Year 12 physics anyone?

[olly]

Hey guys,
My lil bro asked me for help for his physics subject and I don't really remember much, actually I hardly remember none haha..
So I was wondering if anyone here could help us out, it's doing my head in.. look at the pic below

The arrow shows the way the current is flowing and the blue things are light bulbs.
If the switch is closed..
what happens to the intensity of bulb A?
the intensity of bulb B?
the potential difference across bulb A?
the potential difference across bulb B?

My opinion was that the switch being on would turn the circuit into a parallel circuit, thus the A and B would become dimmer and C would stay the same?
As for potential difference, i have no idea haha..

 

[iluszn]

from my way of thinking.

Circuit closed
Bulb A has more current and should be brighter
Bulb B should be brighter aswell

Why i think this. i could just be drunk off work but. If you have the circuit open, the current has to travel through the 3 bulbs in series. With the circuit closed, electricity tends to take path of least resistance thus travelling through A and B , more electricity would travel around to bulb A and B..

Also there should be some numbers to accompany that diagram. unless they just want to know how to calculate potential difference.

Heck i could be way off as i have skimmed over the question between doing work.

 

[joshuafreeman]

good luck didnt do yr 12 physics any other subject though could have gien an answer

 

[GSRSOL]

My memory of physics is pretty vague too ....

with the switch open, the 3 globes are running in series, and therefore the 1st globe from the battery will be brightest, followed by globe B and then globe C.

With the switch closed, globe C becomes parallel to globes A and B, which are still in series. So globe A should be dimmer, globe B should be similar and globe C should be brighter than when the switch is open.

dont quote me on that though LOL

 

[jimmyg]

Bulb C will go out and A + B will share the power C was using.

As there is more power being used up buy A & B the potential diff will go up also at the cost of C.

I think

 

[D3bb4]

If the switch is closed..
what happens to the intensity of bulb A?
Become brighter. More current through it.
the intensity of bulb B?
Become brighter, more current through it.
the potential difference across bulb A?
Potential difference being voltage - increases. V = I x R, thus as current rises due to short circuiting C, and the resistance of the bulb remains the same, potential difference (V) increases.
the potential difference across bulb B?
Same as A.

 

[VR-4Squid]

I think someone needs to call a registered electrician - that's some of the weirder wiring i've seen

 

[EVO-00X]

I actually did year 12 physics (frikken 14 years ago now) and I cant remember than shit lol... actually deb looks like to have it right as when youy put theory into practice you get similar results.

 

[to4garret]

C would go out and A and B will be brighter.

electricity travels the path of least resistance, so its easier to go through the switch rather than the bulb (c)

 

[jimmyg]

If the switch is closed..
what happens to the intensity of bulb A?
Become brighter. More current through it.
the intensity of bulb B?
Become brighter, more current through it.
the potential difference across bulb A?
Potential difference being voltage - increases. V = I x R, thus as current rises due to short circuiting C, and the resistance of the bulb remains the same, potential difference (V) increases.
the potential difference across bulb B?
Same as A.

Agree 100%. Well said too

 

[JayRome]

lol bring him over and let him have a chat to my dad lol

 

[Aussie Babsy]

a and b will have the same current flow through them but b will be dimer because of the voltage drop across a, which is coursed be the resistance of A, c will be dimmer then B for the same resin but when the switch is closed C will not light up because the current will take the parth of least Resistance.

 

[D3bb4]

a and b will have the same current flow through them but b will be dimer because of the voltage drop across a, which is coursed be the resistance of A, c will be dimmer then B for the same resin but when the switch is closed C will not light up because the current will take the parth of least Resistance.

wrong.... sorry.. but no.

Voltage drops both across A and B (bulbs use power, thus may as well be classed as resistors). Whatever one has the most resistance will be brightest, as power = current squared x resistance; so higher resistance would be brighter than lower, as would be more wattage (wattage is measure of power)
In this question, assuming that the bulbs are same wattage, would be same resistance, and same light emission.

If the switch is closed..
what happens to the intensity of bulb A?
Become brighter. More current through it.
the intensity of bulb B?
Become brighter, more current through it.
the potential difference across bulb A?
Potential difference being voltage - increases. V = I x R, thus as current rises due to short circuiting C, and the resistance of the bulb remains the same, potential difference (V) increases.
the potential difference across bulb B?
Same as A.

 

[Skankyjoe]

C would go out and A and B will be brighter.

electricity travels the path of least resistance, so its easier to go through the switch rather than the bulb (c)

I agree with this guy. I'm an electrician and that's definatly some funky wiring Though i'm quite a drunk electrician at the moment.


A should be the brightest, B a little bit dimmer and C won't be on. Because its wired in series.

Oh and by the way they are LAMPS not BULBS, bulbs grow in the ground.

 

[D3bb4]

I agree with this guy. I'm an electrician and that's definatly some funky wiring Though i'm quite a drunk electrician at the moment.


A should be the brightest, B a little bit dimmer and C won't be on. Because its wired in series.

Oh and by the way they are LAMPS not BULBS, bulbs grow in the ground.

As i just said; i don't agree. If you have 2 equal resistors (or 2 equally powered lamps); voltage drop across each = half the voltage source through each, thus light emitted is exactly the same.

I've done electrical engineering subjects through uni, and i'm not drunk...

 

[Skankyjoe]

And as i said....i am drunk.

I just have to wire it up, not know how it works Oh wait i should know that too.

 

[D3bb4]

hahaha. well said.

For when the resistors are different; you need to use voltage divider rule.. but i cbb writting that up atm, as would have to explain too much. google it.

 

[Skankyjoe]

It's getting to technical for year 12 physics and a forum now. I learnt it at tafe, a while back but well it was a while and it never happens in my field anyway.

 

[Aussie Babsy]

As i just said; i don't agree. If you have 2 equal resistors (or 2 equally powered lamps); voltage drop across each = half the voltage source through each, thus light emitted is exactly the same.

I've done electrical engineering subjects through uni, and i'm not drunk...

now i think about it that sounds right

 

[MDK87]

That diagram doesn't give enough info dude.
I think all your brother really needs to know is that -

When the switch is closed, lamp C will turn off, and lamps A & B will slighty increase in brightness as a result of increased potential difference accross them.

This would be different if a constant current source was used, but you're just using a cell (battery), which of course isn't.

Like I said, dodgy diagram, but I think that's really all the question is asking for, a general understanding of electron flow...........oh, physics aye.......well in that case, hole flow.

Good luck to him, exams coming up.