Q. The figure shows a network of currents. The magnitude of current is shown here. The current I will be
A) 3A
B) 9A
C) 13A
D) 19A
Explanation
(c) Regarding Kirchhoff’s junction rule, the circuit can be redrawn as
Current in arm, $AB = 10 - 6 = 4A$
Current in arm, $DC = 6 + 2 = 8A$
Current in arm, $BC = 4 + 1 = 5A$
Hence, $I = 5 + 8 = 13A$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.