Correct Answer: Option 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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