A) \[0.027\Omega \]
B) \[0.054\Omega \]
C) \[0.0135\Omega \]
D) none of these
Correct Answer: A
Solution :
Fact: Potential difference across shunt and galvanometer resistance is same. Suppose G be resistance of galvanometer and current ig. If i is the maximum current, then a part \[{{i}_{g}}\] should pass through G and rest \[i-{{i}_{g}}\] through s. Potential across G and S is same hence. \[{{i}_{g}}\times G=(i-{{i}_{g}})\times S\] \[\Rightarrow \] \[S=\frac{{{i}_{g}}\,G}{i-{{i}_{g}}}\] Here, \[G=400\Omega ,\,\,{{i}_{g}}=0.2mA.\] \[=0.2\times {{10}^{-3}}A,\,\,i=3A.\] \[S=\frac{0.0002\times 400}{3-0.0002}=0.027\Omega \] Note: Shunt resistance is small compared to galvanometer resistance.You need to login to perform this action.
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