jrannis
Senior Member
- Location
- Fort Lauderdale/ Miami
Ok lets go back and look at the 240 volt water heater operating on 120 volts.memyselfandI said:
Ohm's Law
- Thread starter memyselfandI
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gar
Senior Member
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- Ann Arbor, Michigan
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- EE
080620-0547 EST
K8MHZ:
What is "most real applications"? Are these incandescent lamps, electric arcs, resistance furnaces, ovens, DC motors with a constant torque load, computers, or AC induction motors?
Your second paragraph has the requirement of a change of resistance. What happens to the current with a constant resistance when you increase the applied voltage?
.
K8MHZ:
What is "most real applications"? Are these incandescent lamps, electric arcs, resistance furnaces, ovens, DC motors with a constant torque load, computers, or AC induction motors?
Your second paragraph has the requirement of a change of resistance. What happens to the current with a constant resistance when you increase the applied voltage?
.
tryinghard
Senior Member
- Location
- California
memyselfandI said:
Not according to Tables 430.247 through 430.250 (as example).
chaterpilar
Senior Member
- Location
- Saudi Arabia
Mivey,
This is the biggest myth...water i.e ground water is abundant here in the desert..and ground water table has not gone down for decades.
Very few trees to soak up the rain water,...vast open spaces and substantial rain keeps the groundwater level ok.
Coming back to OP, another misconception people have is that "voltage" kills....it is actually "current" which kills...and high voltage just has higher potential to drive the current through the body.
Mr Current is the killer and Mr. Voltage is the "usual suspect"...
..
Cheers.
This is the biggest myth...water i.e ground water is abundant here in the desert..and ground water table has not gone down for decades.
Very few trees to soak up the rain water,...vast open spaces and substantial rain keeps the groundwater level ok.
Coming back to OP, another misconception people have is that "voltage" kills....it is actually "current" which kills...and high voltage just has higher potential to drive the current through the body.
Mr Current is the killer and Mr. Voltage is the "usual suspect"...
..Cheers.
Last edited:
LarryFine
Master Electrician Electric Contractor Richmond VA
- Location
- Henrico County, VA
- Occupation
- Electrical Contractor
Yes, but is the current instantaneous?chaterpilar said:
K8 - where do you get this stuff ?
Consider a light bulb attached to a dimmer. When you increase the voltage, the current goes up. Is this not a "real world" load ?
this is commonly called a resistive, or linear, load. Incandescent bulbs, coffee pots, electric baseboard heaters, duct heating coils, hair dryer coils = all resistive loads.
Capacitive and inductive loads react to changes in voltage or current with capacitive and inductive reactance (most notably motors). This can cause instantaneous or near-instantaneous changes to the output voltages and current. Often a voltage increase may cause a current decrease, or vice versa, but it depends on the circuitry.
As I said in my original post, the best place to investigate circuit behaviour is to invest in circuit theory books or courses.
Consider a light bulb attached to a dimmer. When you increase the voltage, the current goes up. Is this not a "real world" load ?
this is commonly called a resistive, or linear, load. Incandescent bulbs, coffee pots, electric baseboard heaters, duct heating coils, hair dryer coils = all resistive loads.
Capacitive and inductive loads react to changes in voltage or current with capacitive and inductive reactance (most notably motors). This can cause instantaneous or near-instantaneous changes to the output voltages and current. Often a voltage increase may cause a current decrease, or vice versa, but it depends on the circuitry.
As I said in my original post, the best place to investigate circuit behaviour is to invest in circuit theory books or courses.
mivey
Senior Member
do-over
do-over
On a general large scale basis, and for particular load types, we can increase the voltage and thus reduce the supply current. The equipment at the end (the final step-down transformers, or a change in equipment voltage levels, etc.) will, for the most part, keep the amount of power we need the same.
On this basis, we can roughly say the amount of power we need will be about the same (for a constant set of tasks). If we can deliver that power at a higher voltage, the supply current will go down, along with system losses.
While increased voltage helps the economics by reducing our losses, some of the trade off is that higher voltages have higher equipment and operating costs.
I think you may have been blending two (or more?) thoughts together and your thoughts & post became scrambled.
Perhaps you would like to get your thoughts together and re-word your post?
do-over
The first time I read this, I did not follow. I read it again and still don't follow. The best I can tell is maybe you are trying blending in the idea of economic line design, where some of the contributing factors are voltage drop and line loss.K8MHZ said:
On a general large scale basis, and for particular load types, we can increase the voltage and thus reduce the supply current. The equipment at the end (the final step-down transformers, or a change in equipment voltage levels, etc.) will, for the most part, keep the amount of power we need the same.
On this basis, we can roughly say the amount of power we need will be about the same (for a constant set of tasks). If we can deliver that power at a higher voltage, the supply current will go down, along with system losses.
