Q.
In the given circuit for ideal diode, the current through the battery is
A) 0.5 $A$
B) 1.5 $A$
C) 1.0 $A$
D) 2 $A$
Explanation
(c) The current through the battery is
$I = \frac{10V}{5\Omega + 5\Omega} = \frac{10V}{10\Omega} = 1A$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.