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????
current.
no explanation, just remember.... current.
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
Originally posted by Peoplethinkimsarcastic: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.
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.
http://www.science.edu.sg/ssc/detailed.jsp?artid=6064&type=6&root=5&parent=5&cat=57
science center website.
Originally posted by skythewood: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.
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
Originally posted by eagle: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.
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.
Originally posted by skythewood: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.
Originally posted by eagle: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.
no it is simple stuff, and in O level syllabus
-----------------------------------------------------------
|
Households
|
-----------------------------------------------------------
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.
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.
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.
not really.
Originally posted by Garrick_3658: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 ;)
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.
Originally posted by skythewood: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.
Originally posted by eagle: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.
Hello i have a qn regarding transformers.
12V-----(step up transformer)x:20V----------------------y: 20V(stepdown transformer)----2V
hello, is the above theoratically correct. In o lvl specimen paper 2008 mcq qn40, voltage at x is higher than volatge at y
Secondly, if V at x is indeed higher, then why is it that in a normal circuit, there is no Pd across a wire.
why is it that in a normal circuit, there is no Pd across a wire.
1) The wires in a normal circuit are too short to have any significant effect on the resistance in the circuit.
2) In O levels, and probably A levels, the wires are also assumed to be of zero resistance. Actual reason is in (1).
In sensitive circuits on circuit boards (those green colour boards), resistance of wires will have a very huge effect. Hence the usage of gold (very good electrical conductor, very low resistance, lower than copper) for electrical contacts to such circuits. Even in your handphone, small amounts of gold are used as 'wires'.
but the Pd does decrese in long cables used to transport electricity after the voltage has been stepped up?
Originally posted by davidche:but the Pd does decrese in long cables used to transport electricity after the voltage has been stepped up?
-----------------------------------------------------------
|
Households
|
-----------------------------------------------------------
E.g.
p.d. across wires = 1V
p.d. across households = 22000 V
something like that...