Hummina Shadeeba
10 MW
If I pass 100 amps at 1 volt through a resistor it’s the same wattage as if 1amp through a much bigger resistor at 100 volts, and heat is measured in watts so should be the same temp produced no? But only amps create heat.
Heat from amps, but what of watts?
- Thread starter Hummina Shadeeba
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markz
100 TW
Power (P) = Volts (V) x Amps (I)
100W = 100V x 1A
100W = 1V x 100A
But you must also keep in mind the resistance of the wire connecting the resistor to the battery.
That wire could very well be thin with high resistance, which would heat up or that wire could be very thick with low resistance which wouldnt heat up as much or as high. Also how long is that wire, thin and very long has very high resistance as opposed to thin and short which would have lower resistance then thin/long. Temperature also plays a role. Higher temperature means higher resistance.
Calculate the current is V=IR
100W = 100V x 1A
100W = 1V x 100A
But you must also keep in mind the resistance of the wire connecting the resistor to the battery.
That wire could very well be thin with high resistance, which would heat up or that wire could be very thick with low resistance which wouldnt heat up as much or as high. Also how long is that wire, thin and very long has very high resistance as opposed to thin and short which would have lower resistance then thin/long. Temperature also plays a role. Higher temperature means higher resistance.
Calculate the current is V=IR
larsb
1 MW
Power losses (in watts) creates heat. It can be a bit of a mind bender as it’s often stated as ri2 losses for motors so it looks to be only current related but it’s power loss.
Power=Volt*current, volts dropped over the resistance of the motor can be calculated as R*I and result is R*I*I
Power=Volt*current, volts dropped over the resistance of the motor can be calculated as R*I and result is R*I*I
Hummina Shadeeba
10 MW
With high voltage power lines I understood the voltage was increased so as to have less losses in the long wires. If I’m calculating the losses in a 10 mile wire and let’s say it’s one ohm total and 10,000 volts and let’s say 100 amp are passed through... how many watts are lost as heat?
goatman
10 MW
you could try putting 100v 1 amp to your tongue
do you still have a tongue?
do you still have a tongue?
larsb
1 MW
Hummina Shadeeba said:
Ri2 for the loss is 1*100*100 so 10kW out of the total 10.000V*100A of power is lost. efficiency of transfer is 99%
Hummina Shadeeba
10 MW
I^2R to decide power lost. voltage isn’t in the equation. Voltage is relevant in that greater voltage will allow more current but the voltage itself isn’t relevant other than that no? I mean in determining the wattage loss to heat. Thanks for your input and I generally know ohms law but as it relates to power lost in a resistor and the heat produced..while V x I is wattage..when figuring the resistive losses as a wattage voltage isn’t there and just relevant in how many amps it pushes.larsb said:
Thanks
100v to the tongue could be deadly and best to stay within seemingly safe voltages.
Hummina Shadeeba
10 MW
goatman said:
Think would either need be 100v and a whole lot of current based on my tongue resistance or some low voltage applied that would limit me to one amp due to the tongue resistance
goatman
10 MW
if amps kill you not volts
does that mean 1 volt 100 amps will kill you
but 100 volts 1 amp wont
if a taser is 50,000 volts there cant be any amps
at 0.1a its considered lethal, 5000 watts
https://taserguide.com/how-many-amps-in-a-stun-gun/
if you got tazed you get a couple burn holes in your skin
but what happens if you get shot with 50,000 amps and 0.1volts?
does that mean 1 volt 100 amps will kill you
but 100 volts 1 amp wont
if a taser is 50,000 volts there cant be any amps
at 0.1a its considered lethal, 5000 watts
https://taserguide.com/how-many-amps-in-a-stun-gun/
if you got tazed you get a couple burn holes in your skin
but what happens if you get shot with 50,000 amps and 0.1volts?
Ianhill
1 MW
- Joined
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- Messages
- 2,918
Dielectric breakdown VOLTAGE.
