[{{i}_{g}}=10mA=0.01A,r=1\Omega ,i=1A\] As, \[{{V}_{A}}-{{V}_{B}}={{i}_{g}}r=(l-{{i}_{g}})S\] \[\Rightarrow \]\[S=\frac{{{i}_{g}}r}{(l-{{i}_{g}})}=\frac{0.01\times 1}{1-0.01}=\frac{1}{99}\Omega \] (shunt is small resistance in parallel) TEST Edge Galvanometer can be converted into voltmeter as well, for this a high resistance is connected in series with the galvanometer. \[R=\frac{V}{{{I}_{g}}}-{{R}_{g}}\]