A) \[5050\,\,\Omega \]
B) \[5550\,\,\Omega \]
C) \[6050\,\,\Omega \]
D) \[4450\,\,\Omega \]
Correct Answer: D
Solution :
[d] Current through the galvanometer |
\[I=\frac{3}{(50+2950)}={{10}^{-3}}A\] |
Current for 30 divisions \[={{10}^{-3}}A\] |
Current for 20 divisions \[=\frac{{{10}^{-3}}}{30}\times 20\] |
\[=\frac{2}{3}\times {{10}^{-3}}A\] |
For the same deflection to obtain for 20 divisions, let resistance added be R |
\[\therefore \] \[\frac{2}{3}\times {{10}^{-3}}=\frac{3}{(50+1R)}\] |
or \[R=4450\,\Omega \] |
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