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NOTE: THD
R
= THD
2
or 46.2%
THD
2
0.462
=
THD
2
I
3
2
I
5
2
+
I
7
2
+
I
9
2
+
⎝
⎠
I
rms
:=
using the rms current to find THD, we find
I
rms
41.72 amp
=
I
rms
I
1
2
I
3
2
+
I
5
2
+
I
7
2
+
I
9
2
+
⎝
⎠
:=
The rms of the current is the square root of the sum of the squares of the
rms magnitudes of the components:
Problem 35:
NOTE: THD
F
= THD
1
or 52.1%
THD
1
0.521
=
THD
1
I
3
2
I
5
2
+
I
7
2
+
I
9
2
+
⎝
⎠
I
1
:=
Based on the fundamental component, the total harmonic distortion is
Problem 34:
Note that the neutral current is larger than the phase current for this load!
I
n
54.213amp
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This note was uploaded on 03/26/2012 for the course ECET 231 taught by Professor Piller during the Spring '12 term at Purdue.
 Spring '12
 Piller

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