Interesting result. How did you measure voltage and current?
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To that comes why the wavetops for the yellow graph is apruptly cut of, compared to the round tops of the orange and the red graph.
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The yellow graphs follows the power supply - probably similar to a sine wave (minus some offset for the diode).
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The peak values of the voltage increases with the capacitance, so that kind of explains it
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No, as the additional current is needed to charge the capacitor. It discharges itself via the lamp when the power supply is at a negative voltage.
One guess: The capacitor increases current and therefore the average load of the lamp - it gets hotter, and its resistance might increase. If the power supply is not perfect and its voltage depends on the current, this might lead to a higher peak voltage. I would not expect this, but something has to explain your observations and I don't see anything better at the moment.