breaking inductive currents
breaking inductive currents
(OP)
as far as i know, it is more difficult to break lagging currents than break resistive currents. is it true? if it is true why? does it matter the lag of the current,?
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breaking inductive currents
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RE: breaking inductive currents
Anyhow, it is true. It is also true that using capital letters in the beginning of a sentence is a good habit.
Gunnar Englund
www.gke.org
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100 % recycled posting: Electrons, ideas, finger-tips have been used over and over again...
RE: breaking inductive currents
RE: breaking inductive currents
RE: breaking inductive currents
Another consideration is phase relationship between voltage and current as it relates to interrupting at a current zero. There are more details to that subject than I know. Without looking it up in a book to refresh my memory (an ahead-of-time excuse for not being exact), it is easier to interrupt at the natural current zero when the voltage is also zero at that time (resistive current) since in that case no voltage will appear accross the arc/contacts when they reach open circuit condition. That is why it can be more difficult to interrupt a capacitive ac current than a resistive ac current... even though there is no L*di/dt voltage in the capacitive current. The lack of natural current zero's is also the reason that it is tougher to interrupt a dc inductive current than an ac inductive current. I'm sure other forum members can add more if they care to.
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RE: breaking inductive currents
With capacitive circuits (leading), this can in fact be even worse. The capacitors will hold some charge in them from the time of trip and slowly deteriorate over time. However this charge can cause restrike in your breaker since the voltage differential across the interrupter is higher than nominal.
One more thought, NEVER try to interrupt a DC circuit with an AC breaker. There is no zero crossing and thus it is likely not to clear the circuit.
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If it is broken, fix it. If it isn't broken, I'll soon fix that.
RE: breaking inductive currents
Understanding how ignition coils and magnetos work would be very helpful to understanding the concept.
RE: breaking inductive currents
Any other comments on HV cap breakers would be appreciated as well.
Thanks,
Mark
RE: breaking inductive currents
Regards
Marmite
RE: breaking inductive currents
Also, when you start that other thread, please do consider the difference between returning voltage in HV switching and the inrush current when closing low inductance and low resistance capacitive circuits. The two problems have very little with each other to do. So, two threads are recommended.
Gunnar Englund
www.gke.org
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100 % recycled posting: Electrons, ideas, finger-tips have been used over and over again...
RE: breaking inductive currents
If CB has single pole operation, maybe there's just a "Controlled point-on-wave switching device" that makes the difference.
For inductive load switching some CB use resistors to damp out transient overvoltages caused when switching transformer magnetising or shunt reactor currents.
For capacitive current switching the problem is to build up quickly sufficient dielectric strength between the contacts to withstand twice normal voltage one half cycle after interruption. Switching of capacitors or open-ended lines is especially susceptible to switching surges. When switching a capacitive element, the CB could end up with opposite polarity voltages on either side of the switch. This could lead to twice nominal voltage across the circuit breaker. The voltage that appears across the open switching device is called the recovery voltage.
The higher the recovery voltage, the greater the likelihood of a restrike occurring.
When the circuit breaker interrupts the capacitive current the voltage is at a maximum value. One half cycle after the CB opens, the capacitor is still charged to a minimum value but the system voltage has reached a maximum. The circuit breaker now has twice system voltage across it. The probability of a restrike has increased.
When the current is zero the voltage is at a maximum in a capacitive circuit.
RE: breaking inductive currents
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RE: breaking inductive currents
capacitor deenergization was that it demonstrates di/dt voltage is not the whole story by itself.... phase angle
between voltage and current plays a role separate from di/dt voltage. That is why deenergization of an ac
capacitive circuit poses greater challenges than a ac resistive circuit of same current.
http
"Transmission and Distribution Electrical Engineering", 3rd Ed, by Bayliss and Hardy
http://b
Scroll up a little bit to figure 13.2. You can see that at the moment of current zero, the recovery voltage jumps
to the normal system voltage (which is maximum at natural current zero for inductive and capacitive
circuits). There is also an L-C ringing which causes transient overshoot as Marmite has alluded.
All you have to do is read the bolded sentence above to see that this author singles out as the "basic reason"
for the challenge of interrupting ac inductive (and capacitive) currents is the phase relationship between
voltage and current and it's effect on the voltage at the time of current zero. There is very little about L*di/dt
voltage in this discussion.
V = L*di/dt is the most important factor for dc circuits.
For ac circuits the angle between voltage and current at the time of natural current zero also plays a very big role as described above.
Lead and lag are very relevant to the question as shown above. DC circuits such as ignition coils are not representative of the principles of ac circuit interruption because DC circuits do not have a natural current zero.
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