- #1

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Thanks

Richard.

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In summary, the conversation discusses the behavior of reactances in a series circuit with a capacitor and inductor connected to an alternating supply. The reactances of the two components are subtracted from each other due to their opposite phases. The formula for calculating the reactance in this circuit is Z=iωL-1/iωC. The conversation also mentions a generator that needs a new capacitor and the details of its specifications. The conversation ends with the speaker admitting to not understanding the formula and being a licensed electrician who needs to refresh their knowledge.

- #1

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Thanks

Richard.

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- #2

Homework Helper

- 2,593

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Yes, reactance behaves just like resistance in resistive circuits.

- #3

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Ummm, they add like this 5 & 4 = 1.

Capactitive reactance is 180 degrees out of phase from inductive reactance, so you actually subtract one from the another.

It's good to go back to the definitions. In series,

[tex]Z= i\omega L + \frac{1}{i\omega C}[/tex]

where [tex]X_{L}=Im(Z)[/tex]

Capactitive reactance is 180 degrees out of phase from inductive reactance, so you actually subtract one from the another.

It's good to go back to the definitions. In series,

[tex]Z= i\omega L + \frac{1}{i\omega C}[/tex]

where [tex]X_{L}=Im(Z)[/tex]

Last edited:

- #4

Homework Helper

- 2,593

- 5

- #5

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KW: 2.65 Amps:22 KVA: 2.65 Phase:1

voltage: 120 RPM:3600

HZ: 60

would appreciate anyons help!

russhart70

or give me a formula,? PLZ.

- #6

- 686

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[tex] = i\omega L - \frac{i}{\omega C} [/tex]Phrak said:[tex]Z= i\omega L + \frac{1}{i\omega C} [/tex]

because: 1/i = -i

So, they subtract.

- #7

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- #8

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goin to get my books out real quick!

- #9

- 7

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do not understand wat L = wat I know is P=wats I = amps E=volts R=omes

- #10

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I guess I AM

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