Series Circuit Question

[Haji]

In real world, the voltmeter connected across the power supply will read 119.20V instead of 0.80V. Correction may be made accordingly to the figure.

 

[dicklaxt]

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. ) It is based on Kirchhoff's Second Law. The showing of the 16A in two places is unnecessary. Kirchhoff's First Law.

Right on,Thats a big 10-4 from the days of the CB radio

dick

 

[Strathead]

I am having a hard time cyphering through the "Engineer speak" on this thread. Whether the power supply is drawn with a circle and a sine wave, a little battery symbol, or (what I consider) misrepresented with a resistor in a circle, this drawing has no bearing on knowledge that any reasonable person needs. I agree with you George that the meter should read a nominal 120 volts. For any practical purpose that is all that is important. No one drawing this diagram would have any reason to get in to the properties and resistances of the short length of leads from the "generator" of power to the output terminals.

My conclusion is that the drawing is just plain wrong. Other than Mr. Wessel, and you George, was there anyone else who actually said this?

 

[mivey]

My official position is that the loss is real, but is essentially seen as a loss due to transformer inefficiency.

Not correct. It is not just the transformer inefficiency as you proposed.

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.

I stopped before by just saying you were introducing more technically incorrect information. Now it appears the chickens have come home to roost in the post below so I thought I would add another 2 cents.

I am having a hard time cyphering through the "Engineer speak" on this thread.

Let me see if I can simplify it: The sum of voltages around a loop equals zero. If you sum the voltages in the diagram you don't get zero you just get all the voltage drops. What is missing is the EMF source (the voltage rise) that produces these voltages across the resistors displayed.

Whether the power supply is drawn with a circle and a sine wave, a little battery symbol, or (what I consider) misrepresented with a resistor in a circle, this drawing has no bearing on knowledge that any reasonable person needs.

Wrong. Most reasonable persons dealing with supply issues need to understand voltage regulation. Source impedance is a significant concern and should not be ignored. You will have to face transformers and burdens and adequate voltage supply sooner or later.

I agree with you George that the meter should read a nominal 120 volts. For any practical purpose that is all that is important. No one drawing this diagram would have any reason to get in to the properties and resistances of the short length of leads from the "generator" of power to the output terminals.

That is where George's "official position" has led to misunderstandings. It is not just the short lead length. It is the impedance even further upstream as well. In a simple representation, we use a Thevenin equivalent circuit model to represent what is upstream of the load. The Thevenin representation is a voltage and resistor in series.

My conclusion is that the drawing is just plain wrong. Other than Mr. Wessel, and you George, was there anyone else who actually said this?

I did. At least twice.

 

[Strathead]

mivey, I am speaking as an electrician and I make certain assumptions regarding this diagram. First is that it is intended for electricians and its prupose is to represent the relationship between voltage, amperage and resistance in series circuits. From that assumption, I based my response. Putting a meter across the output terminals of a 120V power supply in the real world will produce voltages at 120v plus or minus depending on a multitude of factors. To cloud a series circuit representation by showing the internal resistance of the power supply as a factor in the circuit is unwise in my mind. If I were teaching a group of future electricians this subject matter, I would be thrilled if they actually grasped the concept that a switch with a high resistance is in series with a light load and as such will cause the light to dim and they understood why. That would be a good day! Frankly, I have done my share of troubleshooting and repair and supervision of others and have never cared about upstream losses or resistance yet.

 

[mivey]

To cloud a series circuit representation by showing the internal resistance of the power supply as a factor in the circuit is unwise in my mind. If I were teaching a group of future electricians this subject matter, I would be thrilled if they actually grasped the concept that a switch with a high resistance is in series with a light load and as such will cause the light to dim and they understood why. That would be a good day!

Point taken. I guess it would depend on the audience. But seeing as how the graphic was going to be used, it should be corrected so as not to create a stumbling block for those wanting to advance in the future.

Frankly, I have done my share of troubleshooting and repair and supervision of others and have never cared about upstream losses or resistance yet.

