current
According to Ohm's Law, if the resistance is zero, the power dissipated by a resistor is unmeaning.
$$P_d = R \times I^2$$
$$I = \sqrt{ \frac {P_d}{R} }$$
Can one calculate the resistance given that power is defined? Only by knowing power, how can one calculate the maximum current of a zero-Ohm resistor?
I'm a bit confused as I've seen there is zero-Ohm resistors with specified power in the market.
Since this is homework, I'm not just going to give you the answer, but I will show you a method for attacking this problem. I'll use the same designators rawbrawb did. Whenever you show a schematic, add designators to make it easy to talk about. We don't want to be saying "the second from top resistor next to the switch" or something.
Immediately you should be able to see that this circuit has two parts, the mess of stuff formed by R1-R5, and R6. These two parts are in series. R6 is already as simple as it gets. You should be able to see that R1-R5 is equivalent to a single resistor from the rest of the circuit's point of view. Once you have the equivalent formed by the R1-R5 mess, you have two resistances in series and the solution should be trivial.
The problem now gets down to solving the equivalent resistance of the R1-R5 circuit. This you can break down into pieces. Pretend a 1 V source is applied to this circuit. If we can find the current drawn from that 1 V source, we can find the equivalent resistance by Ohm's law.
Start out by taking away R3. Now you just have two voltage dividers, R1-R4 and R2-R5. Each of these can be simplified into a Thevenin source. From inspection, you can see that the R1-R4 source is 1/2 V and 2 Ω. You can also see that the R2-R5 source is 2/3 V, and 2 seconds with a calculator tells you the resistance is 1.5 Ω. Now you can put R3 back between the two Thevenin sources:
At this point you should be able to solve for the voltages at each side of R3 easily. Once you have those, you go back to the R1-R5 circuit with 1 V applied and figure out the total current. From the total current, you can find the equivalent resistance. with the equivalent resistance of R1-R5, you go back to the original circuit and solve the total current.
I am confused about how pressing the switch actually changes the current being delivered to the [74HC04].
The 74HC04 has a high-impedance input. It will source or sink only miniscule currents, whether the input voltage is high or low (possibly it will sink more current at some in-between voltage, which is why we try to avoid driving it there).
So when the switch is open, there is essentially no current through the resistor. Therefore the voltage across it is 0, and the voltage at the input to the 74HC04 is VCC.
When switch is closed, it shorts the 74HC04 to ground. Since there's now 5 V across the resistor, some current (5 mA) will flow through it and everything is consistent. Still no current flows in or out of the 74HC04 input.
Best Answer
The power rating (and the tolerance) of a so-called 0\$\Omega\$ resistor is sort of vestigial- it comes from the series of resistors that the jumper resembles.
The actual current rating does not necessarily represent the same power dissipation as a similar resistor (the limit may be something like current density rather than power dissipation). For example, the Rohm MCR03ERTJ000 is part of the MCR03 series rated at 100mW, but the jumper version is 50m\$\Omega\$ maximum and rated at 1A max, so only 50mW.
So, it is not valid or safe in general to calculate the current rating, you should look it up in the data or contact the manufacturer, especially if high peak currents will occur, or your jumper could act as a fuse.