Basic misunderstanding about dropping down voltage going in a TTL IC using resistors
Looking at the resistor in isolation doesn't make sense. You need to consider the transistors on either end:







simulate this circuit &ndash; Schematic created using CircuitLab

Q1 represents the TTL output driving the data bus. Q2 represents the TTL input of a chip. Ignore the transistor part numbers, this is an example. R2, R3 and Q2 make up the input of a TTL chip.

When Q1 is switched off, a small current flows through R1, R3 and the base of Q2. This turns on the transistor and the input is considered "on". Because of the nature of transistor junctions, the base will tend to have a voltage not exceeding 0.8V. The other end of R3 can rise higher; the exact value of R3 is very dependant on the design of the TTL input.

When Q1 is switched on, the current through R1 is diverted entirely through Q1 and the bottom end of R1 has a voltage close to zero. No current flows in Q2 and it is "off".