question_answer

 The current I in the given circuit is  

A)  8.3 A              

B)  6.8 A

C)  4.9 A              

D)  2.0 A

Correct Answer: D

Solution :

In the given circuit resistance \[{{\operatorname{R}}_{2}}\,\,and\,\,{{R}_{3}}\] are in series (Fig. I) So, equivalent resistance between them is \[{{R}_{2}}+{{R}_{3}}=\,\,6+6=12\,\,\Omega \] Which is in parallel with \[{{R}_{1}}=3\,\,\Omega \] So, equivalent resistance between A and B is \[{{R}_{AB}}=\frac{3\times 12}{3+12}=\frac{36}{15}\]   Using Ohm?s law, \[I=\frac{V}{{{R}_{AB}}}=\frac{48}{36/15}\] \[=\,\,2\,A\]