In the circuit shown in figure, if the diode forward voltage drop is 0.3 V. the voltage difference between A and B is.
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 2.3 V
From Kirchhoff’s voltage law
$V_A-0.2 \times 10^{-3} \times 5 \times 10^3-0.3$
$-0.2 \times 10^{-3} \times 5 \times 10^3=V_B$
$V_A-V_B=1+0.3+1=2.3 ~V$