question_answer

A galvanometer has resistance of \[400\,\Omega \] am deflects full scale for current of 0.2mA through it. The shunt resistance required to convert into 3 A ammeter is

A)  \[0.027\,\Omega \]                                       

B)  \[0.054\,\Omega \]

C)  \[0.0135\,\Omega \]                     

D)         none of these

Correct Answer: A

Solution :

As shown if \[i\]the maximum current is, then a part \[{{i}_{g}}\] should pass through \[G\] and rest\[i-{{i}_{g}}\] through\[S\]. Potential across \[G\] and Ammeter \[S\] is same, therefore                 \[{{i}_{g}}\times G=(i-{{i}_{g}})\times S\] \[\Rightarrow \]               \[S=\frac{{{i}_{g}}G}{i-{{i}_{g}}}\] Given,\[G=400\Omega ,\,\,{{i}_{g}}=0.2\,\,mA=0.2\times {{10}^{-3}}A\],                 \[i=3A\] \[\therefore \]  \[S=\frac{0.0002\times 400}{3-0.0002}=0.027\,\,\Omega \]