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You are assuming cables and load to be one "unit". But this does not make sense in this case.
Typically, your load requires a specific power and for that it needs some specific current (calcualted by Ohms Law). This current now creates additional losses in the cables and these losses are lower when the resistance is smaller.
Take, for example, a 500 W ...
5
You are basically correct.
For any device that can be approximated as a resistor, using higher resistance wires will decrease power consumption.
This is usually not the primary goal of electronic devices. They need to use power to perform their function, and they are internally designed to use the power they need based on assumptions about the power ...
5
The higher the internal resistance of the cable, the higher the voltage drop will be across the cable. We can calculate this with Ohm's law and a voltage divider.
So we have a Vin than a cable with resistance Rcable and than we have the load resistance Rload
This gives us Vin = I * (Rcable + Rload) = Vcable + Vload This is fixed, can't change it.
And ...
5
The load will only draw the current it requires.
Taking something a bit more modest than a submarine, the starter in my car will draw a few hundred amps from the 12 volt battery when starting the engine, but the headlights will only draw 5 amps each, and the interior light will probably draw less than 1 amp from the same battery.
To control the speed of an ...
3
Power is seldom controlled. Power has two components. Electrical power from a battery is voltage multiplied by current. You can control voltage or current relatively easily, but it is difficult and generally not desirable to control both at the same time.
Mechanical power from a motor is speed multiplied by torque. Here again, you can control one or the ...
3
A good way is the voltage follower circuit. Select an opamp with a low voltage offset, and rail to rail. Make sure the current of the opamp is higher than what the mass flow controller input needs. Vcc is set to near the DAC output (you could go higher if the mass flow controller output will allow.
Make sure bypass capacitors are used on the Vcc terminals.
3
Generally, one would like to minimize power losses via Joule heat.
No, not really. Example: space heater. You want the joule losses. It's not about total "loss", it's about where it happens. You want the power at the heater in your room, but not at the cables on their way to power plant.
Power in the load is "gain", power in cables is loss.
So, I am ...
2
Let's try a practical example
Let's suppose we want to power a string of pole lights at the end of a very long drive, say 2000 feet. We need 60,000 lumens to get the job done. That will require 600 watts of energy.
Supply is constant-voltage 120W, North American supply.
Got the parameters so far? 600 watts. 2000 feet (600m) with 10% deviation ...
1
I = V/R
so the current is limited by the resistance, both internal (all batteries have some) and external: the wires and device or motor connected to the battery terminals (which all have a non-zero resistance, unless they are extremely cold superconductors).
Motors (inductors moving under a load) generate reverse EMF, which counteracts some of a batteries ...
1
Moving itself on grass and dragging something on grass are two different things. For moving itself you can use bigger wheels to deal with the rougher surface, but need gear down for to compensate for the increased torque requirement to maintain the same speed. But dragging something...you need a new motor. No way around that.
Coefficient for rolling ...
1
Use a spring scale to measure the drag force of the rabbit on "beton" (concrete?). Do the same thing for the grass.
The ratio of these two numbers is the same as the power ratio you'll need for the motor, all other things (speed, slope, flatness, etc.) being equal.
1
You're overthinking this. Just unplug the device. That is the exact equivalent of increasing the cable resistance to infinity. Yes, it minimizes power losses. In fact, it minimizes overall power consumption: the device no longer consumes any power at all. It also no longer does anything, but… planet saved.
1
Thanks to the comments I got, I decide to answer my own question.
I am told that I should consider the resistance of the wires and of the load separately. When I do that, I get that the power dissipated in the wire(s) is worth \$ P_\text{Joule, wire} = \frac{V^2R_\text{wire}}{(R_\text{load} + R_\text{wire})^2} \$ which can be thought as a function of \$ R_\...
1
The relays are designed and manufactured to be operated with the stated nominal voltage across the coil when energized.
If you want to make a relay to be operated from a 5V power source, you will use fewer turns of thicker wire than if it is to operate from 12V or 24V, all other things being equal.
For the relay to work mechanically, the ampere-turns of ...
1
Well, a general \$\sin\$ wave is given by:
$$\text{V}(t)=\hat{\text{a}}_1\cdot\sin\left(\omega t+\varphi_1\right)\tag1$$
Using linear components we get a current that is also a \$\sin\$ wave but with a different amplitude and phase with the same frequency, so:
$$\text{I}(t)=\hat{\text{a}}_2\cdot\sin\left(\omega t+\varphi_2\right)\tag2$$
The power that is ...
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