So after a certain voltage a substance conducts and flows current.
50vdc across dry skin no breakdown no current flow i feel fine
65vdc wet skin i start to conduct as we now Volts over restlstance equals amps more the volts increae the more current im capable of flowing.
Then theres the frequency or ac voltage to consider our body is capacitive and theres a limit to how fast you can flip our polartiy nit sure on the driving forces of this it maybe our shape etc like a tesla coil output breakdown with spikes but hold more charge as a sphere not sure but 50 hz will shock you while much higher hertz give more of a burning sensation.
As for 100amps 1v its tesla vs edison again, the form of the resistor would be much different one very low resistance the other very high i can bet on which one will be the cheapest to make.
So after a certain voltage a substance conducts and flows current.
50vdc across dry skin no breakdown no current flow i feel fine
65vdc wet skin i start to conduct as we now Volts over restlstance equals amps more the volts increae the more current im capable of flowing.
Then theres the frequency or ac voltage to consider our body is capacitive and theres a limit to how fast you can flip our polartiy nit sure on the driving forces of this it maybe our shape etc like a tesla coil output breakdown with spikes but hold more charge as a sphere not sure but 50 hz will shock you while much higher hertz give more of a burning sensation.
As for 100amps 1v its tesla vs edison again, the form of the resistor would be much different one very low resistance the other very high i can bet on which one will be the cheapest to make.
Hummina Shadeeba
10 MW
goatman said:
Assuming the voltage is high enough to pass that current you explode. Don’t think at .1v it will be anywhere near 50,000 amps even if electrodes under the skin
How a taser can be a huge voltage but low current...how? Is the power supply limited?
Ianhill
1 MW
- Joined
- Sep 25, 2015
- Messages
- 2,918
https://www.ncbi.nlm.nih.gov/pmc/articles/PMC2763825/
Reading that 500v looks to be the magic number for a dry human body to start to really conduct some deadly amps and find themselfs climbing the stairway to heaven.
Not saying theres circumstances with barefeet water and alike that can vastly reduce this number but with decent rubbee shoes dry day most people survive house hold shocks only when theres water and lots of sweat on your skin and a good ground path no rubber shoes or chilling with your 7kw ring circuit on an extension cord looping it round like a cowboy in the bath then your in trouble.
Reading that 500v looks to be the magic number for a dry human body to start to really conduct some deadly amps and find themselfs climbing the stairway to heaven.
Not saying theres circumstances with barefeet water and alike that can vastly reduce this number but with decent rubbee shoes dry day most people survive house hold shocks only when theres water and lots of sweat on your skin and a good ground path no rubber shoes or chilling with your 7kw ring circuit on an extension cord looping it round like a cowboy in the bath then your in trouble.
larsb
1 MW
Hummina Shadeeba said:
voltage is in the equation.
U=R*I is the first part of the ri2 as (R*I)*I
This voltage is the voltage drop at 100A due to the 1ohm resistance.
Hummina Shadeeba
10 MW
Yea I get it but voltage isn’t in the power lost equation directly and it’s just current x resistance. that current is earlier decided by the voltage and resistance. Nitpicking. But still amps cause heat not volts volts just push the amps. Lost wattage to heat is a measurement of amps ultimately.larsb said:
larsb
1 MW
No, that’s not really true. The way to calculate it is following the energy balance equation with volt*amps representing energy lost.
Why it is calculated in ri2 form is because these are known figures in the equation. You might also take the worlds longest multimeter leads and measure that voltage drop between the start and the end of the wire. Voltage drop will be R*I and will be a 100V in this case.
Why it is calculated in ri2 form is because these are known figures in the equation. You might also take the worlds longest multimeter leads and measure that voltage drop between the start and the end of the wire. Voltage drop will be R*I and will be a 100V in this case.
