The circuit is powered with 3V a battery and contains a black box device who's power consumption we're measuring with a shunt. The rated required voltage for the device is 3V. Whit a 1 Ohm shunt, the voltage drop on the shunt is 50mV meaning that the most of the voltage drop is caused by the black box device. Image name: Case 1.
However, the device power consumptions needs to be measured in an environment where we can't physically place the measuring equipment so we have to use long wires that add 10 Ohms of resistance in the circuit. Image name: Case 2
After the installation we were expecting to measure a decreased current and to probably conclude that doe to additional resistance this doesn't represent the power consumption of the device. However, we've experienced that the current in the circuit has increased almost twice. The only explanation for this is that the voltage drop has for somehow decreased.
Is it a reasonable assumption that the device has detected a voltage decrease on its ends and switched to a lesser resistance mode to be able to consume more power? The reasoning for this assumption is that given that the device is battery powered, it is prepared for a voltage drop and has a mechanism to cope with it.