A) 1025\[\Omega \] in series
B) 1025\[\Omega \] in parallel
C) 975\[\Omega \]in series
D) 975 \[\Omega \] in parallel
Correct Answer: C
Solution :
Given \[{{I}_{g}}=6\,mA=6\times {{10}^{-3}}A,\] \[G=25\Omega ,V=\,6\text{volt}\] To convert galvanometer into ammeter, the required resistance \[R=\frac{V}{{{I}_{g}}}-G\] \[=\frac{65}{6\times {{10}^{-3}}}-25\] \[=1000-25\] \[=975\Omega \] When galvanometer is converted into voltmeter, resistance is connected in series.
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