Q. In the arrangement shown in figure, the current through $5\Omega$ resistor is
A) 2A
B) Zero
C) $\frac{12}{7}A$
D) 1A
Explanation
(a) The circuit may be redrawn as shown in the adjacent figure
Here $E_{eq} = 12V,r_{eq} = \frac{2 \times 2}{2 + 2} = 1\Omega$
$i = \frac{E_{eq}}{R + r_{eq}} = \frac{12}{5 + 1} = \frac{12}{6} = 2\Omega$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.