If I do understand your question correctly, I think you can just use one relay, but "the other way around" (instead of 1 output serving 2 outputs, using it as 2 inputs serving 1 output):
simulate this circuit – Schematic created using CircuitLab
An "easy enough" method is to on/off PWM the supply to the motor with a variable duty cycle square wave.
The 2SD2390 transistors are Darlingtons rated at 10A each.
So use say 4 in parallel.
Apply variable duty cycle square wave drive to the bases.
You'll need a flyback diode/ diodes.
A bunch of 1N400x (1A low frequency diodes) may be the cheapest / ...
Yes. Increasing the load on your battery, either by adding more fans and pumps or replacing the existing fans/pumps with fans/pumps that require more current, will lower the voltage available to all of these devices, including the cooler module.
The reason is that your battery is not an ideal voltage source, so as the load current increases the terminal ...
So we have some ITS4060S-SJ-N, so I decided - why not just test it? Well, it passed up to 40V straight through to the OUT pin when the load drew around 20mA. That indicates that it won't disconnect the load if the supply voltage is too high. I suppose I would need something like http://www.ti.com/product/TPS2663 and the increased cost if I wanted that ...
The generated power is, among other things, a function of the number of photons that strike the panel per square meter per second (the photon flux), AND the wavelengths (the "colors") of the photons that are striking the panel.
A photon's wavelength determines its energy. When a photon strikes the solar panel's surface (the silicon material), if the photon ...
PV panels act like constant voltage sources. For example, if you have a 12V panel, it might put out 17V with no load, 12V up to about 0.3A then drop off rapidly as you increase the load above its max power point. The max power point depends on incident light and panel efficiency.
Your solar panel is working totally normally.
Photons in the sunlight hit the panel. As they are absorbed, they knock electrons free, and those electrons form the current you measure. On a cloudy day, there is less light, so fewer photons, and so fewer electrons.
The amount of energy hitting the panels is dissipated as it comes through the atmosphere, so clear days produce more power than cloudy days - you should have noticed this.
Try recording the weather each day and the power output each day and you will see...