Jump to latest postWhich order harmonic would be effectively removed from a power system by applying a single-tuned filter with a 0.55 μF capacitor and a 500 mH inductor?
How do you calculate this?
Follow Us
NETA Level IV
Affiliate Member
Posted April 19, 2019
Pro Subscriber
Which order harmonic would be effectively removed from a power system by applying a single-tuned filter with a 0.55 μF capacitor and a 500 mH inductor?
How do you calculate this?
NETA Level IV
Posted April 23, 2019
I believe the formula is 1/(2PI*SQRT(LC))Originally Posted by ElectricalTestTech
Junior Member
Posted April 23, 2019
Pro Subscriber
Can you walk us through this calculation? I have tried doing it several times and haven't come up with what they say is the correct answer yet.
NETA Level IV
Posted April 23, 2019
L*C =0.5*0.00000055=0.000000275
sqrt(0.000000275)=0.0005244
2pi*0.0005244=0.003294
1/0.003294=303.49hz
303.5hz/60hz=5.058 or 5th Harmonic
NETA Level III
Posted June 14, 2020
Pro Subscriber
I've seen a slightly different version of this question:
A harmonic filter bank consists of a capacitor bank at 8uF and 40mH. Which harmonic is this bank eliminating?
f = 1 / (2 * pi * sqrt(L * C))
f = 1 / (2 * pi * sqrt(0.040 * 0.000008))
f = 1 / (2 * pi * sqrt(0.00000032))
f = 1 / (2 * pi * 0.00056568542)
f = 1 / (0.00355430631942971339162580115316)
f = 281.35
Harmonic @ 60 Hz = 281.35 / 60 = 4.689 = 5th harmonic
