A person wants to attach a variable resistor/potentiometer to an appliance with power output of “P” to dim it’s light, what is the maximum power to be used by the variable resistor in terms of "P"?
HINT: the answer is independent of the voltage
Have fun!
>>8047065
No.
Isn't it P/2
If we reduce the appliance to a Thevenin equivalent, power consumption across the load resistor occurs when the variable resistor equals the source, so equal power loss over both means P/2
I'm just a chemistry student so I may not remember my circuits right
P/4
>>8047154
Isn't that the RMS though ?
I lied, the answer is dependant on the source voltage, and is equal to Vrms^2/(4R)
>>8047163
P is Vrms^2/R so that works out to P/4
>>8047166
yes
>>8047154
also yes
>>8047057
OP, that's a really poor way to dim a light.
when the knob is turned to where resistance = 0, everything's fine, and the bulb is as bright as ever.
And as resistance increases, the light will dim, but...
Once the potentiometer's resistance equals the bulb's (assuming the potentiometer's resistance goes that high), the total resistance is doubled, current flow is cut in half, and the voltage drop across the bulb is halved.
Since P=IE, power output of the bulb is divided by 4.
BUT now the potentiometer is also seeing the same current flow, and voltage drop as the bulb, and is now also dissipating the same amount of power as the bulb.
Since most potentiometers like the one in your pic are 1/4 watt or 1/2 watt, unless your original bulb output was 1-2 watts, the potentiometer will likely melt.
If the potentiometer's resistance doesn't go that high, it won't be able to dim as well.
In the real world, most AC light dimmers use a semiconductor device (a triac) to "clip" the AC waveform, cutting off the voltage during part of each sine wave.
This works OK for incandescent lights, but not florescents, which is why they take a special dimmer.
>maximum power to be used by the variable resistor in terms of "P"?
Finally, this is 100% troll.
"maximum power to be used" is determined by the power rating of the potentiometer, and the amount of power drawn relates to its resistance (which varies as the knob is turned) as compared to the bulb's resistance.
Yes, it's a really long answer for a troll question, but explaining this kind of thing is my Achilles heel.
>>8047193
this was probably a high school physics question that doesn't care about real world constraints
