A) \[2500\,\Omega \] as a shunt
B) \[245\,\Omega \] as shunt
C) \[2550\,\Omega \] in series
D) \[2450\,\Omega \] in series
Correct Answer: D
Solution :
| [d] To convert a galvanometer into voltmeter, high resistance should be connected in series with it. Let R is the resistance connected in series with the galvanometer. Galvanometer current |
 |
| \[{{i}_{g}}=\frac{V}{G+R}\] |
| or \[R=\frac{V}{{{i}_{g}}}-G\] |
| Given, \[G=50\Omega \] |
| \[{{i}_{g}}=25\times 4\times {{10}^{-4}}={{10}^{-2}}A,V=25V\] |
| \[\therefore \] \[R=\frac{25}{{{10}^{-2}}}-50=2500-50=2450\,\Omega \] |
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