# The NØPU Big Quiz

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

Not open for further replies. 1. I thought it would be good for me to work thru a couple ‘frinstances:

R = 50 Ohms
.01mH = 0.000010 or 10*10^-6
.01uF = 0.000000010 or 10*10^-9
ƒr = 503000Hz or 503kHz

At resonance: Z = (31.6-31.6)+50 = 50 Ohms
At 12V input, current would be 12/50 or 240mA
EL = .240*31.6 = 7.584V
EC = .240*31.6 = 7.584V
ER = .240*50 = 12V
Power Out = 12*.240 = 2.88W
0W dissipated as heat by reactances

Shifting operating frequency to 400kHz:
XL = 25.133 Ohms
XC = 39.789 Ohms
Z = (39.789-25.133)+50 = 64.656 Ohms
At 12V in, current would be 186mA
EL = .186*25.133 = 4.675V
EC = .186*39.789 = 7.4V
ER = .186*50 = 9.3V
Power Out = 9.3*.186 = 1.73W
0.5W dissipated as heat by reactances

Shifting frequency to 800kHz:
XL = 50.266 Ohms
XC = 19.894 Ohms
Z = [50.266-19.894]+50 = 80.372 Ohms
At 12V in, current would be 149mA
EL = .149*50.266 = 7.490V
EC = .149*19.894 = 2.964V
ER = .149*50 = 7.45V
Power Out = 7.45*.149 = 1.1W
0.67W dissipated as heat by reactance

So, we see that both voltage and current through the circuit both decrease – of course producing less power at the output – as the operating frequency moves away from the circuit’s resonant frequency. And, more power is dissipated as heat in the circuit. Direction doesn’t matter.

2. </span><table border="0" align="center" width="95%" cellpadding="3" cellspacing="1"><tr><td>Quote (ae4fa @ June 09 2003,17:32)</td></tr><tr><td id="QUOTE">I thought it would be good for me to work thru a couple ‘frinstances:

R = 50 Ohms
.01mH = 0.000010 or 10*10^-6
.01uF = 0.000000010 or 10*10^-9
ƒr = 503000Hz or 503kHz

At resonance: Z = (31.6-31.6)+50 = 50 Ohms
At 12V input, current would be 12/50 or 240mA
EL = .240*31.6 = 7.584V
EC = .240*31.6 = 7.584V
ER = .240*50 = 12V
Power Out = 12*.240 = 2.88W
0W dissipated as heat by reactances

Shifting operating frequency to 400kHz:
XL = 25.133 Ohms
XC = 39.789 Ohms
Z = (39.789-25.133)+50 = 64.656 Ohms
At 12V in, current would be 186mA
EL = .186*25.133 = 4.675V
EC = .186*39.789 = 7.4V
ER = .186*50 = 9.3V
Power Out = 9.3*.186 = 1.73W
0.5W dissipated as heat by reactances

Shifting frequency to 800kHz:
XL = 50.266 Ohms
XC = 19.894 Ohms
Z = [50.266-19.894]+50 = 80.372 Ohms
At 12V in, current would be 149mA
EL = .149*50.266 = 7.490V
EC = .149*19.894 = 2.964V
ER = .149*50 = 7.45V
Power Out = 7.45*.149 = 1.1W
1.79W dissipated as heat by reactance

So, we see that both voltage and current through the circuit both decrease – of course producing less power at the output – as the operating frequency moves away from the circuit’s resonant frequency.  And, more power is dissipated as heat in the circuit.  Direction doesn’t matter.[/QUOTE]<span id='postcolor'>
Right ...
FA: 10 points
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Question 249

How does the ideal series resonant circuit differ from a one with 'some' R?

3. An “ideal” circuit has no R – and therefore, at resonance, infinite current flows. At resonance, the reactances cancel – and, since there is no resistance, Z=Ø and I=E/Z yields infinite current. Probably not a good situation, really, given other limitations - but impossible in any case, since every circuit has resistance.

If XL and XC are equal, say 31.6 Ohms each, the circuit is resonant. And, if R=Ø (as in the &quot;ideal&quot , then Z=(100-100)+Ø=Ø Ohms.

In the same situation, with 50 Ohms resistance: Z=(100-100)+50=50.Ohms

At resonance, Z always = R, not less, not more. In other words, the value of R is the lowest impedance figure possible, and it is achieved only when the circuit is resonant.

For demonstration, this can be plotted out in a number of ways. This would clearly show that impedance (Z) is at minimum (and equal to R) only at the point of resonance, where the XL and XC lines cross.

4. Harry is taking his time getting back to this, maybe he is looking for a bit more info?

Lets try:

A series resonant circuit with no &quot;R&quot; will have no losses, it will not dissipate any power.
At resonance it will be a true short circuit, zero ohms impedance.

The &quot;Q&quot; of the circuit will be infinite, and the bandwidth will be zero!

Consider the formula:

Q = (resonant freq)/Bandwidth

turn it around

Bandwidth = (resonant freq)/Q

if Q is infinite, Bandwidth = 0

Jim GM3ZMA

5. BOTH OF YOU ARE RIGHT...