The above problem was asked in GATE 2018 Electrical paper. I got the answer as 50. But I am confused if my answer is correct. The funny part is all top 3 coaching institutes has given 3 different answers for this problem. Please help me with this problem.
Note the average inductor current is shown here. The average switch current is half that. That is, 12.5A per switch.
So Kreatryx got the right answer, despite dyslexia.
Intuitively you would think that Made Easy's point that the load is always connected to the source means the output voltage must be the input voltage, but that is not correct.
If you consider just one half, what is happening is you are creating a square wave voltage and passing it through a filter to retain the average DC voltage component. Since the wave is 50-50 mark-space, the average output voltage from both sides is 50V.
When you put them in parallel that voltage does not add. It is just the same as putting two 50V batteries in parallel, you still have 50V.
What does change, just like putting two batteries in parallel, is the current taken from each is divided by two.
In a multi-phase continuous buck converter (which this is) the output voltage (ideally) is the input voltage times the duty cycle.
In this case the duty cycle is 50%. You can think of the LC filter as averaging the input voltage which is a 50% duty cycle 100V square wave. So Vout = 50V.
Output power is then 50^2/1 = 2500W.
Input power = output power (ideally) so input current is 25A.
Each switch will carry the input current for 50% of the time, so the average switch current is 12.5A.