I'm asking this question again because I think my first attempt may have been a little confusing.

My confusion is stemming from how lowering the resistance leads to lower IR drop?

I have a simple circuit I've drawn below where we have a voltage supply in series with a resistor (used to model the resistance of the lines leading up to the CPU blocks) and 2 CPU blocks (just as an example).

Let's say the supply voltage is 5 V, that would mean we have I = V/R = 5 / 4k = 1.25 mA of current flowing through the circuit. This would lead to a voltage drop of V = I * R = 1.25 mA * 2kOhms = 2.5 V through the first resistor causing CPU block 1 to see a voltage of 2.5 V and then the same drop through the second resistor causing the second CPU block to see 0 V.

Now let's say we reduce the resistance to lower the IR drop (let's assume we lower both resistors to 1k Ohms), we'd now have I = V/R = 5/2k = 2.5 mA of current flowing through the circuit which would cause a potential drop of 2.5 mA * 1K = 2.5 V through the first resistor and the same through the second resistor, so both CPU blocks see the same voltage.

How did lowering the resistance cause IR drop to go down? Lowering the resistance simply increase the current according to Ohm's law, so I'm a little confused.

Any help would be greatly appreciated:

I'm asking this question again because I think my first attempt may have been a little confusing.

My confusion is stemming from how lowering the resistance leads to lower IR drop.

I have a simple circuit I've drawn below where we have a voltage supply in series with a resistor (used to model the resistance of the lines leading up to the CPU blocks) and 2 CPU blocks (just as an example.)

Let's say the supply voltage is 5 V, that would mean we have I = V/R = 5 / 4k = 1.25 mA of current flowing through the circuit. This would lead to a voltage drop of V = I * R = 1.25 mA * 2kOhms = 2.5 V through the first resistor causing CPU block 1 to see a voltage of 2.5 V and then the same drop through the second resistor causing the second CPU block to see 0 V.

Now let's say we reduce the resistance to lower the IR drop (let's assume we lower both resistors to 1k ohms,) we'd now have I = V/R = 5/2k = 2.5 mA of current flowing through the circuit which would cause a potential drop of 2.5 mA * 1K = 2.5 V through the first resistor and the same through the second resistor, so both CPU blocks see the same voltage.

How did lowering the resistance cause the IR drop to go down? Lowering the resistance simply increases the current according to Ohm's law, so I'm a little confused.