A) \[4450\,\Omega \]
B) \[5050\,\Omega \]
C) \[5550\,\Omega \]
D) \[6050\,\Omega \]
Correct Answer: A
Solution :
The current through the galvanometer \[=\frac{3}{2950+50}={{10}^{-3}}A\] To reduce deflection from 30 divisions to 20 divisions, the required current \[=\frac{20}{30}\times {{10}^{-3}}=\frac{2}{3}\times {{10}^{-3}}A\] The required resistance R is given by \[\frac{3}{R+50}=\frac{2}{3}\times {{10}^{-3}}\] \[R=4450\,\Omega \]You need to login to perform this action.
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