Q. The diode shown in the circuit is a silicon diode. The potential difference between the points $A$ and $B$ will be
A) 6 $V$
B) 0.6 $V$
C) 0.7 $V$
D) 0 $V$
Explanation
(a) In the given condition diode is in reverse biasing so it acts as open circuit. Hence potential difference between $A$ and $B$ is $6V$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.