Step 1: Decide the diode state first:
The positive terminal of the 15 V cell feeds node A through the 5 ohm resistor. Current flows in the direction of the diode arrow, from A to D. So the diode conducts, and we treat it as a wire with no drop.
Step 2: Use a voltage divider:
Between A and B the 10 ohm resistor and the branch (diode plus 10 ohm) are equal, so they act as one 5 ohm resistance. The 5 ohm series resistor and this 5 ohm resistance share the 15 V in the ratio of their resistances.
\[ V_{AB} = 15 \times \frac{5}{5 + 5} = 7.5\ \text{V} \]
Step 3: Check with Kirchhoff's current law at A:
Let $V_A = V$ with B as ground. Current in from the 5 ohm resistor equals the sum of the currents in the two 10 ohm resistors.
\[ \frac{15 - V}{5} = \frac{V}{10} + \frac{V}{10} = \frac{V}{5} \]
So $15 - V = V$, which gives $V = 7.5\ \text{V}$. The result is the same.
Step 4: Consistency of the diode:
The diode branch carries $7.5/10 = 0.75\ \text{A}$ in the forward direction, so our assumption that it conducts is right.
