Q. Current in the circuit will be
A) $\frac{5}{40}A$
B) $\frac{5}{50}A$
C) $\frac{5}{10}A$
D) $\frac{5}{20}A$
Explanation
(b) The diode in lower branch is forward biased and diode in upper branch is reverse biased
$\therefore i = \frac{5}{20 + 30} = \frac{5}{50}A$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.