A) \[8.1%\]
B) \[10.1%\]
C) \[11.1%\]
D) \[9.1%\]
Correct Answer: D
Solution :
Suppose, \[G\] is the resistance of galvanometer and \[{{I}_{g}}\] is the current for full scale deflection. If \[I\] is the maximum current, then \[{{I}_{g}}\times G=(I-{{I}_{g}})S\] Given, \[G=50\Omega \]and\[S=5\Omega \] \[\therefore \] \[\frac{{{I}_{g}}}{I}=\frac{S}{G+S}=\frac{5}{50+5}=\frac{5}{55}=\frac{1}{11}\] \[\Rightarrow \] \[{{I}_{g}}=\frac{I}{11}\]or\[\frac{{{I}_{g}}}{I}=\frac{1}{11}\] or \[\frac{{{I}_{g}}}{I}%=\frac{1}{11}\times 100%\approx 9.1%\]You need to login to perform this action.
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