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I was thinking along Ohm's Law as well, combined with P = U x I (Power equals Voltage times Current). Those two combined, give P = U^2 / R (Power equals Voltage squared divided by Resistance). In other words, a resistance twice as high will result in half the power output. ;)
Ummmm? What? I mean Watt? And yet, your 'real world' test shows that the volume output remains [noticeably] the same. In other words, best check the battery drain. ;)
Remember I have not been able to establish the impedance of the nüvi 2495LMT speaker, only the one from the nüvi 2595LMT (4 ohm). From a point of standardisation across nüvi's not likely, but maybe the original speaker from the nüvi 2495LMT was 8 ohm too...
As for battery drain: if using an 8 ohm speaker instead of a 4 ohm speaker would only reduce the battery drain (but also reduce the volume output). Right?
That's not necessarily correct in regard to the volume as you've observed anecdotally. Think of the relative electrical resistance in the same way as a water pipe's diameter will effect the flow of water. A larger pipe [less resistance] allows more water to flow, but can the pump [amplifier] cope with the demands required by the smaller pipe [HIGHER resistance] is the question. Hence my comment in Post #10 in regard to the output source.
But hey Mate, if it works ok in the nuvi then it's all good ..... ;))