A) \[1.25\,\Omega \]
B) \[2.5\,\Omega \]
C) \[5.0\,\Omega \]
D) \[10.0\,\Omega \]
Correct Answer: A
Solution :
Key Idea: Potential difference across galvanometer and shunt is same. For G being resistance of galvanometer and \[{{I}_{g}}\] the current across it, the current across S is \[\left( I-{{I}_{g}} \right)\]. Since, \[G\] and \[S\] are in parallel, potential across them is same \[{{I}_{g}}\times G=\left( I-{{I}_{g}} \right)\times S\] \[\Rightarrow \] \[S=\frac{{{I}_{g}}G}{\left( I-{{I}_{g}} \right)}\] Given, \[{{I}_{g}}=0.2\,A,\,G=10\,\Omega ,\,I=1.8\,A\] \[\Rightarrow \] \[S=\frac{0.2}{1.6}\times 10\] \[=1.25\,\,\Omega \]You need to login to perform this action.
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