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#128869 - 03/02/04 04:28 PM inductive reactance
james S Offline
Member

Registered: 05/13/03
Posts: 102
Loc: West England
in a series circuit which is supplied by constant current regulator at 50 HZ(keeps current at 6.6 amps no matter what load is introduced or taken away with respect to the size of regulator)which has say ten transformers connected in the series loop and a lamp across the secondary side of each t'former.i was recently told by a lecturer that by 1 lamp open circuiting, then it would cause the inductive reactance to increase? is this true?

(THIS IS THE WAY AIRFIELD LIGHT ARE SUPPLIED)
The way i understand it is, inductive reactance is caused by a change in current due to the frequency (50Hz) leading to a change in flux within the t'former, but if the ccr maintains its regulation there would not be a change in the value of current so there would not be a diffrent value of current changing (50Hz) hence a maintained XL value?

I am so bad at trying to explain things, i hope this makes sense!!!!!

[This message has been edited by james S (edited 03-02-2004).]

[This message has been edited by james S (edited 03-03-2004).]

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#128870 - 03/03/04 04:25 AM Re: inductive reactance
Scott35 Offline

Broom Pusher and
Member

Registered: 10/19/00
Posts: 2724
Loc: Anaheim, CA. USA
James;

I think I understand what you are asking about here. If I am wrong, please let me know, and I'll try to repost as needed.

The Constant-Current Regulating Supply mentioned, sounds quite similar to the method of powering the older Series Street Lighting Systems, found in Residential Tracts back in the late 50's thru early 70's (the ones in my area were upgraded in 1976 to normal parallel circuit Lighting, using Mercury Vapor HID lamps in Cobra Heads, then these were upgraded to HPS Lamps in 1980).

These were Incandescent Lamps, connected in a Series String.
Load Amperes on the Series String was 6.6 Amps. Unsure of the Voltage rating per individual Lamps, but judging from their brightness at stable operation, must have been around 50VAC - as they appeared similar to some 300 Watt large sized "A" type Incandescent Lamps in output Light.

On these, if one Lamp failed, a "Blow Out" link would automatically fall in place across the Lamp's Terminals, effectively reclosing the Circuit and regaining Continuity.
When this occured, the Regulating Transformer would adjust output Voltage, by means of a movable core (I think - need to verify and reply later).
The way this Transformer "Found" out there was a change in the load Circuit was due to an increased load current - or a change in the Reflected Impedance of the Secondary Circuit.
Since there was a lower total connected Impedance from losing one fixed Resistance in the Series String (the Lamp which failed and was bypassed), the current drawn by the circuit increased while at the same Voltage.
This resulted in an imbalance within the Reg. Transformer, so it "Fixed" this problem by adjusting its self, until the output Voltage matched what was required to push 6.6 Amperes through the connected load.

OK, so much for that!

Basically, a higher Reactance is like a higher Resistance - higher value will result in lower current flow at a fixed Voltage.
That's one of the things related to a given Reactance (X is the symbol for a Reactance).
One of the most useful results on an AC is how a Reactance effects the Frequency.
Lower Hz pass easier through an Inductive Reactance (XL), whereas higher Hz pass easier through a Capacitive Reactance (XC).

BTW; having XL + XC, or one or both Reactances with a pure Resistance "R" in a given circuit is where the term Impedance "Z" comes into play for AC.

As explained in your example with the series connections of Isolated Transformers (10 Transformers connected to the Regulating Transformer), if the load on the Secondary side is high, the XL on the Primary side will also be low. It will reflect the Impedance of the Secondary side's load.
If there's no load on the Secondary side, the Primary side will reflect a large Impedance - and result in a high XL on the Primary winding.
Lowering the Voltage to the Primary will reduce the XL on that Primary winding, resulting in a higher current flowing within the Series circuit - yet it would be difficult to push the needed 6.6 Amps through this Primary winding, as it would act like a Linear Reactor or "Choke" coil, plus result in a very high Secondary Voltage with an open Secondary circuit.

