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Physics Question!

E

efript

Senior member
You have four light bulbs, light bulbs one, two, three, and four. All have the same intensity. Bulb one is unscrewed causing only bulb one to go out. Now, bulb two?s intensity is 30% greater than it was originally. Bulbs three and four each have 2/3 of their original intensity. Based upon this information, how are the four bulbs wired? Support your answer.

my guess is that 1 and 2 are part of a parallel circuit and 3 and 4 are on a series connected right after bulb 2.
 
sounds about right for bulbs 1 and 2. but according to your setup, if 1 is removed, then 2, 3, and 4 are all in series. assuming they're the same resistance, they should all see an equivalent change in intensity. you're not quite there yet.
 
I'm giving it a bit of a try right now... Before doing any actual work, I'm thinking that 1 and 2 are in parallel with each other, and 3 and 4 are in parallel with each other. With 1 and 2 BOTH going to 3 and 4. Easiest way is to draw 2 diamonds connected at a point.

For simplicities sake, I chose 20 ohms for the Resistance of each bulb, and a starting voltage of 120 volts. Thus, the net resistance across 1 and 2 would be 10 ohms, across 3 and 4 would be 10 ohms, total resistance of the circuit 20 ohms, total current = 6 amps, with 3 amps going through each bulb. After removing bulb 1, total resistance would be 20 + 10 =30 ohms, total current = 4 amps, with all 4 going through bulb 2, and 2 amps going through each 3 and 4.

Unfortunately, it's my amps that are in agreement with your question. Not the power nor intensity of the bulbs. Therefore either my answer is wrong, or else I'm having a brain fart.
 
Yeah, the circuit has to have the loads both in series and in parallel like this.

Man, I'm twenty nine years removed from first year engineering.
 
Originally posted by: Squisher
Yeah, the circuit has to have the loads both in series and in parallel like this.

Man, I'm twenty nine years removed from first year engineering.
Click to expand...

methinks your circuit is short-circuited. Look again. 🙂
 
Originally posted by: DrPizza
Originally posted by: Squisher
Yeah, the circuit has to have the loads both in series and in parallel like this.

Man, I'm twenty nine years removed from first year engineering.
Click to expand...

methinks your circuit is short-circuited. Look again. 🙂
Click to expand...

you're right, I bow my head in shame. I guess I was too fast for my own good.

I did my final in one of my engineering classes by making a circuit like this with resistors on the outside circle too with like 8 legs, twenty nine freaking years ago.
 
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