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KVCG202/EN M/H11
Figure 25:
LDC inputs
2I
= I1 + I2
L
V = I1 R
LDC
V = (2I
– I1) (2R
L
I1 RLDC = (2I
L
2I
(2R
L
I1 =
R
+ (2R
LDC
Simplifying
2R
L
2I
L
R
LDC
I1 =
R
L
2
R
LDC
And
(2X + 1)
I1 = I
(X + 1) where X =
L
Ideally I1 should equal IL (also I2 = IL), but since RL is not zero, I1 will exceed IL.
The required value of X to bring I1 down to 1.05IL will be determined by:
(2X + 1)
1.05I
= I
L
L
(X + 1)
1.05X + 1.05 = 2 X + 1
0.05 = 0.95X
X
= 0.0526
Therefore, we require X < 0.0526 for I1 < 1.05IL
Equivalent circuit diagram for two KVGC202 relays with paralleled
+ R
)
L
LDC
– I1) (2R
+ R
)
L
LDC
+ R
)
L
LDC
+ R
)
L
LDC
+1
+1
R
L
R
LDC
Technical Manual
KVGC202
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