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.

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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