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 \[I=\frac{V}{R}=\frac{4}{8}=0.5\,A\]
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