Calculating Series Resonance at 50Hz: Inductance Question

[ForgottenMemo]

Homework Statement

An AC generator produces a voltage of 230<45º volts. It is connected across an impedance of 17.4<-33º ohms.

The supply frequency = 50Hz


I got my assignment back and this question wrong, maybe you can guide me, please because I’m not sure entirety. Help is appreciated.

Question: Calculate the value of inductance that will make the circuit become series resonant at 50Hz?

Homework Equations

1/2π fc = 2 π fl

The Attempt at a Solution

I= 17.4 sin33 = 9.47
C= 1/2 π (50x9.47) = 3.36x10^-6
Would it be: 2 π fl > 2 π(50) = 314 ?

 

[marcusl]

Hmm, check your math. If

$$X_c=\frac{1}{j\omega C}$$

then C = 3.36e-4. When I plug C into

$$L=\frac{1}{\omega^2 C}$$

I get L=0.030.

 

[ForgottenMemo]

Thanks for the responce, but I don't undestand why you say check my math, what wrong with it. You've confused me now.


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Also where did those two other equations come from cause my lecturer has not used them, there not the ones i was using?

1/2π fc = 2 π fl

--

 

[marcusl]

C= 1/2 π (50x9.47) = 3.36x10^-6

The values are correct but the answer is wrong. Try it again.

1/2π fc = 2 π fl

so in your notation,

$$l=\frac{1}{(2\pi f)^2 c}$$

I get l=0.030.

 

[ForgottenMemo]

Hello,

I re-calculated it and got this for C:

C= 1/2 π (50x9.47) = 743.7

--

Using : $$l=\frac{1}{(2\pi f)^2 c}$$



1/(2πx50)^2 = 1.013x10^-5

(1.013x10^-5) x C (743.7) = 7.5x10^-3

Which is utterly wrong.

I broke it down into chucks thinking tit would help, either way i get the same answer above.

 

[marcusl]

Hello,

I re-calculated it and got this for C:

C= 1/2 π (50x9.47) = 743.7

No, that's why I wrote down the equations before. The correct equation in your notation is
$$C=\frac{1}{2\pi f I}.$$

Take a look in your book! Physics Forums is not a replacement for studying.

 

[ForgottenMemo]

Im sorry my limited amount of knowleadge, i find it hard to understand even on a good day (It takes me a couple of times, before it all sinks in, esp, when the lecturer is not even good, nor textbooks or websites make it clear).

$$C=\frac{1}{2\pi f I}.$$ => $$C=\frac{1}{2\pi 50 9.47}.$$ = 3.361*10^-4

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$$l=\frac{1}{(2\pi f)^2 c}$$ => 1 / (2pi x 50)^2 x 3.361x10^-4) = 0.030155 = 30155uH


So basically,that will make the circuit become series resonant at 50Hz, is that it marcus.

 

[marcusl]

Yes, that's it!
Watch out for the number of digits in your answer, though. Since the problem specified three significant digits (17.4 ohms), your answers should also have 3 significant digits: 0.030 H and 3.36e-4 F.

 

[ForgottenMemo]

Thanks for the help Marcus, it was most kind.