Hummina Shadeeba
10 MW
Wait.. I^2R reveals the wattage lost to heat and that formula is enough and no other math need be done right? I don’t need to figure the voltage drop as long as I already know the resistance. It’s the current alone that is causing the heat..voltage and resistance revealing that current
Wattage lost to heat being a measure of just an amperage squared across a resistance?
Wattage lost to heat being a measure of just an amperage squared across a resistance?
serious_sam
10 kW
V=IR
P=VI
P= V^2/R
P=I^2R
P=VI
P= V^2/R
P=I^2R
I^2R to decide power lost. voltage isn’t in the equation. Voltage is relevant in that greater voltage will allow more current but the voltage itself isn’t relevant other than that no? I mean in determining the wattage loss to heat. Thanks for your input and I generally know ohms law but as it relates to power lost in a resistor and the heat produced..while V x I is wattage..when figuring the resistive losses as a wattage voltage isn’t there and just relevant in how many amps it pushes.Hummina Shadeeba said:
[/quote]
You have two issues/questions embedded in your query. First, if you are on the west coast US, the high voltage lines where this matters most, run at 500kV AC. There's also a DC 500kV line than runs down Nevada. On DC, it's simple. The 1 ohm in your question is actually 1 ohm per some distance. Once you input that into your considerations, you can see why the longest transmission lines, carrying the largest amount of power, would benefit from a higher ratio of voltage to resistance over the distance it needs to flow. More resistance, more loss.
The AC case is compounded by other factors (phase shifting, etc.), which just complicate the main reason above.
For your other question, the standard measure of heat transfer is watts, period, whatever way you want to get there.
R is the real enemy. If R were zero, you'd have no loss or heat generated over the conductor, no matter what the length. It's a factor in battery performance, or the efficiency of our motors. Zero resistance, infinite current, no heat transfer. AC adds complications as well, but the heat loss due to resistance would be zero.
I offer my understanding of the matter:
If you have the same piece of wire (ie same resistance), you cannot pass twice the voltage at the same amps .
Increasing the volts will increase the flowing amps.
If you want to keep the amps the same while increasing the source voltage, you have to increase the resistance accordingly. Otherwise more amps will flow.
As per your example, "10,000 volts and let’s say 100 amp" will never be 1 Ohm total resistance.
U/I=R thus 10.000 / 100 = 100 Ohms total, including whatever load is at the end of the power lines. Total resistance limits you to 100 Amps, which at 1 Ohm resistance of the power lines, produce 100V voltage drop across (within) the power lines.
Now decrease the input voltage by half while keeping the resistance.
5.000V across 100 Ohm total resistance will only flow 50 Amps. Which produce a 50A/1R = 50V voltage drop across the power lines.
Now you have less Voltage drop across the power lines, AND ALSO less amps flowing through the Voltage drop.
Thus less heat from flowing Amps AND less heat from dropped Volts.
And this voltage drop is hidden in the I^2R = I*I*R ... I*R is the volt drop, *I is the flowing Amps
Also the voltage drop will be the same "percentage of total volts" for a given resistance, because you are dividing/multiplying by the very same resistance. So the efficiency of transfer is the same 99% (1*100*100 / 10.000*100 = 1% loss , 1*50*50 / 5.000*50 = 1% loss). You input less total power, you lose less total power but same percentage.
Because the wire resistance is 1% of the total load, thus inducing 1% power loss regardless of the total power. And the rest 99% is dissipated at the load/appliances.
Going way back to your power lines question...
The power dissipated/heat generated will divide according to the voltage drop across each component.
Imagine: source voltage 1 MegaVolt , wire resistance 1 Ohm , load resistance 1 MegaOhm.
You get 1 Amp of current, generate 1 MegaWatt at the load, but only 1 Watt within the power line.
Thus you need high voltage in the lines, to be able to operate high-resistance load at the end of low-resistance wires. The greater the difference of resistance between wire and load, the better transfer efficiency. If those two get close together, you get heat problems.
As E-HP said, no resistance = no heat generated.
