# In a basic series-wound generator, what is "load voltage"

1. Jun 25, 2017

### Ocata

Because of the way that series-wound generators are constructed, they possess poor voltage regulation capabilities. For example, as the load voltage increases, the current through the field coils also increases. This induces a greater emf that, in turn, increases the generator's output voltage. Therefore, when the load increased, voltage increases; likewise, when the load decreases, voltage decreases.

Questions:

What is meant by load voltage and output voltage? Is load voltage simply the voltage drop in the circuit?

Is it saying if you at resistance, like an additional light bulb, to the circuit. The total voltage drop across all the resistance (light bulbs) in the circuit will increase?

As I understand a Dc circuit with a battery, increasing resistance in the circuit does not change the voltage drop nor the battery's output voltage according to ohms law. And current decreases linearly as resistance increases.

Then, in a generator, how does the current increase when resistance (light bulbs) in the circuit increase? Or how does the magnetic field in the field coil increase when light bulbs are added?

One possibility I am guessing is that increasing the resistance in the circuit, actually lowers the circuit's current as well (just as in a Dc circuit with a battery as the source voltage). However, when the current is reduced in the field windings, it changes the inductive reactance in the field winding coils in some way that actually allows for a stronger magnetic field, which increases the amplitude of the output voltage and also forces the generator rotor to spin at a faster rate, thus producing a higher frequency, thus more voltage output.

Are the events I just described anything similar to what the statement in the provided text is saying?

Any and all feedback would be greatly appreciated.
Thank you. P.S. If the concepts could be described in non calculus terms, I would be ever more appreciative. Much thanks.

2. Jun 25, 2017

### Ocata

In the next paragraph, it states that output voltage is the difference between voltage drop and induced voltage.

... however, still not sure where in the single wound series generator system the induced and output voltage are occuring. And how increased resistance induces increased end in the field coils. And what comes first after increasing resistance? Is it increased current which causes increased emf or is it increased emf, which causes increased current?

Confused as to how increased resistance causes increased voltage drop..

3. Jun 25, 2017

### Averagesupernova

Any non-perfect voltage source will have an internal resistance. Start with the most basic which would be a dry cell. Measure it's voltage with no load, then again with a load. These measurements will not be the same. The lost voltage is across the internal resistance. From your two voltage readings and the resistance of a known load you can calculate this internal resistance. This principle applies to the generator ad well. After all, the windings are not superconductors so they have some resistance. This should make things a bit clearer for you.

4. Jun 25, 2017

### Ocata

Agreed,

As you increase the resistance in the circuit, internal resistance stays the same, so you measure a greater voltage across the battery, because the ratio of voltage drops increases in the direction of external resistance.

However, with a DC series wound generator, it says "as the load voltage increases, the current through the field coils also increases. This induces a greater EMF output voltage, that, in turn, increase the generator's output voltage."

Therefore, while the increased load increases output voltage, the causes seem different. For a DC battery series circuit, it's the measure across the battery due to the ratio change in voltage drops. For the DC generator, the increased output is said to be due to the increased emf, which is due to the increased current.

In the DC battery circuit, the current decreases as load increases. In the DC generator, the current is increasing, according to the text.

What is causing the current to increase instead of decrease?

5. Jun 25, 2017

### Averagesupernova

Correct.
Sounds reasonable. However, there are more qualified people on this forum to answer this so I will not comment any farther.
In the battery circuit the current most certainly will increase as the load increases. By that I mean the load current increases not the load resistance. Generally when we refer to increasing the load we think of it this way.

6. Jun 25, 2017

### jim hardy

Something isn't right about that statement.
That describes a shunt field motor not a series field one. Series field is oblivious to terminal voltage.

Change it to this
and it describes a series field motor.

Did you add that light bulb in series or in parallel with the other light bulbs that make up the load?
and you seem to criss-cross those thoughts later in your questions.

Go back to the beginning,
draw a wiring diagram
and convince yourself that "increasing load" is an ambiguous term,
you should complete the thought by saying instead
or

Then you're ready to tackle the series field DC machine one thought-step at a time.

By improving our words we reason better.

old jim

7. Jun 26, 2017

### jim hardy

"Load" is whatever is connected to the machine's output terminals.
"Load Voltage" would be the voltage across the load, which is connected to the output terminals.
So, unless author has described something very unconventional, "Load Voltage" and "Output Voltage" are just the same exact voltage called by different names.

Averagesupernova hit the nail on the head with this one.
"Increasing the load" means causing it to draw more current and power , usually you'd accomplish that by decreasing its resistance.
"Increasing the load resistance" will cause it to draw less current and power.

i think that's your source of confusion.
Straighten that out before tackling what happens inside the generator.

.