Diode V-i Characteristics

NCERT notes that a forward-biased silicon diode drops a roughly constant voltage of about 0.7 V once it is conducting, because the current rises steeply beyond the knee voltage while the junction voltage stays nearly fixed. Applying Kirchhoff's voltage law to the series loop, the battery voltage equals the diode drop plus the resistor drop, so $V_R=V_{battery}-V_{diode}=5-0.7=4.3$ V. The current is then $I=\frac{V_R}{R}=\frac{4.3}{100}=0.043$ A, which is 43 mA. The value 50 mA is wrong because it ignores the diode drop entirely and uses the full 5 V across the resistor. The value 57 mA is wrong because it wrongly adds the diode voltage to the supply instead of subtracting it. The value 5 mA is wrong because it misplaces a decimal, treating the resistor as 1000 ohm rather than 100 ohm. As stated in NCERT Class 12, Chapter 14, the forward drop of a silicon diode is close to 0.7 V. A magnitude check confirms that 43 mA across 100 ohm produces 4.3 V, which together with the 0.7 V diode drop restores the 5 V supply.