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There is a parallel LC circuit with unknown values for the inductor and the capacitor. How exactly can I identify the value of the inductor and capacitor?

I know I can apply an sinusoidal function to the circuit and calculate the product of LC by identifying the resonance frequency. However I cannot figure out how exactly one could identify the value of the individual components in the parallel LC circuit. I can only use an oscilloscope and a signal generator.

Note: I cannot remove any components from the LC circuit as it is in a "black-box."

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  • \$\begingroup\$ asdf_man: If all that's in the black box is the parallel LC circuit, then connecting a capacitor across the two wires going into/coming out of the black box is equivalent to connecting the capacitor in parallel with the LC circuit. \$\endgroup\$ – EM Fields May 21 '14 at 19:06
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  1. Identify the resonant frequency (f1)
  2. Add a parallel capacitor of known value (Cx).
  3. Measure the resonant frequency again (f2). If it doesn't oscillate try with other capacitor value.
  4. \${f_2 \over f_1} = \sqrt{ C \over {C + C_x}}\$. The only unknown in this is C. You can calculate it.
  5. Determine L using the old frequency and the determined value of C.
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  • \$\begingroup\$ Thanks for the quick reply. The LC circuit is within a "black-box" so I cannot take out the capacitor or the inductor. Sorry for not mentioning that in the question. \$\endgroup\$ – honey.mustard May 21 '14 at 15:50
  • \$\begingroup\$ @asdf_man why do you want to know LC values? Isn't the frequency enough? \$\endgroup\$ – Cornelius May 21 '14 at 15:51
  • \$\begingroup\$ This is part of an assignment at my school. I'm required to identify the contents of a black box as well as specific value of the components. \$\endgroup\$ – honey.mustard May 21 '14 at 15:54
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A parallel combination of ideal parts would (theoretically) result in a resonant effect with zero bandwidth. Of course, you can measure the existing bandwidth, which is determined by the loss resitanc(es). But you should know how the equivalent circuit diagram looks like - that means: R(loss) in parallel to LC or in series to L or two lossy parts?

On the other hand - as suggested already, why not adding another cap at the output? It is not necessary to "take out" the internal capacitor. I suppose, you have access to the output, don´t you?

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