Electrical Question From A Student

[Charlie Bob]

150 uses 1.30 A, R is 88.5 ohm
10 uses .087 A, R is 1322

current of bulbs in series is high enough to light 10, not enough current to light 150 W


In a series circuit the current it's the same anywhere across the circuit.So both bulbs will use the same current regardless of bulb Watts.

 

[Besoeker]

I don't have the facilities to run an experiment. But you have the facilities to check my physics and my math. Any issues there?

You're assuming constant resistance. Whoever set the question almost certainly expected it to be taken that way.
But, as gar and others have pointed out, filament lamps don't have constant resistance.
Working on the basis that the filament operates at about 3000 degC (about 5400F) and an ambient of 20C (68F) the ratio of hot to cold for tungsten is about 14 to 1.
That results in a voltage across the 150W lamp of about 0.5V and a dissipation of around 50mW.

 

[charlie b]

So we can all agree that it was a badly designed question. That said, if a person is taking a test, and needs to pass the test in order to get a license, then the person has to get the question "right." What is "right," then? In this particular case, I would answer the question under the assumption of a constant resistance. Then I would issue a complaint to the licensing department, regarding the bad question.

 

[Besoeker]

[So we can all agree that it was a badly designed question. That said, if a person is taking a test, and needs to pass the test in order to get a license, then the person has to get the question "right." What is "right," then? In this particular case, I would answer the question under the assumption of a constant resistance.

Charlie

I'm an old fellow.
Had I been asked that question in a test 40 years ago I think I'd have made that same assumption and given the answers expected by examiner.

Had I been asked it yesterday and my future career depended on it, I'd still have given the same expected answers.

I might have felt inclined to include the caveate that it assumes constant resistance. But it wouldn't get a better score.

 

[erickench]

I might be able to help you with the voltage calculation assuming the light bulbs are in series.

First compute the total power: 150+10=160 Watts
Now calculate the current: 160/115=1.39 Amps
Now determine the resistance for each light bulb:

R1=150/1.39=107 Ohms
R2=10/1.39=7 Ohms

Total Resistance=107+7=114

Now the Voltage Drops:

V1=115x107/114=108 Volts

Maybe you were off by one volt?

 

[gar]

091028-2243 EST

eric that is an interesting take on the question, and I assume resistors instead of light bulbs..

The I needs to be squared in the R1 and R2 calculations.

So R1 = 150/1.94 = 77.3
R2 = 10/1.94 = 5.2

and the voltage drops are

V1 = 115*77.3/82.5 = 107.8
V2 = 115*5.2/82.5 = 7.2

These add to 115.0 as a check.

If these were special light bulbs, then the brightness would be a function their design voltages.

.

 

[erickench]

Well you could've used the power numbers to figure it out.

115x150/160=107.8