# The NØPU Big Quiz

Discussion in 'General Technical Questions and Answers' started by N0PU, Jan 9, 2003.

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1. ### AE4FAHam MemberQRZ Page

Inductive reactance increases with frequency (2PiƒL), while capacitive reactance decreases with frequency (1/2PiƒC). That’s why there’s only one frequency where a given inductor and capacitor in a circuit will cancel each other’s reactance. ƒ has no effect on resistance.

2. ### N0PUHam MemberQRZ Page

</span><table border="0" align="center" width="95%" cellpadding="3" cellspacing="1"><tr><td>Quote (ae4fa @ June 08 2003,21:05)</td></tr><tr><td id="QUOTE">Inductive reactance increases with frequency (2PiƒL), while capacitive reactance decreases with frequency (1/2PiƒC).  That’s why there’s only one frequency where a given inductor and capacitor in a circuit will cancel each other.  ƒ has no effect on resistance.[/QUOTE]<span id='postcolor'>
Right...
FA: 10 points
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Question 425

What is the formula for frequency of a tank circuit. [Give both forms!]

3. ### AE4FAHam MemberQRZ Page

ƒ = 1/2Pi(SQRT LC) and ƒr = 1/2Pi(SQRT LC). There are some others, too, with some small modification of the way “resonant frequency” is designated.

That’s probably not what you meant, though. There’s a shortcut that involves substituting the actual value of 1/2Pi, since it never changes – it’s always 0.159.

So, the second form of the formula becomes ƒr = 0.159/(SQRT LC).

4. ### AE4FAHam MemberQRZ Page

Okay, got lost in the notation – again. Harry knows that’s a weak point with me. Let’s try this:

ƒr = 1/(2Pi*(sqrt(L)*(sqrt&copy)

And, substituting 0.159 for 1/(2Pi):

ƒr = 0.159/(sqrt(L)*sqrt&copy

5. ### N0PUHam MemberQRZ Page

</span><table border="0" align="center" width="95%" cellpadding="3" cellspacing="1"><tr><td>Quote (ae4fa @ June 09 2003,09:44)</td></tr><tr><td id="QUOTE">Let’s try this:

And, substituting 0.159 for 1/(2Pi):

Right...
FA: 10 points
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Note: Use square brackets to avoid the dumb icon junk on the board... ie. f = 0.159 / sqrt[L * C]

Question 426

If you have a tank circuit with a 0.01 uF cap and a .01 mH choke what would the resonate frequency be?

6. ### AE4FAHam MemberQRZ Page

ƒr = 0.159/[sqrt[L]*sqrt[C]]
ƒr = 0.159/[sqrt[10*10^-6]*sqrt[10*10^-9]] = 502802Hz
ƒr = 502.8Khz

7. ### AE4FAHam MemberQRZ Page

ƒr = 0.159/[sqrt[L]*sqrt[C]]
ƒr = 0.159/[sqrt[10*10^-6]*sqrt[10*10^-9]] = 502802Hz
ƒr = 502.8Khz
And
ƒr = 1/[2Pi*[sqrt [L]*[sqrt [C]]] = 503290.9
ƒr = 1/(6.28*[sqrt[10*10^-6]*sqrt[10*10^-9]]] = 503290.9
ƒr = 503.3Khz
So, let’s round it to 503000Hz or 503kHz

8. ### N0PUHam MemberQRZ Page

</span><table border="0" align="center" width="95%" cellpadding="3" cellspacing="1"><tr><td>Quote (ae4fa @ June 09 2003,10:37)</td></tr><tr><td id="QUOTE">ƒr = 0.159/[sqrt[L]*sqrt[C]]
ƒr = 0.159/[sqrt[10*10^-6]*sqrt[10*10^-9]] = 502802Hz
ƒr = 502.8Khz[/QUOTE]<span id='postcolor'>
Right...
FA: 10 points
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Question 427

If a tank circuit has a given resonant frequency will X be capacitive or inductive as frequency goes up? as frequency goes down? Why?

9. ### AE4FAHam MemberQRZ Page

As ƒ increases, X will be more inductive. As ƒ decreases, X will be more capacitive.
XL = 2PiƒL and XC = 1/2PiƒC
Inductive reactance increases with frequency; Capacitive reactance decreases with frequency.

10. ### N0PUHam MemberQRZ Page

</span><table border="0" align="center" width="95%" cellpadding="3" cellspacing="1"><tr><td>Quote (ae4fa @ June 09 2003,12:55)</td></tr><tr><td id="QUOTE">As ƒ increases, X will be more inductive.  As ƒ decreases, X will be more capacitive.
XL = 2PiƒL and XC = 1/2PiƒC
Inductive reactance increases with frequency;  Capacitive reactance decreases with frequency.[/QUOTE]<span id='postcolor'>
Right...
FA: 10 points
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Question 428

As the input frequency goes away from the resonant frequency of a tank circuit what happens to voltage and current through the circuit? Why?