Electronics question

SuperMatty · 08-10-2002, 10:26 PM

My girlfriend recently applied for a job and she was asked this question during the interview. We are both having some trouble with it.

If you have 2 bulbs in series, one being a 60 watt bulb, and the other being a 120 watt bulb. Which will glow brighter?

We decided the 120 watt bulb would be brighter, for the reason that P = i^2*R...then:
R1 = 60 watt bulb
R2 = 120 watt bulb

60 = i^2*R1
120 = i^2*R2

So from that you can find that R2 = 2*R1. And since we know the current is the same through both bulbs, the 60 watt bulb will have a voltage drop of V = i*R1; and the 120 watt bulb will have a voltage drop of V = i*2*R1. So wouldn't the 120 watt bulb glow brighter?
According to the interviewer, the 60 watt bulb would glow brighter. But she no longer remembers how he justified it.

Can anyone shed some light on this?

mp3z24 · 08-11-2002, 01:18 AM

Quote:

Originally posted by SuperMatty
Can anyone shed some light on this?

AKKKKK

what a bad pun!!!!

So i think you are backwards on the calculations for the resistance of the light bulbs... the lower power bulb shold have higher resistance.

and yeah....this question bugged the $hit out of me for about 15 minutes, but then i found this ...

The Answer To Your Question

Another Answer...

~mike

SuperMatty · 08-11-2002, 12:12 PM

Hmm....yeah those examples are somewhat related to this. The first one is a bit more crude in its explanation, simply citing that the lower power bulb will have a higher resistance, but how can you show this when both are connected in series? I thought my approach was correct. The other example refers to when either are connected alone to a voltage source.
I mean, is that the only way to figure this one out? By first calculating the resistance of each bulb when connected alone in the circuit, and then using those values?

Rob · 08-11-2002, 12:30 PM

Okay, here we go.

The 60 watt bulb has twice the resistance of the 120 watt bulb

V = IR & P = IV

combinding the two equasions gives

P = (V*V)/R (Can't find the squared button)

Hence to double the power you halve the resistance, and vica-vera.

As the to blubs are in series they act as a potential divider.

As the 60 watt bulb has twice the resistance it'll have twice the voltage drop across it.

Hence for the 60 watt buld

P = (2/3 V * 2/3 V) / 2R

and for the 120 watt bulb

P = (1/3 V * 1/3 V) / R

Removing the constants

for the 60 watt buld

P = (2/3 * 2/3 ) / 2 = 2/9

and for the 120 watt bulb

P = (1/3 * 1/3) = 1/9

Which means that there's twice as much power going thru the 60 watt bulb as the 120 watt bulb, so it should be a fair bit brighter.

SuperMatty · 08-11-2002, 12:38 PM

That's all good and well Rob...but is there any way to determine your first assumption, that the 60 watt bulb has 2x the resistance of the 120 watt bulb...what I'm saying is...is there any way to determine that without placing both in a circuit each by itself and calculating the resistance? Or is it just 'common knowledge' that a 60 watt bulb has 2x the resistance of an 120 watt bulb?

Rob · 08-11-2002, 12:54 PM

P = Power (Watts)
V = Voltage (Volts)
I = Current (Amps)
R = Resistance (Ohms)

V = IR & P = IV (Standard equasions)

combinding the two equasions gives

P = (V*V)/R (Can't find the squared button)

P is inversly proportional to R

Hence to double the power you halve the resistance, and vica-vera. So a 60 watt bulb will have double the resistance of a 120 watt bulb

SuperMatty · 08-11-2002, 01:11 PM

aaahhhh...duh

jol · 08-11-2002, 01:20 PM

i tried that, the smaller will shine brigher, since it has a thinner wire inside it = more resistance = more energy transformed into heat+light

Maveric · 08-11-2002, 02:47 PM

And the faster it will burn out!!!