For example, the diagram in my text book: Variable potential divider shows a filament lamp, in series with a uniform resistive wire, which can have its voltage and current varied by moving the sliding contact, e.g., a rotatable wheel. However, why is a potential divider able to reduce the current through the filament lamp to zero, but a variable resistor in series with the filament lamp cannot reduce current to zero?


To answer the question as well as possible the short vertical length of wire below the variable resistor should be absent. In practice it makes no effective difference, but people start asking about it's resistance and affect on the result. SO assume the bottom wire is attached TO the bottom of the resistive element. Thusly -

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The unloaded potentiometer voltage varies with it's position on the resistive element.
At the top $V_\text{out}$ will be $V_\text{in}$.
At the bottom $V_\text{out}$ will be 0.

When loaded with a bulb $V_\text{out} \leq$ the unloaded value.
At the bottom $V_\text{out} = 0$,
so $I = V/R_\text{load} = 0/R_\text{load} = 0$,
so no light.

When the bulb is connected to the battery in series with a variable resistor the resistance will be (Rresistor + Rbulb) and the current will be
Ibulb = Vin / (Rresistor + Rbulb).
For Ibulb to be zero the combined resistance must be infinite.
As the combined bulb and variable resistance is always finite the current will always be non zero. eg for a 10 Megohm combined resistance and 10 volt supply the current will be 10V/10,000,000 Ohm = 1 uA.

Practical issues:

For a tungsten filament incandescent bulb Rbulb drops as Ibulb drops.

If an LED is used then the effective resistance of the LED is extremely high at very low currents.

  • $\begingroup$ Thanks for the response! I know that resistance is proportional to length, and hence the shorter the wire the less the resistance. However even at an extremely small length (I.e., at the top) surely there would still be some resistance, hence still a small voltage/current. $\endgroup$ – John Feb 17 '16 at 13:39
  • $\begingroup$ @John If thisis homework/assignment you should say so. If not also please advise. | You are reading things I did not say into what I wrote. Assuming you are still talking about the potejntial dividier (and you should make clear what you are talking about) - I said - "voltage varies with its position on the resistive element." ie Vout_unloaded = Vin x (% distance freom bottom). When distance = 50% V = Vin x 50%. When distance = 0% Vout = Vin x 0%. Yes? Ony once happy that that is so, move on to next case. $\endgroup$ – Russell McMahon Feb 17 '16 at 13:54
  • $\begingroup$ This is not homework, it's just understanding a topic. I know you did not state that R is proportional to L, but this is why the resistance and voltage varies - correct? I understand that the voltage varies with the movement of the slider, as this affects the resistance. When the slider is at the bottom, we're saying that the voltage out of the bulb = 0, hence the current = 0. However for the voltage to be 0, surely this assumes that the potential divider (at the bottom) has infinite resistance, which is not possible. Must there not be a very small amount of voltage across the bulb? $\endgroup$ – John Feb 17 '16 at 14:15
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    $\begingroup$ Whether the question is homework is irrelevant. $\endgroup$ – DanielSank May 19 '17 at 6:28
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    $\begingroup$ @TausifHossain Your comment about wire resistance is correct BUT that is not what John was talking about. His comment was re divider resistance being large and reflected his misunderstanding of how a divider works. I have updated my answer and hopefully it will help somewhat to clarify the difference in the two cases. $\endgroup$ – Russell McMahon Jun 25 '17 at 13:44

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