| The current \[{{I}_{1}}\] (in A) flowing through \[1\Omega \] resistor in the following circuit is: [JEE MAIN Held on 07-01-2020 Morning] |
 |
A) 0.5
B) 0.4
C) 0.25
D) 0.2
Correct Answer: D
Solution :
| [d] \[{{R}_{eq}}=\frac{(2+0.5)\times 2}{4.5}=\frac{2.5\times 2}{4.5}=\frac{10}{9}\Omega \] |
| \[\therefore I=\frac{1}{\left( \frac{10}{9} \right)}=\frac{9}{10}A\] |
| \[\therefore {{I}_{1}}=\frac{1}{2}\times \frac{2}{2+2.5}\times \frac{9}{10}\] |
| \[{{I}_{1}}=\frac{1}{5}A=0.2\,A\] |
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