The metering functions within your CB are a marketing afterthought or sales pitch usually made as cheaply as possible and hence won't be an accurate instrument. An external meter will usually provide a better reading but this depends on the manufacturer and what measurement circuitry they utilised. The easy way of knowing is price. (A more expensive meter is usually more accurate, the additional components required and calibration adds to the cost.)
Regardless, all RF power
meters are simple RF voltage measuring devices. A voltage reading isn't much use to us and to calculate the power
in watts we also need to know a resistance. The 'resistance' in this case is the resistive impedance Z and as we know the typical value for coax and the output of our radio is a nominal 50 ohms. Therefore the meter is calibrated to 50 ohms.
Taking Ohm's Law and utilising our resistive impedance Z of 50 ohms as R,
V = SQRT ( P * R ) or the square root of Power
in Watts multiplied by the Resistance in ohms, therefore :
V = SQRT ( 8W * 50 ohms ) = SQRT ( 400 ) = 20 Volts
This for your example of 8 Watts.
So when your meter is reading 8 Watts it is actually reading 20 Volts, (in this case RMS Volts).
This will only be accurate if the impedance is a true 50 ohm, a decent dummy load being the only way of accurately measuring your actual power
. If your impedance varies from that nominal 50 ohms we get different results.
So taking Ohm's Law again but for the case of P or power
and utilising your 8 Watt example with what we've already calculated :
P = V^2 / R or the Voltage squared divided by the Resistance, therefore :
P = 20^2 / 50 = 400 / 50 = 8 Watts.
This confirms our first calculations.
If the resistance changes we will get different results, let's say a nominal 15 ohms either way :
P = 20^2 / 35 = 400 / 35 = 11.43 Watts for a 35 ohm load
P = 20^2 / 65 = 400 / 65 = 6.15 Watts for a 65 ohm load
So if our impedance is lower at say 35 ohms the output wattage will be higher and conversely a higher impedance will give us a lower wattage.
Now, is the actual Wattage higher or lower due to actuality, or errors in our meter measurement?
That's where the real fun of uncertainty comes to play......and we haven't even touched base on the reactive components of L and C, (Inductance and Capacitance), that make up our impedance of Z let alone RF current. (The only way of gaining a more insightful answer.)
All gets a little complicated doesn't it?
What you've seen is that when the linear is 'off' your meter is giving a reading based on what comes after the linear as it is bypassed, so I assume your antenna. When the linear is 'on' it's circuitry is now inline and you are reading a higher value. From the mathematical examples above we can assume that the input impedance is lower on the linears than it is on your antenna.
That, is all we know for certain.
Measurement is one thing, knowing what we're measuring is quite another.
Anyway, all of the above was a good bit of fun for my grey thinking matter even if it only proved my brain is still working. If anyone learned anything from my musings to this question then I'm pleased, as that was my only intention.
Besides, despite knowing the mathematics to such things I much prefer working by 'rule of thumb' and 'er' on the side of caution. It's far less problematical and usually gives you less of a headache.
All the best,