As long as the load impedance is large compared to the source (i.e. the load is relatively small) source impedance is not a significant issue. As we know, that is not always the case.

 

[K8MHZ]

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

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

Edit: It's strictly theoretical, right?

Forgive my poor drawing skills, but here is how I imagine the voltage relationships created by internal resistance exist:


A to A = 0 volts
A to B = 0.4 volts
A to C = 0.4 volts
A to D = 59.6 volts
A to E = 59.6 volts
D to E = 119.2 volts

 

[G._S._Ohm]

0.5 ohm is a more reasonable value for the source resistance of a typical wall outlet. You might get 0.05 taking the 120v off right at the load center. For my house it's about 0.01 but the transformer is right outside.

The least I can say is that a knowledgeable proofreader is needed.

 

[K8MHZ]

Now, if we imagine that the ends of the sine wave in my drawing have 120 volts across them, and a load drops that 120 volts to a lesser value, in order to comply with Ohm's Law we have to make all the math work.

Assigning an 'internal resistance' based upon the amount of voltage drop exhibited by a load satisfies that requirement.

In order for a non-regulated 120 volt PS to stay exactly at 120 volts under a load would mean 0 ohms of internal resistance.

 

[K8MHZ]

0.5 ohm is a more reasonable value for the source resistance of a typical wall outlet. You might get 0.05 taking the 120v off right at the load center. For my house it's about 0.01 but the transformer is right outside.

The least I can say is that a knowledgeable proofreader is needed.

It's not so much proofreading as the graphic is not clear.

Look close. The probes that read 0.8 volts are INSIDE the power supply measuring the voltage across a 'resistor'. If the probes were outside the power supply they would see 119.2 volts.

 

[G._S._Ohm]

It's not so much proofreading as the graphic is not clear.

Look close. The probes that read 0.8 volts are INSIDE the power supply measuring the voltage across a 'resistor'. If the probes were outside the power supply they would see 119.2 volts.

If 3/4ths or more of cognizant people seeing this are mislead, it's bad.

"President Chay Knee" had this percentage of the people believing Iraq and 911 were connected without ever actually saying so. Ya' gotta' wonder how that happened. :?

 

[K8MHZ]

Here is another one of my bad drawings.

The graphic does not show internal resistance to be in series (which it is or it would not cause a voltage drop). Look at this sketch and imagine these voltage points for 16 amps.

A to B = 120 volts
A to C = 0.8 volts
B to C = 119.2 volts
D to E = 119.2 volts

View attachment 7754

 

[K8MHZ]

If 3/4ths or more of cognizant people seeing this are mislead, it's bad.

Agreed. It took me more than a couple looks to try to figure out what was going on in that graphic.

 

[K8MHZ]

Bad drawing 3

Internal resistance shown as symmetric just for the purpose of illustration. (119.8 is an error. It should be 119.2)

 

[K8MHZ]

From both a teacher's and student's standpoint, I think the graphic sucks for what it is trying to do.

Would the lesson of current being the reference in a series circuit not be better taught by leaving out the internal resistance of the power supply and just use 120 volts as the supply voltage?

The current would bump to 16.1 amps, but much clearer to see. Internal resistance should be a separate topic.

 

[G._S._Ohm]

And some voltage sources have negative impedance: the voltage increases slightly with increasing current.

Arc, also, have negative incremental impedance.

 

[K8MHZ]

And some voltage sources have negative impedance: the voltage increases slightly with increasing current.

Arc, also, have negative incremental impedance.

So, internal resistance isn't confusing enough, you have to go through internal impedance in as well???

 

[G._S._Ohm]

So, internal resistance isn't confusing enough, you have to go through internal impedance in as well???

It depends on the accuracy you want. Perfect accuracy = infinite labor.

 

[K8MHZ]

It depends on the accuracy you want. Perfect accuracy = infinite labor.

:thumbsup: .

 

[Electric-Light]

What goes in comes out.