When the term loaded is used to describe voltage divider supply, it refers to the application of an element, network, or system to a supply that draws current from the supply. In other words, the loading down of a system is the process of introducing elements that will draw current from the system. The heavier the current, the greater is the loading effect. Recall from internal resistance of voltage source that the application of a load can affect the terminal voltage of a supply due to the internal resistance.
Fig. 1: Voltage divider supply.

Voltage $V_a$ is unaffected by the load $R_{L_1}$ since the load is in parallel with the supply voltage E. The result is $V_a = 120 V$, which is the same as the no-load level. To determine $V_b$, we must first note that $R_3$ and $R_{L_3}$ are in parallel and $$R'_3 = R_3 || R_{L_3} = 30 Ω || 20 Ω= 12 Ω.$$ The parallel combination gives $$R'_2 = (R_2 + R'_3) || R_{L_2} = (20 Ω + 12 Ω) || 20 Ω$$ $$= 32 Ω || 20 Ω = 12.31 Ω$$ Applying the voltage divider rule gives $$V_b = {(12.31 Ω)(120 V) \over 12.31 Ω + 10 Ω} = 66.21 V$$ versus 100 V under no-load conditions.
Voltage Vc is $$Vc = {(12 Ω)(66.21 V) \over 12 Ω + 20 Ω} = 24.83 V$$ versus 60 V under no-load conditions.
The effect of load resistors close in value to the resistor employed in the voltage divider network is, therefore, to decrease significantly some of the terminal voltages. If the load resistors are changed to the 1 kΩ level, the terminal voltages will all be relatively close to the no-load values. The analysis is similar to the above, with the following results: $$V_a = 120 V$$ $$V_b = 98.88 V$$ $$V_c = 58.63 V$$ If we compare current drains established by the applied loads, we find for the network in Fig. 2 that $$I_{L_2} = {V_{L_2} \over R_{L_2}} ={66.21 V \over 20 Ω }= 3.31 A$$ and for the 1 kΩ level, $$I_{L_2} = {98.88 V \over 1 kΩ} = 98.88 mA < 0.1 A$$ As demonstrated above, the greater the current drain, the greater is the change in terminal voltage with the application of the load.