CT with ammeter numerical

Given: N_p = 1 turn, N_s = 180 turns, ammeter resistance R_A = 0.15 Ω, full-scale primary current I_p = 900 A, system voltage 400 V.

(i) Maximum secondary current

For a CT, ampere-turns balance (neglecting exciting current):

 N_p · I_p = N_s · I_s
 I_s = I_p × N_p/N_s
     = 900 × 1/180
     = 5 A

So the ammeter must be a 0–5 A meter giving full-scale at 5 A.

With the ammeter connected, the CT terminal voltage is only

 V_s = I_s × R_A = 5 × 0.15 = 0.75 V
 Primary drop = 0.75/180 = 0.00417 V ≈ 4.17 mV

(ii) Secondary voltage when the ammeter is removed

When the ammeter is removed, the secondary is open-circuited. There is no secondary m.m.f. to oppose the primary, so the CT behaves like a step-up voltage transformer of ratio 1 : 180 connected in the line. In the limit (core not saturated, no other limit), the voltage across the single-turn primary can rise up to the phase voltage of the system.

Assumption: 400 V is the line voltage of a 3-phase system, so phase voltage = 400/√3 = 230.94 V.

 V_s(open) = V_p × N_s/N_p
           = (400/√3) × 180
           = 230.94 × 180
           = 41 569 V ≈ 41.6 kV

(If 400 V were applied directly across the primary, V_s = 400 × 180 = 72 kV.)

Answer: (i) I_s(max) = 5 A (terminal voltage 0.75 V with ammeter). (ii) On open circuit the secondary voltage can rise to about 41.6 kV (theoretical upper limit; 72 kV if 400 V across primary).

In practice core saturation limits the r.m.s. value, but very high peak voltages still appear; this shows why a CT secondary must never be opened under load and must be short-circuited before removing the ammeter.