The value of R in the above circuit to make the galvanometer measure 2A is
The correct answer is: B
Explanation
Given: I\(_g\) = 50mA = 0.05A, I to be measured = 2A, r = 2\(\Omega\), I\(_s\) = I - I\(_g\) = 2 - 0.05 = 1.95A
Shunt(R) = \(\frac{I_g}{I_s}\) x r
R = \(\frac{0.05}{1.95}\) x 10 = 0.2564\(\Omega\)
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