# 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 $\mu$Wb 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.