While increased voltage helps the economics by reducing our losses, some of the trade off is that higher voltages have higher equipment and operating costs.
I think you may have been blending two (or more?) thoughts together and your thoughts & post became scrambled.
Perhaps you would like to get your thoughts together and re-word your post?
Simply Put:
Simply Put:
Ohm's law assumes a constant value of R. In that case, I is proportional to R at any and all times.
We can extend Ohm's law to impedances, but in that case V and I must be RMS values, and for practical purposes changes in V and I occur simultaneously.
Simply Put:
Ohm's law assumes a constant value of R. In that case, I is proportional to R at any and all times.
We can extend Ohm's law to impedances, but in that case V and I must be RMS values, and for practical purposes changes in V and I occur simultaneously.
gar
Senior Member
- Location
- Ann Arbor, Michigan
- Occupation
- EE
080621-1158 EST
rattus:
Ohm's law does not assume anything constant. On an instantaneous basis v = i*r is perfectly valid. Consider a diode which is a non-linear device. We can experimentally determine a resistance vs current curve for a particular device. Suppose the current is 1 A with 1 V applied, then at that point the resistance is 1 ohm. If we increase the voltage to 1.2 V the current might be 10 A, then the resistance is 1.2/10 or 0.12 ohm. At 1 A the power dissipation in the diode is 1 W and 10 A it is 12 W. We can also define a small signal resistance at a given point on the curve and this is defined by the slope of the v/i curve and this is different than v/i.
In an AC circuit to make a meaning use of the word impedance we want to work with a sine wave with low harmonic content. The only reason for using the RMS value of the sine wave is to have a relationship with DC circuits and a standard for defining impedance.
If you took an arbitrary waveform, such as a 2 V 0.01 second positive pulse riding on top of a 1 V DC voltage, measured the RMS voltage and RMS current with a 1 henry inductor with an internal resistance of 1 ohm and calculated impedance, then applied a 1Hz sine wave of of the same RMS voltage to the same load the calculated impedance would be quite different.
With pure sine wave excitation to an RLC circuit the measured impedance using Z = V/I as the definition of im[pedance will be a function of frequency.
.
rattus:
Ohm's law does not assume anything constant. On an instantaneous basis v = i*r is perfectly valid. Consider a diode which is a non-linear device. We can experimentally determine a resistance vs current curve for a particular device. Suppose the current is 1 A with 1 V applied, then at that point the resistance is 1 ohm. If we increase the voltage to 1.2 V the current might be 10 A, then the resistance is 1.2/10 or 0.12 ohm. At 1 A the power dissipation in the diode is 1 W and 10 A it is 12 W. We can also define a small signal resistance at a given point on the curve and this is defined by the slope of the v/i curve and this is different than v/i.
In an AC circuit to make a meaning use of the word impedance we want to work with a sine wave with low harmonic content. The only reason for using the RMS value of the sine wave is to have a relationship with DC circuits and a standard for defining impedance.
If you took an arbitrary waveform, such as a 2 V 0.01 second positive pulse riding on top of a 1 V DC voltage, measured the RMS voltage and RMS current with a 1 henry inductor with an internal resistance of 1 ohm and calculated impedance, then applied a 1Hz sine wave of of the same RMS voltage to the same load the calculated impedance would be quite different.
With pure sine wave excitation to an RLC circuit the measured impedance using Z = V/I as the definition of im[pedance will be a function of frequency.
.
True. As we all know, relationship for instaneous V and I values for reactive loads are:rattus said:
v = Ldi/dt
i = Cdv/dt
Which is why we get a phase shift for sine waveforms: d(sin(wt))/dt = cos(wt) (add appropriate minus signs where needed).
Again, as we all know: Really hasn't got any thing to do with E leading I, or E happening before I - or which causes the other. Just has to do with with what happens when a function (waveform) is differentiated. ELI the ICEman works only with sine waveforms.rattus said:
carl
gar,
""...the resistance R is a constant independent of i, and the potential difference V between the terminals of a conductor is a linear function of the current in the conductor."
[Sears, F. W., Electricity and Magnetism, Addison Wesley, 1954]
Furthermore, pure sinusoids are assumed, and impedances are assumed to have reactive components.
""...the resistance R is a constant independent of i, and the potential difference V between the terminals of a conductor is a linear function of the current in the conductor."
[Sears, F. W., Electricity and Magnetism, Addison Wesley, 1954]
Furthermore, pure sinusoids are assumed, and impedances are assumed to have reactive components.
gar
Senior Member
- Location
- Ann Arbor, Michigan
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- EE
080621-1833 EST
rattus:
Your quote is incomplete and I have no idea what the intent of the statement is without the complete discussion.
