For the unloaded potentiometer, the output voltage is determined by the voltage divider rule, with $R_T$ in the figure representing the total resistance of the potentiometer as shown in Fig. No.1. Too often it is assumed that the voltage across a load connected to the wiper arm is determined solely by the potentiometer and the effect of the load can be ignored. When a load is applied as shown in Fig. No.2, the output voltage $V_L$ is now a function of the magnitude of the load applied since $R_1$ is not as shown in Fig. No.1 but is instead the parallel combination of $R_1$ and $R_L$. The output voltage is now $$V_L = {R' E \over R' + R_2} \text(with R' = R_1 || R_L)$$ If you want to have good control of the output voltage $V_L$ through the controlling dial, knob, screw, or whatever, you must choose a load or potentiometer that satisfies the following relationship: (7.2) $$\bbox[5px,border:1px solid blue] {\color{blue}{R_L >> R_T}} \tag{1}$$ In general, For example, let's disregard Eq. (1) and choose a 1 MΩ potentiometer with a 100 Ω load and set the wiper arm to 1/10 the total resistance, as shown in Fig. no.3. Then $$R' = 100 kΩ || 100 Ω = 99.9 Ω$$ and $$VL = {99.9 Ω(10 V) \over 99.9 Ω + 900 kΩ} = 0.001 V = 1 mV$$ which is extremely small compared to the expected level of 1 V.
In fact, if we move the wiper arm to the midpoint, $$R' = 500 kΩ || 100Ω = 99.98 Ω$$ and $$V_L = {(99.98 Ω)(10 V) \over 99.98 Ω + 500 kΩ}= 0.002 V = 2 mV$$ which is negligible compared to the expected level of 5 V. Even at $R_1 = 900 kΩ$, $V_L$ is only 0.01 V, or 1/1000 of the available voltage. Using the reverse situation of $R_T = 100 Ω$ and $R_L = 1 MΩ$ and the wiper arm at the 1/10 position, as in Fig. no.4, we find $$R' = 10 Ω || 1 MΩ = 10 Ω$$ and $$V_L ={10 Ω(10 V) \over 10 Ω + 90 Ω}= 1 V$$ as desired.
In general, therefore, try to establish a situation for potentiometer control in which Eq. (1) is satisfied to the highest degree possible. Someone might suggest that we make $R_T$ as small as possible to bring the percent result as close to the ideal as possible. Keep in mind, however, that the potentiometer has a power rating, and for networks such as Fig. No.4, $P_{max} = E^2/R_T = (10 V)^2/100 Ω = 1 W$. If RT is reduced to 10 Ω, $P_{max} = (10 V)^2/10 Ω = 10 W$, which would require a much larger unit.