A) \[5050\,\,\Omega \]
B) \[5550\,\,\Omega \]
C) \[6050\,\,\Omega \]
D) \[4450\,\,\Omega \]
Correct Answer: D
Solution :
Given \[{{R}_{1}}=50\Omega ,\,\,{{R}_{2}}=2950\,\,\Omega \] Current through the galvanometer is given by \[I=\frac{V}{{{R}_{1}}+{{R}_{2}}}\,=\frac{3}{(50+2950)}={{10}^{-3}}A\] Current for 30 divisions \[={{10}^{-3}}A\] Current for 20 divisions \[=\frac{1{{0}^{-3}}}{30}\times 20\,=\frac{2}{3}\,\times {{10}^{-3}}A\] For the same deflection to obtain for 20 divisions, Let resistance added be R. \[\therefore \,\,\frac{2}{3}\,\times {{10}^{-3}}\,=\frac{3}{(50+1R)}\]You need to login to perform this action.
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