"I am running a 50W ultrasonic transducer. I have a 1KV sine wave 153KHz across the transducer with a current of approx 0.05A giving me 50W." from your first post, no you have not, you have a VA of 50W, that is current times voltage. This is not real power. Real power is only dissipated in resistors, later on you say that the resistive impedance of the transducer is 20 ohms, this means that the voltage drop across the resistive part is about .05 X 20 = 1V and so the real power is 1V X .05 = .05W. All the other voltage that you measure is across the series capacitance. This gives a series impedance of ~1kV/.05 = 20K ohms. To make this thing work properly you have to add in series an inductive impedance of 20K ohms. When you used your transformer, its leakage inductance made up for this, so the circuit started to resonate (good) and the voltages rose (good) and so you should have brought your driving power down, else you will over dissipate the transducer. Measuring voltages around a circuit that is somewhere near resonance, will only tell you that much, NOT how much real power is involved.
Frank
Frank