A) \[6k\Omega \]in series
B) \[6k\Omega \]in parallel
C) \[500\Omega \] in series
D) \[500\Omega \]in parallel
Correct Answer: A
Solution :
[a] : \[{{I}_{g}}=16\times 30\mu A=480\times {{10}^{-6}}A\] Let G be the resistance of galvanometer and R be the resistance connected in series to convert the galvanometer into voltmeter of range 0 to 3 V. Then,\[G+R=\frac{V}{{{I}_{g}}}=\frac{3}{480\times {{10}^{-6}}}\] \[=6.25\times {{10}^{3}}=5.25k\Omega \] \[\therefore \]A resistance R nearly \[6k\Omega \]. is to be used in series.
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