Inductor has no high inrush current
- Thread starter gar
- Start date
- Status
Well, what is the circuit? Please derive the equation. I think it is a steady state equation..........
The math to work a numerical example is a pita
plus, it actually is an equation for a time varying i(t)
look at 4.2.8 IL(t) ... it is the equation for current in the L for an RLC ckt
we know that IL must be in amps
and that Vc(0) must be in volts
the e terms are unitless (s is in freq or 1/time and cancels time, unitless)
so the crap below Vc(0) must be Ohms
i= v/z
plus, it actually is an equation for a time varying i(t)
look at 4.2.8 IL(t) ... it is the equation for current in the L for an RLC ckt
we know that IL must be in amps
and that Vc(0) must be in volts
the e terms are unitless (s is in freq or 1/time and cancels time, unitless)
so the crap below Vc(0) must be Ohms
i= v/z
that would take pages
buy this book https://books.google.com/books/abou...s_in_Power_Systems.html?id=nuZSAAAAMAAJ&hl=en
I completed this grad course last fall (my prof was taught by the author and is the vp at mitsubishi for hv cb's)
you must convince yourself
obviously I can't lol
So the exponential factor in "e" is in equation 4.2.8 for current ie i not proportional to z and so ohm law does not hold during a transient
Ohm law states I proportional to V. In your case it is not proportional due to presence of " e " component.
Where does the sinφ come from if it's dc?
RL series ckt
assume V, R and L = 1
i(t) = 1 - e^-t
i(1) = 0.63 a
so v (1) across R = i x r = 0.63 x 1 = 0.63 v
vL = 1 - 0.63 = 0.37 v
i(1.5) = 0.78 a
v(1.5)R = 0.78 x 1 = 0.78 v
vL = 1 - 0.78 = 0.28 v
i(3) = 0.95 a
v(3)R = 0.95 x 1 = 0.95 v
vL = 1 - 0.95 = 0.05 v
as t > inf
i = 1 = v supply / (0 ohm L +1 Ohm R) = 1 a
vR = 1 x 1 = 1 v
vL = 0 as expected
Ohms law applies for a transient condition
same for RC AND RLC
math is more complicated
assume V, R and L = 1
i(t) = 1 - e^-t
i(1) = 0.63 a
so v (1) across R = i x r = 0.63 x 1 = 0.63 v
vL = 1 - 0.63 = 0.37 v
i(1.5) = 0.78 a
v(1.5)R = 0.78 x 1 = 0.78 v
vL = 1 - 0.78 = 0.28 v
i(3) = 0.95 a
v(3)R = 0.95 x 1 = 0.95 v
vL = 1 - 0.95 = 0.05 v
as t > inf
i = 1 = v supply / (0 ohm L +1 Ohm R) = 1 a
vR = 1 x 1 = 1 v
vL = 0 as expected
Ohms law applies for a transient condition
same for RC AND RLC
math is more complicated
The transient is due to the switching of an ac signal
it is the ph angle
it is not dc per se
it is the extracted dc mathematical component of the total transient waveform
gar
Senior Member
- Location
- Ann Arbor, Michigan
- Occupation
- EE
170923-1334 EDT
You may choose to call V = I*Z ohms law, I do not. When Ohm did his research it was based on DC thermocouple voltage, current and conductance of various conductive materials. Ohm's simple equation also works with an invariant resistance with a varying voltage or current.
Once you introduce inductance and/or capacitance the equation is not simple anymore except in special cases.
However, my purpose of this thread is to try to disassociate the concept of high inrush current, or high motor starting current from being caused because the load is being called inductive.
When a transformer is "turned on" a random high inrush current is not caused because the transformer looks to be inductive, but rather because you have a coil of wire around a ferromagnetic material that under certain conditions can be driven into core saturation causing the high current.
A high motor starting current does not occur just because the motor looks inductive, but rather from other factors. As examples: a starting coil that gets switched out near full speed, or in all cases the energy to accelerate the rotor.
An AC solenoid or contactor does not have a high inrush current because it looks inductive, but rather because its inductance changed from low to high following turn on.
That solenoid or contactor when used with DC has no inrush current even though the inductance changes. This can be a real advantage in a solenoid value because a stuck spool does not burnout the coil.
Generally you can not use an AC solenoid or contactor on DC because of design differences and get exactly the same results. An AC device can be used on DC at a lower voltage, but not the reverse. AC units have shading coils to produce a shifted magnetic field. The shading coil does not exist in a DC designed unit. Thus, DC units used on AC just chatter. Pull-in and holding currrents will be different. AC units have the advantage of high pull-in force with low holding power.
.
You may choose to call V = I*Z ohms law, I do not. When Ohm did his research it was based on DC thermocouple voltage, current and conductance of various conductive materials. Ohm's simple equation also works with an invariant resistance with a varying voltage or current.
Once you introduce inductance and/or capacitance the equation is not simple anymore except in special cases.
However, my purpose of this thread is to try to disassociate the concept of high inrush current, or high motor starting current from being caused because the load is being called inductive.
When a transformer is "turned on" a random high inrush current is not caused because the transformer looks to be inductive, but rather because you have a coil of wire around a ferromagnetic material that under certain conditions can be driven into core saturation causing the high current.
A high motor starting current does not occur just because the motor looks inductive, but rather from other factors. As examples: a starting coil that gets switched out near full speed, or in all cases the energy to accelerate the rotor.
An AC solenoid or contactor does not have a high inrush current because it looks inductive, but rather because its inductance changed from low to high following turn on.
That solenoid or contactor when used with DC has no inrush current even though the inductance changes. This can be a real advantage in a solenoid value because a stuck spool does not burnout the coil.
Generally you can not use an AC solenoid or contactor on DC because of design differences and get exactly the same results. An AC device can be used on DC at a lower voltage, but not the reverse. AC units have shading coils to produce a shifted magnetic field. The shading coil does not exist in a DC designed unit. Thus, DC units used on AC just chatter. Pull-in and holding currrents will be different. AC units have the advantage of high pull-in force with low holding power.
.
You mean the point on a sine wave at which it was switched in?
So what angle is it because that's what your diagram in post #51 shows.
Last edited:
gar
Senior Member
- Location
- Ann Arbor, Michigan
- Occupation
- EE
170923-1511 EDT
Fig. 3.2 above is a good illustration of the AC current in an RL switched AC circuit.
If you can visualize from this plot what happens if the time constant is very long and turn on occurs at a steady state current peak, then you can see why the upper limit of peak current is just 2 times the steady state peak.
.
Fig. 3.2 above is a good illustration of the AC current in an RL switched AC circuit.
If you can visualize from this plot what happens if the time constant is very long and turn on occurs at a steady state current peak, then you can see why the upper limit of peak current is just 2 times the steady state peak.
.
What do you think it is?
Exactly what I said it is and that's what your diagram in post #51 shows.
Well, it isn't exactly your diagram anyway, is it?
- Status