Wilkinson divider power distribution

For a Wilkinson divider, what would happens on power distribution if port3 is open circuit o/c and port2 is matched. (port1 is input port).

All power delivered to port3 will be reflected, and half of it deliver to port1.

Could someone explain in detail?

If you do this and if the power is relative high, first of all the internal 100 ohms resistor will burn out.

any other detail explanation beside that?

Just a simple simulations for the above situation, it gets:

S11=-6dB (25%)
S21=-3dB (50%)

It means that if the resistor is working, half of the input power from port 1 goes to port 2, 25% of the input power reflected back to port 1.

Where is the rest of the 25%? Because wilkinson power divider actually is not pure lossless, the 100 Ohm resistor consumes the rest 25% power.

Here is the interesting point:
When port 2 and port 3 are both 50 Ohm, the potentials across the 100 Ohm resistor leads are the same, so in this case the wilkinson power divider is lossless if input power from port 1.

However, when port 3 fails, goes to open or short, the potentials over 100 Ohm resistor are no longer the same, so current would flow across it, and the power would become heat.

thank you.