Any wire, power supply, capacitor, etc. has a finite resistance (except superconductors which we'll ignore here). As a result, by Ohm's law, when a current flows through a resistance, a voltage results (which has the effect of appearing as a voltage drop at the end of the wire).
So, if you have a 6V battery, with an internal resistance of 1 Ohm - then if you connect a 5 Ohm resistor to its terminals, 5 V will appear at the terminals of the resistor and 1 Amp will flow. 1 V will appear to have vanished at the battery terminals - in reality, you have created a series circuit with a 1 Ohm and 5 Ohm resistor in series, where the 1 Ohm 'resistor' is actually the carbon electrodes of the battery, and the resistance of the electrolyte solution.
Similarly, wires, PCB traces, etc. all have a finite resistance and will produce similar 'voltage drops' in response to current.
Things are slightly more complex with AC, because you have inductive effects. All wires have inductance, as do alternators and transformers. For brevity I won't expand on this further, but will mention the term 'impedance' as being the combined effect of resistance (which affects DC and AC systems equally) and inductive (or, more strictly, reactive) effects.
So, as an example. My apartment receives 240 V. When I turn on my kettle (which takes 13 A), the lights dim slightly, but perceptibly. I've measured the voltage on the lighting circuit, and the result is that the voltage drops to 238 V. From this, I can work out the impedance of the electrical supply to my apartment, up to the point it reaches the panel (by using Ohm's law: V=IR. I=13, V=2; R = 0.15 Ohms). If, instead, I measure the voltage at the dual receptacle feeding the kettle, the voltage drops to 236 V. I can work out the total impedance as 0.30 Ohms - or in other words, the wiring from my panel to the receptacle has an impedance of 0.15 Ohms. This receptacle is supplied by 4 mm^2 copper cable - which has a resistance of 5 mOhms/meter - and is about 10 meters long (or 20 m round trip). As you can see, the figures are close enough.
Lights are very sensitive to voltage fluctuations, so perceptible dimming is normal when switching large loads. However, the dimming should not be excessive. Hence electrical codes specify the maximum voltage drop that a cable may experience under load - so if you are running long cables, you have to calculate the expected voltage drop and use cables with a higher current rating if necessary. E.g. a 2.5 mm^2 cable is safe for 20 A - but if you run 50 meters, it would have a round trip resistance of 800 mOhms - which would lead to a voltage drop of 16 V at 20 A - which would not be acceptable under most codes - so, if you actually wanted to use 20 A, you would have to use a larger size cable.
Also don't forget that motors tend to take a large current surge - e.g. an AC unit which takes 10 A when running, might take 40 A at startup for a fraction of a second. This can cause brief, but severe, voltage drop and cause visible light flickering.
When it comes to electronic power supplies, things get a bit more complex. Any power supply will have internal resistance (or impedance) - be it in the transformer, electronics, wiring, etc. Many modern electronic PSUs are 'regulated' - they measure the output voltage and adjust power flow to maintain it. Certainly, any PC PSU is regulated.
So why do PC PSUs show a voltage drop? Well, the way the regulation works is that power flow through the supply can be throttled back. The heart of the regulator is the 'error amplifier' this measures the 'error' in output voltage, and this is used to adjust the throttling. E.g. if the error voltage is at target, the throttle may set to 50% power. If error voltage is -0.1V (0.1 V below target), the PSU may throttle up to 75%. At +0.1V, it may throttle back to 25% power, etc. This is designed to compensate for the voltage drops in the PSU electronics. But what it means is that you have some form of voltage drop appearing. Let's say that this PSU is a 12 V, 10 A PSU. At 5 A, the PSU is operating at 50% load, and so the output voltage will be on target. However, let's say you increase demand by adding a 5 Ohm resistor in parallel. The expected current would be 7.4 A, but the PSU is throttled back, so the voltage sags. The error amp sees this and ramps up the power output. Things stabilize when the extra 'boost' ordered by the error amp, exactly matches the extra demand. In this case, the PSU would stabilize at about 75% output at 11.9 V - the error amplifier instructing a 25% boost in power flow in response to a 0.1 V error.
The point here is that there is a ramp up-down of power as error voltage changes. If you simply have a brick wall - voltage below target - 100%, voltage above target - 0% - what will tend to happen is that the voltage fluctuates uncontrollably, but the average is around target. (This is how your central heating furnace might work - it's either on at full power, or its off - it provides an irregular flow of heat, but the average temperature is pretty close to what you set, and because of the thermal inertia of your house, things average out OK).
So, you have to make it a ramp so that the power can be adjusted smoothly. In theory, the steeper your power throttling curve, the better your voltage regulation (you could have you error amp demand 25% boost in response to a 0.001 V error) - but in practice, in gets very impractically difficult to make the regulator produce smooth output the steeper your ramp gets - especially, if you have to cope with wildly fluctuating loads (e.g. CPUs). In fact, on the latest motherboard CPU voltage regulators, the throttling ramp is deliberately flattened (causing a larger than would otherwise be expected voltage drop under load - many people refer to this additional drop as Vdroop). Without this extra 'droop', when the CPU suddenly fires up another core, the sudden change in load can cause uncontrolled deep voltage sags and spikes for a short time (which could cause CPU instability). The added droop provides a stabilizing effect which avoids the wild fluctuations (and ensures that the CPU voltage doesn't drop to unstable levels - but at the cost of allowing CPU voltage to fall slightly below target at high loads).
(The reason for this stabilizing effect is due to delay in the feedback mechanism. The error amp can't detect fluctuations instantly, and the PSU cannot respond to the error amp instantly. The steeper your error amp's ramp, the more problems you run into with instability when load suddenly changes. E.g. if you suddenly double the load, by firing up a 2nd CPU core, you can get a big voltage dip, before the regulator has time to respond. The risk is that if the error amp is too sensitive, it will see the big dip and massively overcompensate, causing a power spike, which in turn causes overcompensation the other way, and a dip...then a spike... then a dip... giving uncontrolled voltage oscillations).
The other issue is that the regulator might only measure the voltage internal to the PSU. You may still get voltage drop on the wires between the PSU and the load. One way of getting around that is to measure the voltage at the load. This is done my CPU regulators on modern motherboards. The CPU actually has dedicated pins connecting direct to the power grids on the die, which are meant to be connected to the VRM's voltage sensor - this allows the VRM to regulate the voltage at the die, fully compensating for the voltage drop in the PCB traces, pins, etc.
Of course, the regulator can only regulate up to the limits of the PSU. If the PSU is throttled to 100%, then adding any further load will result in large voltage drops - because the error amp has no way to command the PSU to increase power further.
Finally, PSUs may provide overload protection. One quite popular way of doing this on bench PSUs is to use a technique called 'foldback' limiting. When the current approaches the maximum limit, the PSU regulator automatically reduces the target voltage to prevent the current from exceeding the maximum limit.
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