Power Supply Circuit ±9V from a 6V centre tap transformer
RMS versus peak: -



RMS is the equivalent DC voltage that would dissipate the same heat in a resistor when the AC voltage is applied. You can do the math\$^1\$ if you want but, for a sinewave, the RMS is 70.71% of the peak OR the peak is \$\sqrt2\$ higher than the RMS.

When ever we talk about AC voltages (without specifying peak or peak-to-peak) we imply RMS hence, a 6V RMS signal will have a peak of 8.49 volts (nearly 9 volts). After passing through a rectifier diode this might be more like 8 volts DC on the smoothing capacitor.

However, transformers unloaded will run at a slightly higher voltage on the output so expect a volt more (or so) on the output.



\$^1\$ Take a 10V peak sinewave. Square the waveform - this produces a double-frequency sinewave between 0V (lowest peak) and 100V (highest peak). Take the Mean - 50V then take the Square Root - 7.071 volts i.e. 7.071 volts is the RMS value of a sinewave having a peak of 10V.

R = root, M = mean, S = square = root of the mean of the squared signal.