A) 0.50A
B) 0.25A
C) 1.00A
D) 1.50A
Correct Answer: C
Solution :
[c] \[2\Omega ,4\Omega \] and \[2\Omega ,\] are in series and their resultant \[(=8\Omega ,)\] is in parallel with 8 \[\Omega ,\]which gives rise to \[4\,\Omega ,\]. Next, \[2\Omega ,\,\,4\,\Omega ,\]and\[2\Omega ,\]are in series and their resultant \[\left( =8\text{ }\Omega \right)\]is in parallel with \[8\Omega \], which again gives rise to \[4\Omega \]. Now, \[3\Omega ,\,\,4\Omega \]and \[2\Omega ,\] are in series and their resultant is \[9\Omega ,\] \[\therefore \,\,I=\frac{9V}{9\Omega }=1A\]
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