But if the Secondary of that Transformer was closed (like what was done in the Series Street Lighting example above), then the Primary winding would reflect a very low Impedance to the source -resulting in a very low XL on the Primary winding - which would allow the required 6.6 Amps to flow through it with little opposition.
The Regulating Transformer would adjust the output Voltage to suit, and the results would be an apparent power - in Volt-Amps, as was drawn previously (and is required by the connected loads).

It's getting late, so let me quit here and wait for your reply.
Let me know if I covered the areas enough or at all, or if you have further questions.

Scott35
_________________________
Scott " 35 " Thompson
Just Say NO To Green Eggs And Ham!

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#128871 - 03/03/04 07:02 AM Re: inductive reactance
james S Offline
Member

Registered: 05/13/03
Posts: 102
Loc: West England
Lowering the Voltage to the Primary will reduce the XL on that Primary winding, resulting in a higher current flowing within the Series circuit - yet it would be difficult to push the needed 6.6 Amps through this Primary winding, as it would act like a Linear Reactor or "Choke" coil, plus result in a very high Secondary Voltage with an open Secondary circuit.

i know the voltage will reduce on the secondary side, but would the voltage not be increased by the ccr compensating for the increased impeadance in the secondary to regulate the 6.6 amps?

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#128872 - 03/04/04 02:04 PM Re: inductive reactance
pauluk Offline
Member

Registered: 08/11/01
Posts: 7693
Loc: Norfolk, England
James,

The inductive reactance of a simple choke is caused by the magnetic flux setting up a counter-e.m.f. in the windings. A transformer with its secondary open circuit behaves in a similar way to a basic choke, but as you place a load on the secondary some of the flux which was previously creating that counter-e.m.f. is now being used to induce current in the secondary circuit. So as you increase the load, the amount of counter-e.m.f. in the primary is less, and thus the reactance is reduced.

The overall impedance reflected by a transformer is the ratio of primary-to-secondary turns squared, i.e. :-

Zp / Zs = (Np / Ns) ^ 2

where Np and Ns are the number of primary and secondary turns respectively.

Now if the primary power is from a "normal" constant-voltage source, then clearly the increase in XL and Z from removing the secondary load causes a corresponding decrease in the primary current.

But where the primary is being fed from a constant-current source, the higher reactance and impedance results in the current source delivering a higher voltage on the primary to maintain the same current flow.

So in short, what the lecturer told you is correct.



[This message has been edited by pauluk (edited 03-04-2004).]

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#128873 - 03/04/04 04:36 PM Re: inductive reactance
james S Offline
Member

Registered: 05/13/03
Posts: 102
Loc: West England
SCOTT35/PAULUK


THANKS FOR YOUR TIME AND KNOWLEDGE MUCH APPRECIATED

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#128874 - 03/06/04 08:58 AM Re: inductive reactance
wocolt Offline
Member

Registered: 12/06/02
Posts: 117
>>
 Quote:
Now if the primary power is from a "normal" constant-voltage source, then clearly the increase in XL and Z from removing the secondary load causes a corresponding decrease in the primary current.
<>,
Then heres a question for you . How does removing the load from the secondary increase the transformers reactance ? Or for that matter changing the primary voltage change the reactance.
According to Faradays Law, FLUX(m) = 1/Np(integral)Vp(t)dt((e{ind} = N * d(phi)/dt))
This equation states that the average flux in the winding is proportional to the integral of the voltage applied to the winding, and the constant of proportionality is the reciprocal of the number of turns in the primary winding 1/Np.
Ignoring for a moment the effects of leakage flux, from that equation we get
Flux = Vmax/(wNp) * Cos(wt) = Flux in Webers.


Hence if the primary voltage is reduced how does this change the Reactance of the transformer ?
Since as stated the 'a' turns ration is Ns/Np then the flux produced in the primary will be reduced by the amount of the applied voltage.
But I dont see how the reactance of the transfromer is changed by changing the voltage, as long as the frequency is constant.
WOC

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#128875 - 03/06/04 01:33 PM Re: inductive reactance
pauluk Offline
Member

Registered: 08/11/01
Posts: 7693
Loc: Norfolk, England
Simply changing the voltage applied across a winding doesn't change its reactance.

However, reducing the load on the secondary winding results in an increase in the primary's inductance value, and that in turn does cause a corresponding increase in reactance: XL = 2*pi*f*L

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