Understanding Lightbulbs: Solving the Confusing Voltage Question in Physics

[Jarfi]

So I got this wrong in my physics test, the question goes: The original voltage is 230 v and the lightbulbs are marked so said voltage is assumed for them. How much does the voltage need to drop for a 60 watt bulb to start glowing like a 40 watt bulb.

I first tried to find the resistance/ohms of each bulb but couldn't since no current is given. I tried to find a current but realized that both the current and ohms could vary and give the same voltage, V=RI. So there is no fixed current or voltage to work with so for all I know this could be 1000 amps and 0,230 ohms... I ended up doing the only thing i could think off which was:

40w/60w*230v=153... 230-153=77 voltage drop.

My test was flawless apart from this and all the teacher wrote was "no" on my answer -_-

Can anybody explain lightbulbs and how this works to mee.. or ill start loosing sleep over this;)

 

[Integral]

You should know the power relationships, use the one that relates power, resistance and voltage to find the resistance of each bulb. Go from there.

 

[rude man]

Theoretically, you should take into account the fact that efficiency (light power output/electrical power input) varies with electrical power. And that current = V/R but R varies with voltage (current) also. So the total picture is pretty complicated.

I'd say a few assumptions need to be made. Depending on how rigoreous your course is, the simplest assumption is constant filament resistance and constant efficiency.

You just took the ratio of voltages = ratio of watts. But if I drop the voltage in half, do I really drop the power in half? I don't think so. So I would look at those ratios again.

 

[Jarfi]

Theoretically, you should take into account the fact that efficiency (light power output/electrical power input) varies with electrical power. And that current = V/R but R varies with voltage (current) also. So the total picture is pretty complicated.

I'd say a few assumptions need to be made. Depending on how rigoreous your course is, the simplest assumption is constant filament resistance and constant efficiency.

You just took the ratio of voltages = ratio of watts. But if I drop the voltage in half, do I really drop the power in half? I don't think so. So I would look at those ratios again.

There is no account of efficiency or such. We are given the law V=AI, so i can simply use that to find the current: 60W/230V=0,26A... then x=230v/0,26=885ohms. Then I know the power has dropped to 40W so I find the voltage corresponding: 40W=0,26A*XV->x=154 volts, that is the volt drop is 230-154=76V

Then I think, oh ofc the current must've dropped with the voltage uh let's use an equation with only watts, voltage and resistance... I=V/ohm so P=V*V/ohm that is P=v^2/ohm so 40w=v^2/885ohms->> v^2=0,0452 so V=0,21? wtf? totally wrong and different results.

I mean this totally and definitely depends on current and resistance.. and neither is give, you could have high current and low resistance or opposite and have the same results, what am i doing wrong?

nevermind, I found out... lol 40w=v^2/885ohms->> v^2=35400(ACCIDENTALLY DEVIDED) so V=188. Case closed.