Series Circuit Question

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George Stolz

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Where would I connect a voltmeter to find that reading though?

(I appreciate the speedy replies, got to be comfortable explaining this tomorrow. :D)

Edit: It's strictly theoretical, right?
 

Hv&Lv

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I'm no expert, but it looks right to me. if you remove one or all the other loads the voltage on the power supply will rise. I see a resistor on the power supply. Normally pictures don't show that, they just show the starting voltage. Here it is showing drop across the power supply.
 

Hv&Lv

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Get an old string of christmas lights and check voltage at each side of an individual light.
 

mivey

Senior Member
Where would I connect a voltmeter to find that reading though?
Nowhere.

The points shown are part of representative drawing, and not a very good one. To make it better, it should show an ideal voltage source in series with the 0.05 ohm resistor. Still would not be a real place inside the power supply. There are other ways to measure this reistance or volt drop but a meter hooked up like this ain't it in the real world.
 

hurk27

Senior Member
The problem with putting the resistance in parallel with the power supply in this equation will only reflect upon the load presented to the supply ahead of this point, it will not reflect on the equation after this point as you will still have 119.2 volts and 16 amps to figure out the rest of the circuit, so I agree with Mivey as the 0.05 resistor can only be calculated if it is in series with the power supply.

here is what I came up with:

Supply:
Et= 120v
It= 16a
Rt= 7.5Ω
Pt= 1920w

Rps
E= 0.8v
I= 16a
R= 0.05Ω
P= 12.8w

R1 & R2 (conductor 1&2)
E= 2.4v
I= 16a
R= 0.15Ω
P= 38.4w

RL
E=114.4v
I= 16a
R=7.15Ω
P=1830.4w

Losses in circuit
E= 5.6v also VD
R= 0.35Ω
P= 89.6w


This is done many times to see if a student is paying attention as it very common to do this on test.
 
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Is this right?

View attachment 7752

Seems to me that the voltmeter on the power supply should be reading 120V, not .8V. Is there something I'm missing here? :?
The graphical representation is badly made for what it attempts to illustrate. The voltage readings around the circuit should be taken from point to point around the circuit and the sum of the individual voltage readings around the circuit should add up to the voltage that is read at the source terminals, so the voltage across the source terminals should be 119.2V instead of 0.8V. (Is this drawing saying that it is completely safe to touch the outgoing terminals of a loaded circuit at 0.8V? My practical experience says otherwise. :D) It is based on Kirchhoff's Second Law. The showing of the 16A in two places is unnecessary. Kirchhoff's First Law.
 
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George Stolz

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My official position is that the loss is real, but is essentially seen as a loss due to transformer inefficiency. Since the secondary is constructed of wire that has resistance, that .8V is lost getting to the very center of the secondary. Essentially, the ends of the conductor extend internally into the transformer to the center of the winding.

Would y'all buy that? :)
 

mivey

Senior Member
The volt meter should reflect the loss across the resistance of the power supply as it should read 119.2v which is the circuit supplied voltage, not 120v.

...and the sum of the individual voltage readings around the circuit should add up to the voltage that is read at the source terminals...

The problem is the meter is not really across the source terminals. It is just a bad graphic. Stating that the graphic is missing a series ideal transformer would be the most appropriate techically correct fix.

Just use a pointer during the presentation and say that just outside the leads of the 0.8V meter there is an ideal 120V source that is not pictured. The combination of the ideal voltage source and series impedance models a simple power supply. If you neglect that, then you face the KVL question about voltages in a loop summing to zero (for conservative fields anyway).

that .8V is lost getting to the very center of the secondary. Essentially, the ends of the conductor extend internally into the transformer to the center of the winding.

Would y'all buy that? :)
No. It is making a bad situation worse because it introduces even more technically incorrect information.
 

ActionDave

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My official position is that the loss is real, but is essentially seen as a loss due to transformer inefficiency. Since the secondary is constructed of wire that has resistance, that .8V is lost getting to the very center of the secondary. Essentially, the ends of the conductor extend internally into the transformer to the center of the winding.

Would y'all buy that? :)
No. But here is what I would buy- "What you need to know to get you through the day in the real world is that amps are the same any where in a circuit. Keep pestering me about the flaw in this illustration and I am going to get a physics professor and a EE in here. By the time they are done with you, you won't know which way to open a door to get out of a room. AND there will be a test. Flunk it and you won't get any PDU's"

Go get 'em teach.
 
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