question_answer

In the circuit shown, the total current supplied by the battery is

A)  2 ampere                           

B)  4 ampere

C)  1 ampere                           

D)  6 ampere

Correct Answer: B

Solution :

 \[6\Omega \] and \[2\Omega \]are in parallel combination Hence, \[{{R}_{AB}}=\frac{6\times 2}{6+2}=1.5\Omega \] \[V=IR\Rightarrow \frac{60}{15}=1\Rightarrow 1=4A\]\[[\because {{\operatorname{R}}_{eq}}=\frac{3\times 3}{3+3}=\frac{9}{6}=1.5\Omega ]\]