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:
Answer & explanation
Correct answer: option 1
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]
$=0.2×5×10^{-3}×10^{3}$
$=1V$
$∴V_{AB}=V_1+V_2+V_3=1+1+0.3=2.3V$