# The NØPU Big Quiz

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

Not open for further replies. 1. 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. </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. 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. 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. </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. ƒr = 0.159/[sqrt[L]*sqrt[C]]
ƒr = 0.159/[sqrt[10*10^-6]*sqrt[10*10^-9]] = 502802Hz
ƒr = 502.8Khz

7. 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. </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. 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. </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?