0
10 volt
5 volt
15 volt
The diode acts as a 'short' here, since it is in a forward biasing. The equivalent resistance across AB is (10)(10)/(10+10) kΩ = 5 kΩ. Thus the voltage gets divided in the ratio 2:1 across the first resistance and the equivalent resistance across AB. Thus potential difference across AB is (1/3)Vin = 10V.