Ah yes, you have to split the corner nodes to do that. Since the loops of V2 and V4 certainly contain no flux or voltage.
Tim
-- Deep Friar: a very philosophical monk. Website: http://webpages.charter.net/dawill/tmoranwms
I've always wanted a Really Big Inductor, so I could charge it with, say, a rectifier on the 120V line, then disconnect it from the rectifier (drawing a big nasty arc at ~constant current) so I can short-circuit it (equivalent to open-circuiting a capacitor), then open it at will, or apply it to loads, to play with the energy contained within.
The required L/R time constant to play with it by hand is more than a few seconds long. If we assume a 100V arc for fractional seconds (~0.2s) doesn't discharge it appreciably (~10%), we're looking at more than a few henries. If the "full charge" current is 1A, a change of 0.1A is 200H, and a few seconds is under 100 ohms. That's a pretty fair amount of 18AWG wire on a big stack of steel plates (with dB/dt fairly low, eddy currents at least don't matter).
1A into 200H is 200 Wb flux. Silicon steel saturates at ~1.2T, so you need at least 167 m^2 cross sectional core area per turn. 1500 turns on a ~foot square core would get in the right area; adjust airgap to achieve the desired inductance (= amperage at this flux level). 1500 turns with an average wire length of maybe 6' per turn is 1.7 miles of wire; a resistance under 100 ohms requires under 9 ohm/kft, which fits pretty closely to 19 AWG. Typical for this current is 24AWG, so it's a big, overratead inductor indeed...Tim
-- Deep Friar: a very philosophical monk. Website: http://webpages.charter.net/dawill/tmoranwms
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1R, , t e so V1=3D
certainly v1 =3D -v3 and because v =3D n d=CE=A6/dt and n =3D half loop =3D 0.5 v =3D .5 x 1/1 =3D 0.5 volts for half loop the two halves of the full loop are not independent, superseding resistance difference effects.
The two resistors are in parallel, not in series!
assume ohms : r1 =3D 1 r2 =3D 10 parallel 1/r =3D 1/r1 + 1/r2 =3D 1.1 r =3D .90909 ohm v3 =3D .5v i =3D .5/.90909 =3D .55 amp =3D .5 + .05 parallel i1 + i2
Tom
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Hello,
They are not in parallel (for magnetic induction). Please make one of the resistors zero Ohms, when in parallel, all meters should read zero, but this isn't the case. Imagine yourself a generator winding. There is only one significant resistance: the load, wire resistance is kept to a minimum (efficiency issue).
The indication of voltmeters under varying magnetic field depends on how the wires are routed. As Blackhead said that the voltmeters don't enclose flux, there can only be a reading because of voltage drop due to resistance. See it as when the voltmeters are in a coaxial structure.
Only in case of a loop with zero resistance, all meters will read zero. This is because the resultant time varying current will fully counteract the driving field.
When the voltmeter wires make loops also, then the situation becomes complicated (as Tim allready said).
With kind regards,
Wim PA3DJS
wim said: "Sorry, I missed the correct resistor values, v1 (across 1 Ohms) will read 0.091V, and V3 will read -0.909V. All others will read zero. Wim
It is not true that v1 is not equal to v3. They are connected. There is no paradox. They are parallel for Thevenin Equivalent circuit: any combination of voltage sources, current sources, and resistors with two terminals is electrically equivalent to a single voltage source V and a single series resistor R.
Emulate a 20H choke with a DSP feeding a switchmode amplifier driving direct from the rectified and PFC boosted mains? Possibly less effort than building a winder and making the real thing.
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v1 1 volt v2 0 v3 1 volt v4 0 because the two resistors are in parallel and
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Yowzer, that's a big inductor. I haven't tried putting any numbers to the problem, but say a 10 second ramp up and a d(flux)/dt that will give maybe 10mV of total emf (for once around the loop). So you can use a DMM on the mV scale and see things.
George H.
"Connected" doesn't matter. As Wimpie said, it depends on the routing. Consider two RF voltmeters connected to a transmission line, one at a 1/4 lambda from the source, the other at 1/2:
source========M1=========M2
What will the meters read, using Vs as the source? Feel free to leave the line unterminated, or to terminate it somewhere to the right of M2.
Ed
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Hello,
Would you be so kind to answer this question: What happens (what will be the voltages) when one of the resistors is zero, leaving all other conditions unchanged?
Wim PA3DJS
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,When one resistor is zero, a short circuit is created. This can be done with a superconductor, like mercury at 1K. Bang! The loop explodes.
Back to the original schematic. There are two nodes, B and C. Let's give them new names that are meaningful: GEN and GND, generator and ground. The resistors are lumped circuit elements and geometrically, little informatoin is specified, just a loop around a flux. Imagine the following geometry. The ground node has tiny resistors down close to ground and a big loop of wire goes up to surround a flux. The loop is a generator, r1 is a 1 ohm generator internal resistance and r2 is the load resistor. If the load becomes an open circuit, the loop still exists in space and 1 volt is generated across the 1 ohm generator wire. The wire loop becomes a C shaped generator with 1 volt causing a backflow of current from the open tip, through the 1 ohm R1 and into GND. Adding a 10 ohm load at the tip of the C shaped wire makes it look like a circle O, but electrically, the lumped resistors are in parallel to make a double CC shaped generator.
Conclusion
The new schematic for the circuit has one loop : a voltage source is in series with two parallel resistors. The original schematic did not show the voltage source. It is a 1 volt DC generator. The new schematic is called the Thevenin Equivalent circuit where all resistors and sources are replaced by one voltage source in series with one resistor Req. Req is the equivalent resistor of the parallel R1 R2 combination. Kirkoffs voltage law is satisfied since the total voltage around the loop is zero. a Clockwise addition is -1v +1v=0 volts because the -1 volt is across R1||R2 and +1 volt is a voltage source called a generator.
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1R, ,As far as I can see, I can't see a shorted loop as there is still one resistor present. See this resistor as the load of a single turn generator or transformer winding.
You seem to suggest that the induced EMF is between B and C. From my experience the induced EMF is in series with te loop, not parallel to it.
I do not agree, when you remove R2, there will be no current in the loop, hence the voltage across R1=3D0 and all the induced EMF appears between B and C. The source representing the induced EMF is not parallel to R1, but in series.
With kind regards,
Wim PA3DJS
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