Q. Consider the circuit shown in the figure. The current $I_{3}$ is equal to
A) $5\ amp$
B) $3\ amp$
C) $- 3\ amp$
D) $- 5/6\ amp$
Explanation
(d) Suppose current though different paths of the circuit is allows :
After applying $KVL$ for loop (1) and loop (2)
We get $28i_{1} = - 6 - 8 \Rightarrow i_{1} = - \frac{1}{2}A$
and $54i_{2} = - 6 - 12 \Rightarrow i_{2} = - \frac{1}{3}A$
hence$\ i_{3} = i_{1} + i_{2} = - \frac{5}{6}A$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.