A) \[45\,\Omega \]
B) \[96\,\Omega \]
C) \[80\,\Omega \]
D) \[40\,\Omega \]
Correct Answer: B
Solution :
Given \[G=4\,\Omega V=12V,{{I}_{g}}=12\,mA\] Let a resistance R be connected in series with the galvanometer then \[{{I}_{g}}=\frac{V}{R+G}\] \[\Rightarrow \] \[12\times {{10}^{-3}}=\frac{12}{R+4}\] \[\Rightarrow \] \[R+4=\frac{12}{12}\times {{10}^{3}}=100\] \[R=96\,\Omega \]You need to login to perform this action.
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