Question

A 2200/220 V, 50 Hz, single-phase transformer has exciting current of 0.6 A and a core loss of 361 watts, when its h.v. side is energised at rated voltage. Calculate the two components of the exciting current.

Answer 1

The two components of the exciting current are: Magnetizing current (Im) = 0.424 A, Hysteresis current (Ih) = 0.424 A .

Here is the calculation of the two components of the exciting current:

Given: ,Exciting current (Ie) = 0.6 A ,Core loss (Pc) = 361 W ,Voltage (Vh) = 2200 V ,Frequency (f) = 50 Hz

Calculation:

Calculate the reactance of the core (Xc):

Xc = Pc / (Ie^2)

Xc = 361 W / (0.6 A)^2

Xc = 1002.778 Ω

Calculate the impedance of the core (Zc):

Zc = √(R_c^2 + Xc^2)

where R_c is the resistance of the core.

Calculate the resistance of the core (R_c):

R_c = |Zc|

R_c = √(1002.778 Ω)^2

R_c = 1002.778 Ω

Calculate the magnetizing current (Im):

Im = Ie * R_c / √(R_c^2 + Xc^2)

Im = 0.6 A * 1002.778 Ω / √(1002.778 Ω)^2

Im = 0.424 A

Calculate the hysteresis current (Ih):

Ih = √(Ie^2 - Im^2)

Ih = √(0.6 A)^2 - (0.424 A)^2

Ih = 0.424 A

Therefore, the two components of the exciting current are:

Magnetizing current (Im) = 0.424 A

Hysteresis current (Ih) = 0.424 A