A) 1.25 A
B) 1 A
C) 0.75 A
D) 0.5 A
Correct Answer: D
Solution :
Here, \[2\Omega \] and \[2\Omega \]are In parallel \[\therefore \] \[\frac{1}{R}=\frac{1}{2}+\frac{1}{2}\] \[R=\frac{2\times 2}{2+2}=1\Omega \] Now, internal resistance \[(1\Omega ),2\Omega ,4\Omega \]and resistance R are in series. \[\therefore \] \[{{R}_{net}}=1\Omega +2\Omega +4\Omega +1\Omega \] \[=8\Omega \] Hence, current\[I=\frac{V}{R}=\frac{4}{8}=0.5A\]
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