If I have a dissipating device (meaning power, and no energy storage) then r = v/i makes good sense even if resistance is not constant.
A 100 W incandescent lamp with 120 V excitation has a resistance of 144 ohms and a current of 0.83 A at that operating point, and also has a small signal resistance of 144 ohms at that point. Does it make sense to describe the resistance under these conditions?
This same lamp has a resistance of 9.6 ohms at very low current. Fluke 27 measurement. Does it make sense to define the resistance at this point?
Subject this lamp to some infrared excitation, a heat lamp, and the low current resistance measurement is 12.5 ohms. Again does it make sense to define resistance at this point?
Are you saying that we can not define resistance as v/i under these different conditions for the same device?
The resistance of the lamp when cold will define the initial inrush current from a DC source, and the steady-state resistance when 120 V is applied will define the power consumed over a long term average and thus your electric bill. Certainly I would not use 9.6 ohms at 120 V to determine power consumption as it relates to my electric bill. However, it is true that the instantaneous power at turn-on with 120 V DC is 14400/9.6 = 1500 watts.
.
rattus:
Your quote is incomplete and I have no idea what the intent of the statement is without the complete discussion.
If I have a dissipating device (meaning power, and no energy storage) then r = v/i makes good sense even if resistance is not constant.
A 100 W incandescent lamp with 120 V excitation has a resistance of 144 ohms and a current of 0.83 A at that operating point, and also has a small signal resistance of 144 ohms at that point. Does it make sense to describe the resistance under these conditions?
This same lamp has a resistance of 9.6 ohms at very low current. Fluke 27 measurement. Does it make sense to define the resistance at this point?
Subject this lamp to some infrared excitation, a heat lamp, and the low current resistance measurement is 12.5 ohms. Again does it make sense to define resistance at this point?
Are you saying that we can not define resistance as v/i under these different conditions for the same device?
The resistance of the lamp when cold will define the initial inrush current from a DC source, and the steady-state resistance when 120 V is applied will define the power consumed over a long term average and thus your electric bill. Certainly I would not use 9.6 ohms at 120 V to determine power consumption as it relates to my electric bill. However, it is true that the instantaneous power at turn-on with 120 V DC is 14400/9.6 = 1500 watts.
.
ELA
Senior Member
- Occupation
- Electrical Test Engineer
Rattus,
Nice quote from 1954 I am almost that old!
As a strict definition, many definitions for ohms law can be found , even today , that state the choice of R as the constant of proportionality.
Would you agree that like many math equations you can also choose another quantity to be the "given" or as some might call, the constant, and then calculate to determine the "unknown" quantity?
I myself have used ohms law often to calculate voltage or current in circuits with variable resistances? Or used a constant current source and a variable resistance to generate a varying voltage.
There are resistances that are not independent of current , of course these are non linear. While ohms law is generally described as being used in a linear circuit, or the linear portion of the curve, does it not still apply at any given instant along the E or I curve of a non linear resistance device?
Nice quote from 1954
I am almost that old!As a strict definition, many definitions for ohms law can be found , even today
, that state the choice of R as the constant of proportionality. Would you agree that like many math equations you can also choose another quantity to be the "given" or as some might call, the constant, and then calculate to determine the "unknown" quantity?
I myself have used ohms law often to calculate voltage or current in circuits with variable resistances? Or used a constant current source and a variable resistance to generate a varying voltage.
There are resistances that are not independent of current , of course these are non linear. While ohms law is generally described as being used in a linear circuit, or the linear portion of the curve, does it not still apply at any given instant along the E or I curve of a non linear resistance device?
gar said:
Point is that Ohm's Law, as formulated by Georg Simon Ohm, assumes an ideal resistance--constant that is. In many cases, this law applies very well, e.g., in the case of low TC resistors dissipating low power.
Certainly there are many situations where resistance varies greatly for various reasons. In these cases, Ohm's Law may not be applicable at all, or special considerations must be applied.
The fact that resistance often varies greatly does not change the assumptions of the law.
ELA said:Rattus,
Nice quote from 1954
As a strict definition, many definitions for ohms law can be found , even today
Would you agree that like many math equations you can also choose another quantity to be the "given" or as some might call, the constant, and then calculate to determine the "unknown" quantity?
I myself have used ohms law often to calculate voltage or current in circuits with variable resistances? Or used a constant current source and a variable resistance to generate a varying voltage.
There are resistances that are not independent of current , of course these are non linear. While ohms law is generally described as being used in a linear circuit, or the linear portion of the curve, does it not still apply at any given instant along the E or I curve of a non linear resistance device?
