- #1

jaus tail

- 615

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Hi,

Sorry this question may be silly but I have an exam on 11th Feb on machines, maths and electronics.

Book had two diagrams for alternator external characteristics. But then which one of 2 diagrams is right?

Book says:

For zero pf loads, it is seen V(t) = E(f) +/- I(a)X(s)

It's minus for lagging loads, and plus for leading loads.

For lagging loads as load increases terminal voltage drops.

And for leading loads as load increases terminal voltage will rise.

But then book says that left diagram is for no load, then why is I(a) varying as in (on a scale). For no load, I(a) = 0. So terminal voltage must be constant horizontal line.

Then for right diagram book says:

With alternator running at rated V(t) and current, what happens if I(a) is reduced to zero(This means open circuit right?)

Book says that no load voltage may rise or fall depending on the load pf. But why?

If I(a) is zero then load pf must have no effect on excitation voltage.

Book says that E(f) > V(t) at lagging load. And E(f) < V(t) for leading loads. But I'm not understanding why the above two graphs are different?

Could this be correct theory:

Like for left graph. I have three alternators. One has lagging pf, one has unity pf, one has leading pf.

I run them on no load and adjust excitation (field current, same field current for three machines) such that terminal voltage is constant.

Now I add load to three machines (lagging, unity and leading) to and see the change in current.

For lagging loads, the addition of load reduces terminal voltage, whereas for leading load the addition of load increases terminal voltage.

But how does this experiment give right graph results?

Sorry this question may be silly but I have an exam on 11th Feb on machines, maths and electronics.

Book had two diagrams for alternator external characteristics. But then which one of 2 diagrams is right?

For zero pf loads, it is seen V(t) = E(f) +/- I(a)X(s)

It's minus for lagging loads, and plus for leading loads.

For lagging loads as load increases terminal voltage drops.

And for leading loads as load increases terminal voltage will rise.

But then book says that left diagram is for no load, then why is I(a) varying as in (on a scale). For no load, I(a) = 0. So terminal voltage must be constant horizontal line.

Then for right diagram book says:

With alternator running at rated V(t) and current, what happens if I(a) is reduced to zero(This means open circuit right?)

Book says that no load voltage may rise or fall depending on the load pf. But why?

If I(a) is zero then load pf must have no effect on excitation voltage.

Book says that E(f) > V(t) at lagging load. And E(f) < V(t) for leading loads. But I'm not understanding why the above two graphs are different?

Could this be correct theory:

Like for left graph. I have three alternators. One has lagging pf, one has unity pf, one has leading pf.

I run them on no load and adjust excitation (field current, same field current for three machines) such that terminal voltage is constant.

Now I add load to three machines (lagging, unity and leading) to and see the change in current.

For lagging loads, the addition of load reduces terminal voltage, whereas for leading load the addition of load increases terminal voltage.

But how does this experiment give right graph results?

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