how to work out resistance

CandleLight

When a 100 watt bulb is lighting in a 230V socket how do you work out the resistance?

http://en.wikipedia.org/wiki/Electrical_resistance#DC_resistance is V divided by current. what is the current in above situation

DublinDilbert

p = v i and v = i x r

Combining the above:

p = v x v / r

r = v x v / p

brightspark

Two basic formulas expressed in various ways
(this is just the simple DC formulae no allowance for power factor etc.)

V=IxR
R=V/I
I=V/R

W=VxI
V=W/I
I=W/V

so from W=VxI and I=V/R

we get W=VxV/R (Vsquared / R)
or R = Vsquared / W

Does that help??

(Another consideration is that most materials change resistance as they heat/cool so the resistance of a lightbulb isn't that straight forward!)


(Dublin Dilbert is a faster typist!!)

CandleLight

DublinDilbert wrote: »

p = v i and v = i x r

Combining the above:

p = v x v / r

r = v x v / p

I do not understand what is p?

cml387

529 ohms.

Power(P) is Voltage squared divided by resistance.

230 X 230 = 52900.
52900 divided by R is 100
Therefore R is 529

CandleLight

cml387 wrote: »

529 ohms.

Power(P) is Voltage squared divided by resistance.

230 X 230 = 52900.
52900 divided by R is 100
Therefore R is 529

cheers thanks

the higher the wattage of the bulb the lower the resistance is it?
a 60 watt bulb would be 529 divided by 60?

cml387

CandleLight wrote: »

cheers thanks

the higher the wattage of the bulb the lower the resistance is it?
a 60 watt bulb would be 529 divided by 60?

Yes higher watts lower resistance.

60 watt would be 52900 divided by 60 i.e 881 ohms

CandleLight

cml387 wrote: »

Yes higher watts lower resistance.

60 watt would be 52900 divided by 60 i.e 881 ohms

yes i meant 52900, thanks for your help;)

M cebee

if you measure the tungsten lamp with an ohmeter the cold resistance will be lower than the calculated value

CandleLight

brightspark wrote: »

Two basic formulas expressed in various ways
(this is just the simple DC formulae no allowance for power factor etc.)

V=IxR
R=V/I
I=V/R

W=VxI
V=W/I
I=W/V

so from W=VxI and I=V/R

we get W=VxV/R (Vsquared / R)
or R = Vsquared / W

Does that help??

(Another consideration is that most materials change resistance as they heat/cool so the resistance of a lightbulb isn't that straight forward!)


(Dublin Dilbert is a faster typist!!)

this is just the simple DC formulae no allowance for power factor etc.

what is power factor? House is AC 230 isn't it?

M cebee

ya 230 nominal voltage

cml387

CandleLight wrote: »

what is power factor? House is AC 230 isn't it?

May I ask what your interest is in finding out these answers, apart of course from the obvious advantages in the quest for knowledge?

Power factor is a complicated calculation based on the effect of inductive loads, but is not necessary in simple calculations of AC current and resistance.

Bruthal

Yes 528 ohms is for the 100w bulb. You may find measuring it when in hand with an ohm meter gives you a very much lower result in the range 30 to 40 ohms.

The 100w/230v gives 0.435 amps
V/I gives the 230/0.435 = 528 ohms as has been said.

But measure the bulb with an ohm meter like i said above will give a resistance far lower, this is because when the bulb filament heats to white hot, its resistance increases drastically to the 528 ohms. So at the instant of switch on, the bulb will have up to 15 times its normal operating current for a very short period of time until it reaches operating temperature.

This can be significant for large banks of incandescent bulbs, although this scenario probably does not really happen now.

Power factor has little bearing on incandescent bulbs which are mostly resistive loads.

Bruthal

brightspark wrote: »

Two basic formulas expressed in various ways
(this is just the simple DC formulae no allowance for power factor etc.)

V=IxR
R=V/I
I=V/R

W=VxI
V=W/I
I=W/V

so from W=VxI and I=V/R

we get W=VxV/R (Vsquared / R)
or R = Vsquared / W

Does that help??

(Another consideration is that most materials change resistance as they heat/cool so the resistance of a lightbulb isn't that straight forward!)


(Dublin Dilbert is a faster typist!!)

A but you typed far more, so he cheated:)

CandleLight

cml387 wrote: »

May I ask what your interest is in finding out these answers, apart of course from the obvious advantages in the quest for knowledge?.

was just curious , read something about resistance and wanted to understand it in a practical way

Bruthal

CandleLight wrote: »

cheers thanks

the higher the wattage of the bulb the lower the resistance is it?
a 60 watt bulb would be 529 divided by 60?

Divided by 60 but then multiplied by the 100 watts because the resistance in the 60 watt would be 100/60 times higher than a 100 watt bulb.

60 watts would be a higher resistance than 100 watts, so you could say
529 ohms for the 100 watt, divided by 60 x 100 would give you the resistance of the 60 watt bulb. (or 529 x 100/60)

So 529x100/60 = 881 ohms