Certainly, we can use the principle of Ohm's law to compute DC resistance and/or dynamic resistance at a point on a curve, however, Ohm's Law as given does not apply to the large signal behavior of such non-linear devices.
gar
Senior Member
- Location
- Ann Arbor, Michigan
- Occupation
- EE
080621-2100 EST
rattus:
Certainly a steady state large AC signal applied to a non-linear resistance is not likely to calculate a very meaningful value for resistance of the load from the current. However, I may very well get meaningful and useful resistance information from the calculation of the instantaneous value of v/i.
.
rattus:
Certainly a steady state large AC signal applied to a non-linear resistance is not likely to calculate a very meaningful value for resistance of the load from the current. However, I may very well get meaningful and useful resistance information from the calculation of the instantaneous value of v/i.
.
mivey
Senior Member
Georg Simon Ohm was studying the flow of current in metallic conductors and described a linear relationship between voltage and current.
Today, we know that Ohm's law for a substance is E = p*J where p is the resistivity of a material, E is the electric field, and J is the current density. If E is not proportional to J, then you have a nonlinear substance.
While Ohm's law was describing a linear relationship, his original law has been modified to work in modern times.
A resistor is characterized by its voltage-current curve where the resistance R is the slope of the curve (and G is the conductance and G = 1/R).
The linear time-invariant (LTI) resistors have a linear v-i curve with constant slope and do not vary with time and are the ones seen in the Ohm's law equation: v(t) = R*i(t) or i(t) = G*v(t) where R & G are constants.
As for others:
A linear time-varying (LTV) resistor has a straight v-i curve but the slope varies (gradually or abruptly) with time: v(t) = R(t)*i(t) or i(t) = G(t)*v(t).
A nonlinear time-invariant (NLTI) resistor does not have a straight v-i curve and are called nonlinear (NL) resistors, like an incandescent lamp. A diode is another such device with a v-i characteristic of: i = io*[e^(v/VT)-1] but is not bilateral like the lamp.
For a physical nonlinear device to have a non-positive slope (nonmonotonic) they can have either the voltage or current defined by the other: v = f(i) for current controlled or i = f(v) for voltage controlled (like a tunnel diode), but can be both current and voltage controlled.
The final case is the nonlinear time-varying resistor (NLTV) but I don't know of an example of a real one off hand.
Today, we know that Ohm's law for a substance is E = p*J where p is the resistivity of a material, E is the electric field, and J is the current density. If E is not proportional to J, then you have a nonlinear substance.
While Ohm's law was describing a linear relationship, his original law has been modified to work in modern times.
A resistor is characterized by its voltage-current curve where the resistance R is the slope of the curve (and G is the conductance and G = 1/R).
The linear time-invariant (LTI) resistors have a linear v-i curve with constant slope and do not vary with time and are the ones seen in the Ohm's law equation: v(t) = R*i(t) or i(t) = G*v(t) where R & G are constants.
As for others:
A linear time-varying (LTV) resistor has a straight v-i curve but the slope varies (gradually or abruptly) with time: v(t) = R(t)*i(t) or i(t) = G(t)*v(t).
A nonlinear time-invariant (NLTI) resistor does not have a straight v-i curve and are called nonlinear (NL) resistors, like an incandescent lamp. A diode is another such device with a v-i characteristic of: i = io*[e^(v/VT)-1] but is not bilateral like the lamp.
For a physical nonlinear device to have a non-positive slope (nonmonotonic) they can have either the voltage or current defined by the other: v = f(i) for current controlled or i = f(v) for voltage controlled (like a tunnel diode), but can be both current and voltage controlled.
The final case is the nonlinear time-varying resistor (NLTV) but I don't know of an example of a real one off hand.
glene77is
Senior Member
- Location
- Memphis, TN
to Larry Fine:
I like your comments.
Try this quote from someone else in the forum:
< Two turtles crossing the road were killed in a head-on collision.
My understanding is that the Fields of Force on Copper wire
are nearly instantaneous,
from one end of the wire to the other.
My understanding is that electrons move in Copper
at a rate of some millimeters per second,
from one end of the wire to the other.
I invite your comments.
I like your comments.
Try this quote from someone else in the forum:
< Two turtles crossing the road were killed in a head-on collision.
My understanding is that the Fields of Force on Copper wire
are nearly instantaneous,
from one end of the wire to the other.
My understanding is that electrons move in Copper
at a rate of some millimeters per second,
from one end of the wire to the other.
I invite your comments.
glene77is
Senior Member
- Location
- Memphis, TN
Bob,
A Dead Short is a drastic reduction of Load Impedance!
BUT, the Source Impedance is still there,
which is what limits the Power Transfer,
and kills the Voltage.
A Dead Short is a drastic reduction of Load Impedance!
BUT, the Source Impedance is still there,
which is what limits the Power Transfer,
and kills the Voltage.
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