Physics Question Urgent 2

Peoplethinkimsarcastic SG

Brightness of a bulb depends on its current or potential difference (voltage)?

If it is current, P=IV which means that voltage is low when a bulb is bright? o.o

If it is voltage, P=IV again which means that current is low when a bulb is bright? o.o

Makes no sense. However when P= I^2R or P=V^2/R it makes sense. Can someone explain????

skythewood SG

current.

no explanation, just remember.... current.

Zandermccoll SG

V = p.d
potential difference, according to Wikipedia

"The difference in electrical potential between two points is known as voltage."

So I guess V is just the difference between the two points, not the actual charge moving through the wire (current). Meaning... that's probably what causes the bulb to be bright. When the current kena resisted by the filament, light + heat energy produced.. err.. UM..

AHH NOT SURE LA

skythewood SG

o level don’t ask you question like, why is the current the one that cause the bulb to be brighter. you just need to know current is the one.

eagle SG

Brightness of a bulb depends on its current or potential difference (voltage)?

If it is current, P=IV which means that voltage is low when a bulb is bright? o.o

If it is voltage, P=IV again which means that current is low when a bulb is bright? o.o

Makes no sense. However when P= I^2R or P=V^2/R it makes sense. Can someone explain????

it depends on both

i.e. it depends on power, which is energy per unit time.

skythewood SG

more likely than not... you will come across this question... why do the power cable transfer electricity at a very high voltage?

the answer is to keep current low, which will keep heat loss low.

arigatoast SG

http://www.science.edu.sg/ssc/detailed.jsp?artid=6064&type=6&root=5&parent=5&cat=57

science center website.

eagle SG

more likely than not... you will come across this question... why do the power cable transfer electricity at a very high voltage?

the answer is to keep current low, which will keep heat loss low.

This confusion is extremely common among students. It's not because heat loss or power loss is solely due to current.

That's because the voltage across the wires is at a lower fraction as compared to the whole circuit.

Current is used because I²R is the current in the wires * resistance.
If you use V²/R, you have to take the potential difference across the wires only, and not the total voltage that is being transmitted. They will work out to be the same.

skythewood SG

not the best source, but read this.

http://en.wikipedia.org/wiki/Electric_power_transmission#Losses

warning: will contain stuff that is definitely out of your syllabus

skythewood SG

This confusion is extremely common among students. It's not because heat loss or power loss is solely due to current.

That's because the voltage across the wires is at a lower fraction as compared to the whole circuit.

i like the E= V²/R formulae. R should be taken as constant. according to this formulae, the voltage should be low to keep E, the heat lost, low.

eagle SG

i like the E= V²/R formulae. R should be taken as constant. according to this formulae, the voltage should be low to keep E, the heat lost, low.

Students normally don't use the voltage given in power transmission because the voltage given includes the potential difference across the households, and not solely the voltage across the wires.

skythewood SG

Students normally don't use the voltage given in power transmission because the voltage given includes the potential difference across the households, and not solely the voltage across the wires.

not sure what you are talking about, but i guess you are right. must be pretty deep stuff.

eagle SG

no it is simple stuff, and in O level syllabus

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When you increase your transmission voltage, I decreases. And V=IR across the wires also decreases. So in actual fact, when you raised the voltage for transmission, the potential difference across the wires drop. So the V²/R will give you a lower value.

skythewood SG

yup. when using the V²/R, consider the P.D .

when using the I²R, consider the actual current.

simple stuff. now i just need to find a kid taking O level to agree with you.

Garrick_3658 SG

In short, exclude the potential difference of other appliances and use only the e.m.f. supplied to the circuit, that is travelling in the wires?

When using the E = V^2/R formula, that is.

skythewood SG

not really.

eagle SG

In short, exclude the potential difference of other appliances and use only the e.m.f. supplied to the circuit, that is travelling in the wires?

When using the E = V^2/R formula, that is.

use only the p.d. across the wires

it's not called e.m.f. travelling in the wires ;)

skythewood SG

do take note that the power supply is not directly connected to your home, but has to go through some giant transformer somewhere to be converted to 240V. and it is not exactly connected in series.

so for o level kids, high voltage ==> lose less energy.

stop digging.

eagle SG

do take note that the power supply is not directly connected to your home, but has to go through some giant transformer somewhere to be converted to 240V. and it is not exactly connected in series.

so for o level kids, high voltage ==> lose less energy.

stop digging.

It would clearer if you say high transmission voltage means less energy loss during transmission

because high voltage does not mean losing less energy, even for O levels.

skythewood SG

It would clearer if you say high transmission voltage means less energy loss during transmission

because high voltage does not mean losing less energy, even for O levels.

nvm.