Proof That the Fault Current in a Capacitor is Limited to Three Times the Line Current for an Ungrounded Wye Capacitor Bank
First we will look at what happens to the voltage when a capacitor fails in one phase just before the fuse clears the faulted capacitor.
Figure 1 shows a balanced floating WYE capacitor connection. Let 1 PU (per unit) equal VaN, VbN and VcN.
Assume a capacitor is failing in leg (a). Just before the fuse attached to this capacitor clears, the neutral (N) shifts to Va or, N’. See figure 2.
There is an Increase in Voltage Across the Capacitors in Lines VbN’ and VcN’. As the Capacitor Fails and Prior to the Capacitor Fuse Clearing the Neutral Will Shift From N to N’. Calculating the Legs of the Triangle Created by Sides b =(VbN), a=(VaN) and b’= (VbN’) We Can Use The Law of Cosines
(b’)^{2} = a^{2} + b^{2 }–2ab(CosN)
(b’)^{2} = (VbN’)
But a=VaN =b=VbN=NN’=1PU, then a^{2} = b^{2} = 1
N = 120^{0
}CosN = (0.5)
Then (VbN’)^{2} = 1^{2} + 1^{2} – 2(1)(1)(0.5) = 3
(VbN’) = sqrt(3) or, approximately 1.73205
Using this knowledge for our vector analysis we will let Side a = Side b = sqrt(3). Therefore, the voltage on these legs will be at linetoline potential.
The shift in the capacitor current will have the same effect as the voltage shift. Under normal conditions the current leads the voltage by 90^{o} . See figure 3.
Using Kirchhoff’s law, Under Balance Conditions, We Have
I_{N} = IØa + IØb + IØc = 0
When the failure in leg (a) occurs as described above, the current will change accordingly.
See Figure 4.
In Figure 3, I_{N} = 0. However, I_{N} Shifts to IØa Position and is Now Considered I_{N’}. Again, Using Kirchhoff’s law We Have the Following
I_{N} = IØb + IØc
Under normal conditions IØa = IØb = IØc, but as shown in the voltage calculation when unit in leg (a) begins to fail, Vb = Vc = (sqrt(3) * 1pu.) The same calculations will apply in the current calculations and IØb = IØc, = (sqrt(3) * normal line current).
Using Polar and Complex Number Calculations From Figure 4, We Have the Following
I_{N’} = (sqrt(3) * ∠330^{O}) + (sqrt(3) * ∠30^{O})
= (1.5  j0.866) + (1.5 + j0.866)
= (3 ± j0)
Therefore, I_{N’ }= 3 times normal line current.
See figure 5 for a vectorial solution. You will note that the figure is a Rhombus. The diagonals of a Rhombus intersect at right angles. Also, the diagonals divide the Rhombus into four equal and congruent triangles. Then A = B, let A and B equal (sqrt(3)).
Then We Have the Following
Vector A + Vector B = 2C (Resultant)
C = (Cos30^{0})(sqrt(3))
C = (0.866)(sqrt(3))
C = 1.5
∴ 2C=3
