Question:

Published on: 27 June, 2022

**In a certain current transformer, the following data is obtained. Nominal ratio = 25/5A, Turn ratio = 3, primary terns = 40, secondary turns = 120, secondary resistance = 0⋅16z, secondary reactance = 0⋅195Ω, secondary burden = 15 VA, Burden power factor = 0⋅7, secondary terminal voltage = 3V. **

**Find ratio & phase angle errors. The magnetising and loss ampere turns corresponding to an emf of 4⋅26V are13 & 10⋅1 respectively. **

Answer:

Secondary burden=15VA, secondary winding current=5A=\(I_s\)

Secondary circuit impedance =\(\frac{15}{5^2}=0.6\Omega\)

Secondary circuit reactance=\(0.195\Omega\)

Phase angle of secondary circuit \(\delta=\sin^{-1}{\left(\frac{0.195}{0.6}\right)=0.325}\)

\(\therefore\sin{\delta=0.00567,\ \cos{\delta}}=0.9999\)

Primary turns=40=\(N_P\) , secondary turns =120=\(N_S\)

Turns ratio,n=\(\frac{N_S}{N_P}=\frac{120}{40}=3\), Nominal ratio=\(K_n=\frac{25}{5}=5\)

Magnetizing current \(I_m=\frac{magnetizing\ mmf}{primary\ winding\ turns}=\frac{13}{40}=0.325\)

Loss component \(I_e=\frac{excitation\ for\ loss}{primary\ winding\ turns}=\frac{10.1}{40}=0.2525\)

Actual ratio \(R=n+\frac{I_e\cos{\delta}+I_m\sin{\delta}}{I_s}=3+\frac{0.2525\ast0.9999+0.325\ast0.0567}{5}\)

\(=3+\frac{0.2524+0.001848}{5}=3.0508\)

Ratio error =\(\frac{nominal\ ratio-actual\ ratio}{actual\ ratio}\ast100=\frac{5-3.0508}{3.0508}\ast100=+63.889%\)

Phase angle \(\theta=\frac{180}{\pi}\left[\frac{I_m\cos{\delta}-I_e\sin{\delta}}{nI_s}\right]=180\left[\frac{0.325\ast0.9999-0.2525\ast0.0056}{3\ast5}\right]\)

\( =180\left[\frac{0.3249675-0.0014316}{15}\right]=3.8824°\)

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