Bias a valve on the hot side, moves it nearer class A. Poor eficiency but low distortion. Think Fender.
Bias cold and you move to class B almost class C if biased off. Good eficiency but high distortion. Think Marshall.
To quote 41mA dissipation means not much without the anode to cathode voltage reading as well. Then you can work out the idle power being dissipated.
This may help to explain it;
The maximum output from an EL34 is about 25 watts. So measure the anode voltage of the amp, for this exercise we will assume that to be 445 volts. Divide the 25 by the 445 then multiply by 0.7 the answer in ma. is the 70% plate load bias setting required for that amplifier.
The 0.7 means 70% Hot anode / plate dissipation. Therefore you can use 0.5 (50%) as a Cool AB setting, 0.6 (60%) as a Warmer AB setting the one I would prefer. Or go for the 0.7 or even 0.75 (70 - 75%) if you like it hot.
Example: 25 / 445 X 0.7 = 39.3 ma. Bias
The bias figure will depend on the anode voltage which will vary according to the current drawn by the valve. More current less volts, less current more volts. So you may need to do the calculation several times if large adjustments are made.
Example 2. 25 / 480 X 0.7 = 36.4 ma. Bias
Example 3. 25 / 410 X 0.7 = 42.6 ma. Bias
If looking for the Class A requirement, that's 100% plate dissipation. The figure to multiply by would be 1 but check the valve is operating within recommended plate voltages. Here is an EL34 example with 300 volts on the plate / anode.
25 watts max o/p of the valve divide by plate volts 300 multiply by 1 = bias current requirement for class A operation. Example. 25 / 300 X 1 = 0.083 = 83ma. or 25 / 350 X 1 = 0.071 = 71ma.
Not rocket science - dead easy elementary stuff.
If you have a cathode biased amp, you can measure the bias current drawn by dividing the voltage drop across the cathode resistor, by its resistance. Example Cathode resistor 470 ohms voltage measured across it 25 volts.
25 / 470 = 0.053 = 53ma
53 ma flowing through the cathode resistor is the sum of the anode and the screen grid currents together. Deduct about 5% screen current 2.65 subtracted = 50.35ma anode current.