On the circuit shown here, if the voltage drop across forward baised diode is 0.3 V, the voltage difference between A and B would be:

2.3 V

The correct answer is Option (1) → 2.3 V

$V_2$, Voltage drop across forward biased diode = 0.3 V

$V_1$, Voltage drop across AC = I × R [By ohm's law]

$=0.2×5×10^{-3}×10^{3}$

$=1V$

$V_3$, Voltage drop across CD = I × R [By ohm's law]

$=0.2×5×10^{-3}×10^{3}$

$=1V$

$∴V_{AB}=V_1+V_2+V_3=1+1+0.3=2.3V$