Does a resistor reduce current and voltage?
Yes. this reduction depends on the connection scheme. The series resistor is used to limit the voltage by making the current constant, but in parallel, r is used to limit the current to make the voltage constant.
Let’s say you have a source of voltage and a load. the voltage v on the load is the voltage source; and the current in the load is v / rl, where r1 is the load resistance.
Now you insert a resistor r in series with the load; the voltage on the load is now reduced from v to v * rl / (rl + r); and the current is reduced from v / rl to v / (rl + r).
It depends on how the resistance is added to the existing circuit.
The circuit has a voltage or current source and a load. the source and the charge both have resistance, effectively. adding resistance means changing the existing resistance. the requires knowledge of the circuit details.
If I put a wire with almost no resistance to the terminals . I get a huge current through the wire (and the battery can deliver it).
Yet, if I measured the voltage across the terminals, I would find the voltage dropped to 1.2 volts. (It all depends on the capacity of the battery and its internal resistance, a big truck battery will sag a little and melt the wire, the operation of a welder or fuse, or the use of a little battery.
But, I take the same battery and I place a 10m resistor on the terminals. hardly any current will pass through the resistance, and when I measure the voltage, I find that it only drops to 11.99 volts.
So, really a resistor lets a limited amount of current flow through, and with this current, through this given resistance, I have x voltage drop, high resistance, low current, low voltage drop. low resistance, high current flow, high voltage drop.