question_answer

In the circuit shown, the current in the \[1\Omega \] resistor is:                           [JEE MAIN 2015]

A) 0.13 A, from Q to P       

B) 0.13 A, from P to Q

C) 1.3 A, from P to Q

D) 0A55

Correct Answer: A

Solution :

[a]

\[{{E}_{eq}}=\frac{\frac{5}{5}\frac{6}{3}=\frac{9}{5}\times \frac{3}{3}}{\frac{1}{3}+\frac{1}{5}}=\frac{3}{\cancel{15}\times \frac{8}{\cancel{15}}}=\frac{3}{8}\]

\[{{r}_{eq}}=\frac{3\times 5}{8}=\frac{15}{8}\]

\[\]

\[=0.13\]