How Do You Calculate Self-Inductance of a Resistor in an A.C. Circuit?

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In summary, the self-inductance of a resistor in a series AC resistive capacitive circuit is a small property that is usually only significant at very high frequencies. For a wire-wound resistor, the self-inductance is typically around 1 μH. For other resistor types, a formula can be used to estimate the self-inductance based on the length and radius of the resistor. This information may be useful for evaluating the inductive circuit in a later part of the investigation.
  • #1
stevedave12
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does anybody know how to calculate the self-inductance of a resistor in a series A.C resistive capacitive circuit?
(have values for supply voltage, R voltage, C voltage, current)
any help at all would be much appreciated
 
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  • #2
A resistor's self-inductance is a property of the resistor itself, dependent on its type (e.g., wire-wound, etc.), shape, size, etc. Usually a resistor's self-inductance is very small and only significant at very high frequencies. What AC frequencies are you considering?? If at very high frequencies, are you also considering the self-inductance of the connecting wires between components?? What is your intended application?? What are your resistor's specifications (type, physical characteristics, etc.)??


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  • #3
the frequency range I am using is 1-10kHz, I am not considering the self inductance of the wires, i only need the value for a later part of my investigation (to take into account when evaluating in inductive circuit) although the only the only information i have on the resistor is it's actual resistance.
 
  • #4
The self-inductance of a "Wire-Wound Resistor" (usually used for precision purposes) is probably in the neighborhood of 1 μH (1 x 10(-6) Henry). The self-inductance of other resistor types can probably be estimated with that due to a straight wire having the length=("S" in cm) and radius=("R" in cm) of the resistor (plus leads) and given by:
L {in μH = 10(-6) H} = (0.002)*S*{Loge(2*S/R) - (3/4)}
which is presented in Eq #3 of the following reference:
http://www.wireductsales.com/knowledge/docs/noise_appnote.htm


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  • #5
Thanks for your help, much appreciated
 

Related to How Do You Calculate Self-Inductance of a Resistor in an A.C. Circuit?

What is self-inductance of a resistor?

Self-inductance of a resistor is a property that describes the ability of a resistor to store energy in the form of a magnetic field when current flows through it.

How is self-inductance of a resistor calculated?

The self-inductance of a resistor is calculated using the formula L = μrμ0N2A/l, where μr is the relative permeability of the resistor, μ0 is the permeability of free space, N is the number of turns in the resistor, A is the cross-sectional area, and l is the length of the resistor.

What factors affect the self-inductance of a resistor?

The self-inductance of a resistor is affected by the number of turns in the resistor, the cross-sectional area, the length, and the material used. The permeability of the material also plays a significant role in determining the self-inductance.

Why is self-inductance important in a circuit?

Self-inductance is important in a circuit because it can cause a change in current flow when there is a change in the magnetic field around the resistor. This can lead to voltage spikes and can affect the overall performance of the circuit.

How can the self-inductance of a resistor be reduced?

The self-inductance of a resistor can be reduced by using materials with low permeability, reducing the length of the resistor, and increasing the cross-sectional area. Adding a parallel capacitor can also help reduce self-inductance in a circuit.

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