A) \[1\,\Omega \]
B) \[2\,\Omega \]
C) \[\frac{1}{2}\Omega \]
D) \[2.5\,\Omega \]
Correct Answer: B
Solution :
In series , \[{{i}_{1}}=\frac{2E}{2+2r}\] In parallel, \[{{I}_{2}}=\frac{E}{2+\frac{r}{2}}=\frac{2E}{4+r}\] Since \[{{i}_{1}}={{i}_{2}}\] Þ\[\frac{2E}{4+r}=\frac{2E}{2+2r}\] Þ \[r=2\Omega \]
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