Old Ed wrote:
"Couild you explain this concept, and/or provide some references?"

Suppose we adjust a variable d-c supply to full-scale indication on an
external meter. Next, install a chopper (lossless on-off circuit) driven
at a high frequency to produce a square wave interruption of the d-c
with a 50% duty cycle, and insert the chopper contacts in series with
the external meter.

The chopper connects the external meter 50% of the time and disconnercts
it 50% of the time. The meter reads 50% of full scale.

Another way to reduce the scale reading to 50% is to insert a resistor
in series with the meter. If it is an 0-1 ma meter and if it has an
internal resistance of 1000 ohms, insertion of a 1000-ohm resistor in
series with the meter will reduce the meter scale indication to 50%.

The chopper as part of the meter source eliminates current to the meter
50% of the time.

The resistor which has the same effect and produces the same scale
indication as the chopper on the effective output current exacts its
loss of 0.0005 amp x 0.5 volt or 0.25 milliwatt during 100% of the time.

The chopper eliminates the power-losing resistor by substituting
off-time in the power source. The source only supplies the power used by
the load. With a resistor limiting power to the load, the source
supplies its loss and the load power.

The power in the load, a meter in our example, is the same using either
the resistor or the chopper. The resistor is analogous to a Class-A
amplifier. The chopper is analogous to a Class-C amplifier.

The off-time has the same effective opposition to current to a load as a
dissipative resistance. As the time-limited currented opposition to load
current consumes no power, it is called a dissipationless resistance.

Best regards, Richard Harrison, KB5WZI