Relays coils is a current-driven device. If you can apply the specified voltage across the coil, there will be sufficient current and the relay will click. Ohmic losses in the winding is about the only difference.
+5V, +12V, +24V are some of the common supply rail voltages. Of course, the makers of the relays want to make relays for various systems. That's the rationale behind having a family of similar relays with different coil voltages.
P.S. Don't forget the protection diode in parallel with the coil.
All of these solutions will waste exactly the same amount of electrical energy, and generate the same amount of heat. Each of your circuits just changes which component gets hot.
These are all what's called a linear current source. A linear current source works (by definition) by converting excess voltage to heat. If your load requires 2V to reach the desired current of 5A, and the supply is 5V, the heat (power) \$P\$ will be the excess voltage \$E\$ times the current \$I\$:
\$ P = IE = 5A(5V-2V) = 5A\cdot 3V = 15 W\$
No way around that with any linear current source. You can spread it out or move it around different components, but you will never reduce it. Blame physics. The energy has to go somewhere.
If you want to reduce the wasted energy, you probably want a switched mode power supply (SMPS). The design of such is worthy of an entire book, but if you want a quick introduction, I suggest you read How can I efficiently drive an LED? Although you aren't driving an LED, the problem is essentially the same, since LEDs are ideally also driven with a current source.
However, since it sounds like your load is a fixed \$1 \Omega\$ resistor, you don't really need a current source. A voltage source would do just as fine, since a resistor is a current - voltage converter, by Ohm's law:
\$ E = IR \$
If it's allowable in your application, a simpler solution than an SMPS is to just switch the full battery voltage over your load on and off rapidly. If your supply is 5V then this will deliver 5A to your load. You can deliver 5A or 0A with low losses, and if you need something between those, then you switch it on and off rapidly, so the average current is your desired value. For many applications, this is good enough. If not, a SMPS is basically that, with an inductor added to smooth the current out to the average value across switching cycles.
Best Answer
You are getting confused because you are thinking that voltage flows.
Current flows, but voltage does not flow. This means you can block a current so that the current flow is zero. But you can't block voltage, not because voltage is unblockable, but because it doesn't make sense to talk about blocking it since it doesn't flow in the first place.
It is like how you can block a ball from falling off a table, but you can't block it from having \$E=mgh\$ of potential energy while sitting on that table.
simulate this circuit – Schematic created using CircuitLab
What is the current flow in the circuit on the left? It's an open-circuit which means the circuit path is broken and not closed which means no current can flow. Agreed?
If no current can flow, what is the voltage drop across R1? \$V_{R1} = IR_{1} = 0amps \times 100 \Omega = 0V\$ Agreed?
So if the voltage drop across R1 is 0V, what is the voltage \$V_{unknown}\$? It is V1 because V1 has a voltage of 1V, but R1 has a voltage drop of 0V.
The circuit on the right is the same except we have connected a voltmeter VM1 where \$V_{unknown}\$ is so we can actually measure it. The voltmeter is designed to allow no current to flow through it so the circuit on the right is no different from the circuit on the right. In other words, the voltmeter VM1 can be thought of as an infinite Ohm resistor as far as the circuit is concerned.
Don't even worry about output resistance (and more generally, impedance) of voltage or current sources until you understand the above.
simulate this circuit
These are a non-ideal voltage source and a current source with output resistance. Everything inside the dotted box is part of the non-ideal source.
For the non-ideal voltage source, \$R_{parallel}\$ cannot influence the voltage that the ideal voltage source applies to \$R_{load}\$. The ideal voltage source simply applies its ideal voltage to both \$R_{parallel}\$ and \$R_{load}\$. So we do not talk about internal parallel resistances as an output resistance for a voltage source.
But it should be obvious that \$R_{series}\$ can interfere with the voltage that the ideal voltage source is trying to apply to \$R_{load}\$, and the voltage that \$R_{load}\$ actually gets. If current flows, then some voltage will be lost across \$R_{series}\$ and change the voltage seen by \$R_{load}\$. The more current that it needs to supply, the more the voltage drops below ideal. So when we talk about the output impedance for a voltage source, we are talking about a series resistance.
simulate this circuit
For the current source, \$R_{series}\$ cannot influence the output current. It cannot affect the current running through \$R_{load}\$. The ideal current source simply drives whatever current it wants to drive and it goes through both \$R_{series}\$ and \$R_{load}\$. Therefore, we do not talk about series resistances as an output resistance for current sources.
But it should be obvious that \$R_{parallel}\$ can disrupt the current that the ideal current source is trying to push through \$R_{load}\$, and the current that actually flows through \$R_{load}\$. Since the current source outputs a fixed amount of current, some current will be split between \$R_{series}\$ and \$R_{load}\$. That means that as more current is supplied, less and less current reaches \$R_{load}\$. So when we talk about an output resistance for a current source, we are talking about an internal parallel resistance.
Now...what do they mean when they say an IDEAL voltage source has zero output impedance and an IDEAL current source has infinite output impedance? Look at the circuits for V1 and I2.
What does \$R_{series}\$ have to make the circuit for V1 into an ideal voltage source? It needs \$R_{series} = 0 \Omega \$, ergo ideal voltage sources have zero output resistance.
What does \$R_{parallel}\$ have to make the circuit for I2 into an ideal current source? It needs \$R_{parallel} = \infty \Omega \$, ergo ideal voltage sources have zero output resistance.
So by now, you should have realized that the output resistance for a current source is considered to be in parallel, not in series like it would be for a voltage source. If this parallel resistance is infinite it doesn't block any output current from reach the load, rather it stops output current from leaking away before it reaches the load.