Help: need a compact shunt resistor for current measurement
ok, so the size of the current-measurement shunt is largely about how much current your circuit draws.  the current meter works by assuming a shunt of a specific resistance, so that when a specific (maximum) current goes thru it, a specific voltage will be dropped across it (simple Ohms Law, V = I * R), with a certain degree of error according to the tolerance of the shunt resistor.

so for example, a 12V battery driving a 10ohm load = 1.2Amps, &amp; therefore the power delivered to the load P = V*I = 14.4 Watts.  to make a decent measure of that current without the shunt resistor having too much of a negative impact on the voltage seen by your load, you'd probably want a shunt resistor of about one hundredth - say, 100 milli-ohms, and with 1.2 Amps flowing through that it'll drop 120mV - that's the voltage your "amp meter" is measuring and showing as being "1.2 amps".  but only when you use a shunt of 0.1ohms (100mohms), because that's what the meter is calibrated for.

the "problem" with shunt resistors is that they dissipate heat, because they're a simple resistor &amp; you're measuring the voltage dropped across it as a result of the current passing through it.  the power absorbed by the shunt P = (I^2)*R (I-squared x R).  So the power dissipated in the shunt goes up with the square of the current.

so for that 50 Amp shunt example you gave, even though I don't know its resistance (it can't be gleaned from that ebay page), that's a lot of amps!  so it'll get hot.  which means it needs to be big enough to dissipate that heat without destroying itself.

that's why I asked you how much current you're expecting to draw.  I'm concerned by your answer, because I asked you what the max current might be, and you gave me an answer in watts - power.  let's say it's 1.6 amps.  that's not much compared to the 50 amps of the shunt example you've given.  like, 10 times higher measuring capability.

so yes, doing some figures in my head, you could get away with a somewhat smaller shunt, because your currents are less than 1/10th of what that shunt is capable. the trouble now is, not knowing the resistance of that shunt, nor what the range of shunt-voltage that amp-meter is expecting to measure, I can't advise you on what ohm your shunt needs to be to work with that meter.  until we know that, it's almost impossible to point you at something specific.  but generally speaking, at the current you're suggesting, a simple 1 watt resistor with 1% tolerance, of some ohms, is probably all you need.

but my bigger questions are:  why 3 meters?  if you're manually switching between 2 batteries with a switch, in which case only 1 is ever being used at any given time, why not just have 1 meter?  and, what value is the voltage &amp; current reading to a user of this thing?  why do they need to know voltage &amp; current draw of this system, and are they expected to know that 9.0 V means the 3-cells-in-series battery is almost flat, and that 12.6 is fully charged?  why not just have a few LEDs showing battery state based on crude voltage measurements?