What are the readings of the voltmeters in this circuit?

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• ----(v2)------- - | | + --A-----------------B-- + | | | | | # # | | # * * # | (v1) R1 * L1 * R2 (v3) | # + # | | # * * # | | | * * | | | | | | - --D-----------------C-- - | | + ----(v4)------- -

Suppose the solenoid L1 gives a constant rate of change of magnetic flux 1Wb/sec into the screen, all resistances are 0R except R1 =3D 1R, R2 =3D 10R. If the measurement loops are small so they enclose negligible flux, what will be the readings of voltmeters v1, v2, v3, v4?

Cheers, Larry.

Doing other people's homework is boring, and doesn't help anyone in the end. What is this, a take-home exam?

Cheers

Phil Hobbs

--
Dr Philip C D Hobbs
Principal
ElectroOptical Innovations
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Briarcliff Manor NY 10510
845-480-2058

email: hobbs (atsign) electrooptical (period) net
http://electrooptical.net

Hello Larry,

V2 and V4 will read zero (as you mentioned the measurement loops don't catch flux).

V1 and V3 is a tricky one. The current that will flow counter clockwise, will itself generate a flux, but be happy that flux is steady (not time varying), so it will not generate a second EMF in the loop.

As you mentioned 1Wb/s, this means the induced voltage is 1V, and this will divide equally across the resistors (0.5V per resistor). So V1=3D

+0.5V, V3 =3D -0.5V.

Hoping I don't have the signs wrong....

Wim PA3DJS

without abc, PM will reach me.

Don't forget one resistor is 1R, the other 10R ;)

It's a problem showing how two voltmeters in parallel don't always show the same readings.

t- Hide quoted text -

Which two voltmeters are in parallel? They all look separated by finite resistances to me...

inite

v3 and v1 are in parallel. v =3D d=CE=A6/dt =3D 1v total for loop, wimpie is right

0.5v on each resistor

Ah, thanks -- I did miss that.

Wim,

If I told you that R1 and R2 were changed to 0Ohms each (i.e., all four sides are now just perfect conductors of the same length), what would your answers be?

I'd be tempted to say 0.25V on each meter. How about you? :-)

.

finite

No, since R1 =3D 1R and R2 =3D 10R

Larry

,

=3D

Hello Larry,

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 PA3DJS

What is the self and mutual inductance of each segment? Geometry matters now.

The original problem can be solved trivially by inserting a controlled voltage source (dPhi/dt dependent, in this case, but dI/dt is more typical, assuming mutual inductance is known) anywhere within the loop. Equal results are had if the source is placed between A and B, C and D, or in series with R1 or R2.

Tim

--
Deep Friar: a very philosophical monk.
Website: http://webpages.charter.net/dawill/tmoranwms

ides

ers

Hello Joel,

Now you will enter a special situation, you will never reach steady state as the L/R (inductance/resistance) will be infinite.

Due to the zero resistance, the net flux through the loop will be zero, no voltage is no flux. An increasing current will flow, counteracting the driving flux. The actual current depends on size of loop and thickness of the wire).

With kind regards,

Wim PA3DJS

1R, , t e s 1=3D

I forgot the minus sign for V3....

wim

Ah, I should have mentioned that I was sticking with the geometry represented in the original posting: Each leg being the same length.

If you place your dPhi/dt voltage source in the A-B leg, surely V2 is now going to read 1V? Perhaps I misunderstood what you meant...

Ah, good point; thanks.

OK, if we make the resistors very small but equal, we're still at 0.25V per leg (after many L/R time periods), I believe.

"Faraday's Paradox" on Wikipedia makes for interesting reading.

---Joel

t- Hide quoted text -

That's such a cool problem though. My boss always talks about a solenoid with a very slow current ramp, so you can stick your voltmeter here and there and measure the different voltages.

George H.

rs

Yup, it matters how things are hooked up.

Mind you, I've only heard about this problem in theory and never done it in practice.

George H.

r