KCET 2026 Physics Solved Paper

Concept:A diode is reverse biased when the p-side (anode) is at a lower potential than the n-side (cathode).That is, $V_p < V_n$.Explanation:Check the potentials at the anode and cathode for each figure.Option A: $V_p = 5 \mathrm{~V}$ and $V_n = 0 \mathrm{~V}$.Here, $V_p > V_n$, so the diode is forward biased.Option B: $V_p = -20 \mathrm{~V}$ and $V_n = -10 \mathrm{~V}$.Since $-20 \mathrm{~V} < -10 \mathrm{~V}$, we get $V_p < V_n$.So, the diode is reverse biased.Option C: $V_p = 15 \mathrm{~V}$ and $V_n = 10 \mathrm{~V}$.Here, $V_p > V_n$, so the diode is forward biased.Option D: $V_p = 20 \mathrm{~V}$ and $V_n = -5 \mathrm{~V}$.Here, $V_p > V_n$, so the diode is forward biased.Only in option B is the condition $V_p < V_n$ satisfied.Answer:Option B