Merz-Price: unprotected winding (5000 kVA)

For an earth fault at a point x (fraction of winding from the neutral), the voltage driving the fault current is x × phase voltage, and the current is limited mainly by the neutral earthing resistance. The relay operates only if this current exceeds the setting; faults close to the neutral give too little current, so that part of the winding is unprotected.

Data: 5000 kVA, 6600 V, star; R_n = 6.5 Ω; winding impedance 0.5 + j2 Ω per phase; relay operates at 30% of full-load current.

Step 1: Full-load and operating currents

I_FL = 5000 × 10³ / (√3 × 6600) = 437.39 A
I_op = 0.30 × 437.39 = 131.22 A
V_ph = 6600 / √3 = 3810.51 V

Step 2: Unprotected portion (usual textbook method)

Neglecting the impedance of the short faulted part of the winding (small compared with 6.5 Ω):

x × V_ph / R_n = I_op
x = I_op × R_n / V_ph
  = 131.22 × 6.5 / 3810.51
  = 0.2238

So 22.38% of the winding (from the neutral end) is unprotected.

Step 3: Including the winding impedance (check)

The faulted part has impedance x(0.5 + j2) Ω, so:

x × 3810.51 = 131.22 × |6.5 + x(0.5 + j2)|
Solving numerically: x = 0.2283

This gives 22.83% unprotected, only slightly more; the winding impedance has little effect because R_n dominates.

 Neutral ─R_n(6.5Ω)─ N ──x──●F── (1−x) ── line
                             │
                            earth

Answer: about 22.4% of the winding (22.8% if the winding impedance is included) near the neutral is unprotected; about 77.6% is protected.