Active and Passive Harmonic Filters calculation
09 Mar 2024
Tags: Electrical Engineering Power Systems Power Quality Active and Passive Harmonic Filters calculation
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Harmonic Filters in Electrical Engineering
This calculator provides the calculation of active and passive harmonic filters for electrical engineering applications.
Explanation
Calculation Example: Harmonic filters are used in electrical engineering to reduce the harmonic content of a power system. Active harmonic filters use power electronics to inject a current that cancels out the harmonic currents produced by nonlinear loads. Passive harmonic filters use capacitors and inductors to resonate at the harmonic frequencies and provide a low impedance path for the harmonic currents.
Related Questions
Q: What are the benefits of using harmonic filters?
A: Harmonic filters provide several benefits, including improved power quality, reduced equipment losses, and increased system stability.
Q: How do I choose the right harmonic filter for my application?
A: The selection of the right harmonic filter depends on several factors, including the type of load, the harmonic content of the system, and the desired level of harmonic reduction.
Variables
Symbol  Name  Unit 
 ——  —  — 
V  Source Voltage  V 
f  Fundamental Frequency  Hz 
h  Harmonic Order   
X  Source Impedance  ? 
C  Filter Capacitance  F 
L  Filter Inductance  H 
Calculation Expression
Filter Impedance: The impedance of the filter is given by Z = sqrt(R^2 + (2?fL  1/(2?fC))^2)
sqrt(X^2 + (2*pi*f*L  1/(2*pi*f*C))^2)
Filter Current: The current through the filter is given by I = V / Z
V / Z
Harmonic Voltage: The harmonic voltage across the filter is given by V_h = I * Z
I * Z
Calculated values
Considering these as variable values: C=1.0E4, V=230.0, f=50.0, h=5.0, X=5.0, L=0.1, the calculated value(s) are given in table below
Derived Variable  Value 
 ——  — 
Filter Impedance  Sqrt(25.0+(31.415926535897930.0031830988618379067/C)^2) 
Filter Current  230.0/Sqrt(25.0+(31.415926535897930.0031830988618379067/C)^2) 
Harmonic Voltage  I*Sqrt(25.0+(31.415926535897930.0031830988618379067/C)^2) 
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