   Whatsapp  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. 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. 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. 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:
$$\bbox[5px,border:1px solid grey] {R_L >> R_T} \tag{1}$$
In general,
when hooking up a load to a potentiometer, be sure that the load resistance far exceeds the maximum terminal resistance of the potentiometer if good control of the output voltage is desired. Fig. 3: Loaded potentiometer with $RL << R_T$.
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. 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. Fig. 4: Loaded potentiometer with RL >> RT.
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. 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. 4],
$$P_{max} = E^2/R_T = (10 V)^2/100 Ω = 1 W$$
If $R_T$ is reduced to $10 Ω$,
$$P_{max} = (10 V)^2/10 Ω = 10 W$$
which would require a much larger unit.

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