A) 4.9\[\Omega \]
B) 7.9\[\Omega \]
C) 5.9\[\Omega \]
D) 6.9\[\Omega \]
Correct Answer: A
Solution :
Potential gradient \[x=\frac{e}{(R+{{R}_{h}}+r)}.\frac{R}{L}\] Þ \[\frac{0.2\times {{10}^{-3}}}{{{10}^{-2}}}=\frac{2}{(R+490+0)}\times \frac{R}{1}\] Þ R = 4.9 W.
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