The current $I_{1}$ (in A) flowing through $1 Ω$ resistor in the following circuit is

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answer is A.

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Detailed Solution

$R e s i s \tan c e i n p a r a l l e l = \frac{R_{1} R_{2}}{R_{1} + R_{2}} = \frac{1 (1)}{1 + 1} = \frac{1}{2} Ω \Rightarrow 1 / 2 Ω i s i n s e r e i e s w i t h 2 Ω \Rightarrow R e s i s \tan c e i n s e r i e s = R' = \frac{1}{2} + 2 = \frac{5}{2} = 2.5 Ω \Rightarrow 2.5 Ω i s i n p a r a l l e l w i t h 2 Ω \Rightarrow R e s i s \tan c e i n p a r a l l e l = R'' = \frac{R_{1} R_{2}}{R_{1} + R_{2}} = \frac{2 (2.5)}{2 + 2.5} = \frac{5}{4.5} = \frac{10}{9} Ω \Rightarrow V = i R'' \Rightarrow 10 = i \frac{10}{9} \Rightarrow i = 9 A C u r r e n t t h r o u g h u p p e r b r a n c h = I' = \frac{V}{R'} = \frac{10}{2.5} = 4 A c u r r e n t t h r o u g h 1 Ω = \frac{i'}{2} = \frac{4}{2} = 2 A$