A semiconductor diode is connected with its p side at $+5\,\text{V}$ and its n side at $+3\,\text{V}$. Which best describes its state and the approximate diode voltage drop (assume Si, $V_F \approx 0.7\,\text{V}$)?

1. Bias direction: the p side ($+5\,\text{V}$) is at a HIGHER potential than the n side ($+3\,\text{V}$). The p-to-n potential difference is positive — this is the FORWARD-BIAS configuration. 2. In an ideal-diode model with a fixed forward drop, a forward-biased Si diode clamps about $0.7\,\text{V}$ across itself once it conducts. 3. The applied voltage difference $5 - 3 = 2\,\text{V}$ must equal: (diode drop) + (any drop in the rest of the circuit). So if the only other element is a series resistor, that resistor drops $2 - 0.7 = 1.3\,\text{V}$. 4. Option A inverts the bias direction. Option C confuses the magnitudes. Option D ignores the diode clamp altogether. 5. Sanity check: under reverse bias, the diode would NOT clamp at $0.7\,\text{V}$; the entire applied voltage would drop across the diode (until breakdown). _Source: NCERT Class 12 Physics Part 2, Ch 14, §14.6.1 (p-n junction under forward bias — barrier voltage clamp), p. 12._