Now I hope this all convoluted mess can be understood despite my limited English...
Cheers, xFrankie
larsb said:
If you have the same piece of wire (ie same resistance), you cannot pass twice the voltage at the same amps .
Increasing the volts will increase the flowing amps.
If you want to keep the amps the same while increasing the source voltage, you have to increase the resistance accordingly. Otherwise more amps will flow.
As per your example, "10,000 volts and let’s say 100 amp" will never be 1 Ohm total resistance.
U/I=R thus 10.000 / 100 = 100 Ohms total, including whatever load is at the end of the power lines. Total resistance limits you to 100 Amps, which at 1 Ohm resistance of the power lines, produce 100V voltage drop across (within) the power lines.
Now decrease the input voltage by half while keeping the resistance.
5.000V across 100 Ohm total resistance will only flow 50 Amps. Which produce a 50A/1R = 50V voltage drop across the power lines.
Now you have less Voltage drop across the power lines, AND ALSO less amps flowing through the Voltage drop.
Thus less heat from flowing Amps AND less heat from dropped Volts.
And this voltage drop is hidden in the I^2R = I*I*R ... I*R is the volt drop, *I is the flowing Amps
Also the voltage drop will be the same "percentage of total volts" for a given resistance, because you are dividing/multiplying by the very same resistance. So the efficiency of transfer is the same 99% (1*100*100 / 10.000*100 = 1% loss , 1*50*50 / 5.000*50 = 1% loss). You input less total power, you lose less total power but same percentage.
Because the wire resistance is 1% of the total load, thus inducing 1% power loss regardless of the total power. And the rest 99% is dissipated at the load/appliances.
Going way back to your power lines question...
The power dissipated/heat generated will divide according to the voltage drop across each component.
Imagine: source voltage 1 MegaVolt , wire resistance 1 Ohm , load resistance 1 MegaOhm.
You get 1 Amp of current, generate 1 MegaWatt at the load, but only 1 Watt within the power line.
Thus you need high voltage in the lines, to be able to operate high-resistance load at the end of low-resistance wires. The greater the difference of resistance between wire and load, the better transfer efficiency. If those two get close together, you get heat problems.
As E-HP said, no resistance = no heat generated.
Now I hope this all convoluted mess can be understood despite my limited English...
Cheers, xFrankie
bobhurd3d
100 mW
It's understandable, xfrankie.
bobhurd3d
100 mW
ohms law is great, but power isn't always converted into heat.
Take your e-bike motor. It actually converts power into a rotation.
Take an antenna. It can make several 1000W dissapear in the sky as radio waves.
Take a led. It converts power into light.
Not all loads are resistive.
Connect a dc voltage source to a capacitor. At start a large current will flow. The capacitor will charge and the current flow will decrease exponentially. That's why you often see a spark if you connect a controller to a battery.
Connect a dc voltage source to an inductor. At start, no current will flow. The current flow will increase exponentially and finally will be limited by the inductor resistance.
Power transmission wires bring power from A to B. They have a resistance. The higher the voltage, the lower the current needed to transfer that same amount of power, so the less heat losses in the Power transmission wires. The amount of current passing trough the wires is determined by the load at the end of the wire.
Take your e-bike motor. It actually converts power into a rotation.
Take an antenna. It can make several 1000W dissapear in the sky as radio waves.
Take a led. It converts power into light.
Not all loads are resistive.
Connect a dc voltage source to a capacitor. At start a large current will flow. The capacitor will charge and the current flow will decrease exponentially. That's why you often see a spark if you connect a controller to a battery.
Connect a dc voltage source to an inductor. At start, no current will flow. The current flow will increase exponentially and finally will be limited by the inductor resistance.
Power transmission wires bring power from A to B. They have a resistance. The higher the voltage, the lower the current needed to transfer that same amount of power, so the less heat losses in the Power transmission wires. The amount of current passing trough the wires is determined by the load at the end of the wire.
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