3 ms·
"average current through the bulb should be proportional to the average voltage across it" That is exactly correct, the reason they were seeking clarification,
by lux_scintilla 5mo ago
"average current through the bulb should be proportional to the average voltage across it" That is exactly correct, the reason they were seeking clarification, and the core of suggested solution.
V=I*R
If V = Hours and I = Minutes, then by necessity R=Hours/Minutes. Typically a light bulb has mostly fixed resistance (R). Adding a potentiometer to the circuit allows you control the value of R.
- labcomputer 5mo agoSo lightbulbs actually dont have fixed resistance. The tempco is pretty big, and temperature of course depends on power (with an annoyingly large time lag when power is reduced). That being said, the bulb does have a well-defined resistance at a given point in time, so voltage and current are of course not quantities that can be indefinitely controlled. This falls into the same category as “why isn’t my power supply with voltage and current controls working correctly?!?”
- Neywiny 5mo agoOh maybe I misunderstood. I thought they meant separate meters, so I didn't understand what was hard about that
- raphman 5mo agoOh - it didn't occur to me that the original poster might have thought about three different circuits - one with a voltmeter, one with an amperemeter, and one driving the light bulb. Maybe that was their intention. I originally assumed that the bulb would be somehow connected to voltmeter and amperemeter.