Correct Answer: Option D

Explanation

R\(_s\) = \(\frac{(i_gr_g)}{I-i_g}\) = \(\frac{(50 \times 10^{-3} \times 5)}{5 - (50 x 10^{-3})}\) = \(\frac{(0.25)}{4.95}\) = 0.0505 = 0.05Ω

There is an explanation video available below.

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