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					 A 200/1 Current transformer (CT) is wound with 200 turns on the secondary on a toroidal core. When it carries a current of 160 A on t he primary, the ratio and phase errors of the CT are found to be – 0.5% and 30 minutes respectively. If the number of secondary turns is reduced by 1 the new ratio error (%) and phase error (min) will be respectively
 
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- 0.0, 30
 - – 0.5, 35
 - – 1.0, 30
 - – 1.0, 25
 
 
Correct Option: A
| Actual Ratio, R1 = n1 | ![]()  | 1 + | ![]()  | ||
| IP | 
| = 200 | ![]()  | 1 + | ![]()  | ||
| IP | 
Where n1 is the turn ratio,
| n1 = | ||
| NP | 
| Nominal ratio, KN = | = 200 | |
| 1 | 
| Ratio Error, R = | × 100 | |
| R1 | 
Initially ratio error = – 0.5%
| ⇒ -0.5 = | × 100 | |
| R1 | 
⇒ R1 ≅ 201
| Now, n2 (turn ratio) = | = 199 | |
| NP | 
| R2 = 199 | ![]()  | 1 + | ![]()  | = 199 × | ≈ 200 | ||
| IP | 200 | 
| Then ratio, error = | = | = 0 | ||
| R2 | 200 | 
| Phase angle error, θ = | ![]()  | ![]()  | |||
| π | IP | 
that is independent of turn ratio. Hence ratio error, reduces to zero, however phase angle error remains same.

