In the circuit shown in figure, if the diode forward voltage drop is 0.3 V, the voltage difference between A and B is.
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 2.3 V
The current of 0.2 mA flows from A to B through the two \( 5\,\mathrm{k\Omega} \) resistors and the diode, which are in series; the diode points from A to B, so it is forward biased and drops 0.3 V. The capacitor branch carries no steady current, so it does not affect the drop. By Kirchhoff's voltage law, \( V_A - V_B \) is the sum of the drops along the path.
\( I = 0.2\,\mathrm{mA} = 0.2\times10^{-3}\,\mathrm{A} \)
\( IR = 0.2\times10^{-3}\,\mathrm{A} \times 5\times10^{3}\,\Omega = 1\,\mathrm{V} \text{ for each resistor} \)
\( V_A - V_B = 1\,\mathrm{V} + 0.3\,\mathrm{V} + 1\,\mathrm{V} = 2.3\,\mathrm{V} \)
Note: A similar question was also asked in CUET (UG) 2023 on 23-06-2023 (Slot 1).