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,\[G=400\Omega ,\,\,{{i}_{g}}=0.2\,\,mA=0.2\times {{10}^{-3}}A\], \[i=3A\] \[\therefore \] \[S=\frac{0.0002\times 400}{3-0.0002}=0.027\,\,\Omega \]You need to login to perform this action.
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