# Frequency of Natural Resonance with Parasitic Component due to Oscilloscope

So, I've just performed a lab on measuring the resonance of an LCR circuit. I rearranged the circuit so each time the component in question had one lead on the ground plane, and then measured it's frequency response.

Each time, an oscilloscope was attached across the component in question. When the natural frequency was measured for each component, each time it was different.

I'm assuming this is because the oscilloscope itself has an inductance/capacitance and therefore adds a parallel component to each measurement. I want to calculate the value of this parasitic effect. How would I go about doing that?

Inductor Value: 109.4 mH
Capacitor Value: 2.116 nF
Resistor Value: 327.2 Ω

The natural frequency when measured across each component was as follows...

Resonant Frequency of Inductor: 10.1kHz
Resonant Frequency of Capacitor: 10.075kHz
Resonant Frequency of Resistor: 10.425kHz

• schematics please – berto Mar 16 '16 at 21:43
• en.wikipedia.org/wiki/File:RLC_series_circuit_v1.svg – Bryan.Russell.Potts Mar 16 '16 at 21:46
• Just a simple series LCR circuit. But measuring the natural frequency across each component. – Bryan.Russell.Potts Mar 16 '16 at 21:47
• What you're measuring each time is the resonant frequency of the entire circuit, not of the individual components. Resistors don't have resonant frequencies, at least not in the 10kHz range. The input impedance of a scope is usually written on it, and it is typically 1Meg||20pF. – user207421 Mar 16 '16 at 22:29
• I know. But by adding the oscilloscope across each component, I'm adding on additional circuitry that is changing the resonant frequency. The oscilloscope is constituting a parallel impedance with each component as I'm measuring the resonant frequency. I want to be able to calculate the effective impedance from the information given. – Bryan.Russell.Potts Mar 16 '16 at 23:11