Q.
The circuit has two oppositely connected ideal diodes in parallel. What is the current flowing in the circuit?
A) 1.71 A
B) 2.00 A
C) 2.31 A
D) 1.33 A
Explanation
(b) In the given circuit diode $D_{1}$ is reverse biased while $D_{2}$ is forward biased, so the ircuit can be redrawn as
[image:14_image210.emf]
[$\because$ For ideal diodes, reverse biased means open and forward biased means short]
Apply KVL to get current flowing through the circuit.
$-$ 12 + 4$i$ + 2$i$ = 0
$\Rightarrow \ = \frac{12}{6} = 2\ A$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.