Philips Electromagnetic Lamp manual 125, Neutral interruption and resonance

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3.6 Neutral interruption and resonance

Fig. 119. The consequences of interrupted neutral in a phase/neutral network.

L1		IR
		L1
Itotal				IR
				R1
1000	V1	230V			1000
400V		N			IR
		R2	R3	R4
					R5
250	V2	230V			1000
					each
L2		L2
		4 IR

This makes 1000 + 250 = 1250 Ω . So the current will be 400 / 1250 = 0.32 A.

The voltage across R1 will be 0.32 . 1000 = 320 V (V = I . R), so the power in R1 will be 320 . 0.32 = 102 W.

The voltage across the four parallel resistors is 0.32 . 250 = 80 V, so the power in each resistor is 80 . 0.08 = 6.4 W.

Now it is seen that the smaller load (R1) is overloaded by a higher voltage (320 instead of 230 V) and a higher current (0.32 A instead of 0.23 A).The higher load (R2 to R5) is greatly underloaded.

In practice the circuits are not that simple, but the essential aspect is that in case of a floating neutral the smallest load will receive a higher voltage and a higher current and so will be overloaded.

A second possibility of resonance has to do with the employment of inductive and capacitive circuits in the same installation. In the capacitive circuit, the impedance of the capacitor is twice the impedance of the inductive ballast. So when an inductive and a capacitive circuit get in series, the total impedance will be zero, resulting in an unlimited current (resonance).This can happen in a delta-network when one phase is interrupted (Fig. 120) or in a star-network with common neutral when the neutral is interrupted (Fig. 121).

Resonance problems can be prevented with special switch gear. If the neutral in a star-network or a phase in the delta-network fails, such special gear switches off the overall supply for the lighting installation.

B

‘TL’D S

‘TL’D S

B

C

Fig. 120. Resonance in a delta-network.

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