In the figure potential difference between A and B is:

The correct answer is Option (4) → 8 V

The upper terminal of the 24 V battery is positive, so the diode is in forward bias. An ideal diode in forward bias has zero resistance and no voltage drop, so it behaves like a plain wire and A is at the same potential as the junction of the two left-hand resistors.

The lower 8 kΩ resistor on the left and the 8 kΩ resistor on the right (through the diode) are in parallel between that junction and B.

    \( \displaystyle R_{\parallel} = \frac{8\times8}{8+8}\,\mathrm{k\Omega} = 4\,\mathrm{k\Omega} \)

This is in series with the upper 8 kΩ resistor.

    \( R_{\mathrm{total}} = 8\,\mathrm{k\Omega} + 4\,\mathrm{k\Omega} = 12\,\mathrm{k\Omega} \)

    \( \displaystyle I = \frac{24\,\mathrm{V}}{12\,\mathrm{k\Omega}} = 2\,\mathrm{mA} \)

The potential difference across the parallel part, which is \( V_A - V_B \):

    \( V_A - V_B = 2\,\mathrm{mA}\times4\,\mathrm{k\Omega} = 8\,\mathrm{V} \)