Instrument Transformers

Question 1
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 the 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 new ratio-error(%) and phaseerror( min) will be respectively
A
0.0, 30
B
-0.5, 35
C
-1.0, 30
D
-1.0, 25
GATE EE 2006   Electrical and Electronic Measurements
Question 1 Explanation: 
Nominal ratio =k_n=\frac{200}{1}=200
Primary current =I_P=160A
Actual ratio = R_1=n_1\left ( 1+\frac{I_e}{I_P} \right )\;\;...(i)
\% \text{Ratio error} =\frac{k_n - R_1}{R_1}\times 100
-0.5=\frac{200-R_1}{R_1} \times 100
Actual ratio =R_1=201
n_1=\text{turn ratio}=200
R_1=200\left ( 1+\frac{I_e}{I_P} \right )\;\;...(ii)
when number of secondary is reduced by 1
\begin{aligned} N_s &= 199\\ n_2&=199 \\ R_2&=n_2\left ( 1+\frac{I_e}{I_P} \right ) \\ &= 199\left ( 1+\frac{I_e}{I_P} \right )\;\;...(iii) \end{aligned}
Dividing equation (ii) and (iii),
\begin{aligned} \frac{R_1}{R_2} &=\frac{200}{199} \\ \Rightarrow \;R_2 &= R_1 \times \left ( \frac{200}{199} \right )\\ &=201 \times \frac{200}{199}\approx 200 \\ \% \text{Ratio error} &=\frac{k_n-R_2}{R_2} \times 100\\ &= \frac{200-200}{200}\times 100=0 \end{aligned}
Phase angle error =\theta =\frac{180}{\pi}\left ( \frac{I_m}{I_P} \right )
Reduction of one or two turns of the secondary winding, no doubt, reduces the ration error, but it has no effect on the phase angle error.
Question 2
A 50 Hz, bar primary CT has a secondary with 500 turns. The secondary supplies 5 A current into a purely resistive burden of 1\Omega. The magnetizing ampere-turns is 200. The phase angle between the primary and second current is 175.4^{\circ}. The core flux in the CT under the given operating conditions is
A
0
B
45 \muWb
C
22.5 m Wb
D
100.0 mWb
GATE EE 2004   Electrical and Electronic Measurements
Question 2 Explanation: 
Z_s=\text{Burden}=1\Omega
Voltage induced in the secondary
\begin{aligned} E_s&=I_S \times Z_S=5 \times 1=5V \\ E_s&=4.44 f\phi _mN_s \\ \therefore \;\; \phi _m&=\frac{E_s}{1.44fN_s} \\ &= \frac{5}{4.44 \times 50 \times 500}=45\mu Wb \end{aligned}
Question 3
A 50 Hz, bar primary CT has a secondary with 500 turns. The secondary supplies 5 A current into a purely resistive burden of 1\Omega. The magnetizing ampere-turns is 200. The phase angle between the primary and second current is
A
4.6^{\circ}
B
85.4^{\circ}
C
94.6^{\circ}
D
175.4^{\circ}
GATE EE 2004   Electrical and Electronic Measurements
Question 3 Explanation: 
Phase angle between primary and secondary current
\theta =\left ( \frac{180}{\pi} \right )\left ( \frac{I_m}{nI_s} \right )\;\text{degree}
No. of turns in primary =N_p=1
No. of turns in secondary =N_s=500
n=\text{turn ratio} =\frac{N_s}{N_p}=500
I_s= secondary winding current=5A
Magnetizing ampere turn =200
I_m=\frac{\text{Magnetizing ampere turn}}{N_p} =\frac{200}{1}=200A
\theta =\left ( \frac{180}{\pi} \right )\left ( \frac{200}{500 \times 5} \right )\approx 4.6^{\circ}
Question 4
A 500A/5A, 50 Hz transformer has a bar primary. The secondary burden is a pure resistance of 1 \Omega and it draws a current of 5 A. If the magnetic core requires 250 AT for magnetization, the percentage ratio error is
A
10.56
B
-10.56
C
11.8
D
-11.8
GATE EE 2003   Electrical and Electronic Measurements
Question 4 Explanation: 
Nominal ratio =\frac{500}{5}=100
Number of turn in primary N=1 (primary bar)
Magnetizing current,
I_m=\frac{\text{mmf}}{\text{No. of turns}}
\;\;=\frac{250}{1}=250A
Secondary current,
I_s=5A n=\frac{500}{5}=100
Primary current,
I_P=\sqrt{(nI_s)^2+I_m^2}
\;\;=\sqrt{(100\times 5)^2 +(250)^2}
\Rightarrow \; I_P=559.017A
Actual ratio
\;\;=\frac{I_P}{5}=\frac{559.017}{5}=111.803
Ratio error
\;\;=\frac{\text{Nominal Ratio} - \text{Actual Ratio}}{\text{Actual Ratio}} \times 100
\;\;=\frac{100-111.803}{111.803}\times 100=-10.56\%
There are 4 questions to complete.
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