These are the power transistors that are meant to reduce the DC input volts to the output volts. First make a sketch or where the wire go to/from and the transistor pins. One wire will go to the output terminal (+?) perhaps via a very small resistor and the current meter. This is the emitter. With two transistors, each emitter normally goes via its own resistor then combined to the current meter then to the terminal. Another wire will be the input voltage from the diodes and a very large capacitor, I would expect this wire to be commoned to both transistors. This is the collector. The third wire is the base wire, could be commoned or not and it should have on it the set output voltage +a bit (.7V?) and will go to the control board. Do not worry about leaving the 48V on, it would have destroyed anything within a few micro secs, but if anything starts to smoke. . . . Make suitable notes of the above connections. Disconnect one base wire and one emitter wire. try it again, one transistor will handle only half the rated current. if it works the transistor you have disconnected was faulty. Else check C->E with ohm meter should be > 1M both polarities. Put one probe on B, the C should be >100K and E less then 5K, swop them over, results should swop over if transistor good. if transistor looks good. Do the same to the other transistor. If both look good, then switch on and measure the volts on the lead to the Bs, it could be driving the transistors fully on if the control board is faulty.
Frank