Sure, let’s break it down. A push‑button with a typical contact resistance of 0.1 Ω at a 10 mA load drops about 1.0 mV, which isn’t huge, but the real kicker is the repeated dwell time when the button stays depressed. If the button lingers 5 % of the time, that’s an extra 0.5 mV per event, and if you fire 1 000 times a day the cumulative loss can push that 10 % figure you mentioned. A leaner design would use a spring‑loaded reed or a solid‑state latch, or at least a lower‑resistance contact, to shave off that steady drain. Let's model the duty cycle and pick the smallest R that keeps the voltage drop under 0.5 mV per cycle.

Sure thing. For a 0.05 Ω contact at 10 mA you drop 0.5 mV each press. If the button stays down 5 % of the time, that’s an extra 0.025 mV per cycle, negligible on the 0.5 mV budget. Over 1 000 presses, the extra drain stays under 0.025 V—well below the 10 % battery loss threshold. So the cost bump is the only trade‑off. If that extra cost is acceptable, keep the 0.05 Ω. If not, a reed will eliminate the voltage drop entirely, but you’ll need to account for the magnetic field and switch latency. Let's run a quick duty‑cycle script to confirm the numbers and then decide.