---
title: "CT with ammeter numerical"
subject: "Switchgear and Protection"
chapter: 2
key: "semester_6/switchgear_and_protection:ct-ammeter-numerical"
source: https://www.bctengineer.com/q/semester_6/switchgear_and_protection/ct-ammeter-numerical
site: BCT Engineering
---

# CT with ammeter numerical

Switchgear and Protection · Chapter 2: Current and Potential Transformers

- Asked in: 2078 Chaitra, 4 marks
- Video: [Switchgear and Protection Course: Lecture 12e -  Numerical Examples on CT Performance](https://www.youtube.com/watch?v=9ONTPH8BNqg) — Dr. Ahmad Al-Subhi
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_6/switchgear_and_protection/ct-ammeter-numerical
- Whole chapter: https://www.bctengineer.com/md/semester_6/switchgear_and_protection/chapter-2

## Question

A current transformer is connected in 400 V system and has 1 turn on its primary and 180 turns on its secondary. An ammeter with 0.15 Ω internal resistance is connected to CT. The ammeter is required to give a full scale deflection when the primary current is 900 Amps. Calculate: i) The maximum secondary current. ii) Secondary voltage at CT terminals when ammeter is removed.

## Short answer

Using ampere-turns balance: I_s = I_p × N_p/N_s = 900 × 1/180 = 5 A maximum secondary current. With ammeter removed, secondary open-circuit voltage V_s = (400/√3) × 180 ≈ 41.6 kV (phase voltage times turns ratio). This demonstrates why CT secondary must never be opened under load.

## Answer

**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):
```text
 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
```text
 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.
```text
 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.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. A CT has 1 primary turn and 180 secondary turns; primary current is 900 A. What is the secondary current?
   A. 5 A
   B. 162 kA
   C. 0.5 A
   D. 180 A
   Answer: A. I_s = 900 x 1/180 = 5 A. (Key point: Secondary current)

2. Ammeter resistance is 0.15 ohm at 5 A. What is the CT terminal voltage?
   A. 30 V
   B. 0.75 V
   C. 1.2 V
   D. 0.03 V
   Answer: B. V = 5 x 0.15 = 0.75 V. (Key point: Terminal voltage with meter)

3. Why use 400/sqrt3 and not 400 V across the primary?
   A. The CT is delta connected
   B. Because 400 V is DC
   C. The ammeter reduces it
   D. The 400 V is taken as the 3-phase line voltage, so a single-turn primary sees the phase voltage
   Answer: D. If 400 V were applied directly, V_s would be 72 kV; the answer assumes the line value, so phase = 230.94 V. (Key point: Assumption)

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