No-load current components, iron loss and core flux

Given: $I_0 = 10$ A, $\cos\phi_0 = 0.3$ lagging, $V_1 = 400$ V, $f = 50$ Hz, $N_1 = 500$.

$$\sin\phi_0 = \sqrt{1 - 0.3^2} = 0.9539$$

(a) Components of no-load current

Working (core-loss) component, in phase with $V_1$:

$$I_w = I_0\cos\phi_0 = 10 \times 0.3 = 3\ \text{A}$$

Magnetizing component, lagging $V_1$ by 90°:

$$I_\mu = I_0\sin\phi_0 = 10 \times 0.9539 = 9.54\ \text{A}$$

Check: $\sqrt{3^2 + 9.54^2} = 10$ A.

   V1 ^
      |
 Iw=3 +- - - - - - - - - - - - -+  I0 = 10 A
      |               ___...--'
      |      ___...--'
      | _..-'   phi0 = 72.5 deg (between V1 and I0)
      O-------------------------+----> flux
                          Im = 9.54 A (along flux)

(b) Iron loss

At no load the input power is almost entirely iron loss:

$$P_i = V_1I_0\cos\phi_0 = 400 \times 10 \times 0.3 = 1200\ \text{W}$$

(c) Maximum and rms flux in the core

From the EMF equation (taking $E_1 \approx V_1$ at no load):

$$\begin{aligned} E_1 &= 4.44\,f\,\phi_m N_1 \\ \phi_m &= \frac{400}{4.44 \times 50 \times 500} = 3.604 \times 10^{-3}\ \text{Wb} = 3.604\ \text{mWb} \end{aligned}$$

For sinusoidal flux:

$$\phi_{rms} = \frac{\phi_m}{\sqrt2} = \frac{3.604}{1.4142} = 2.548\ \text{mWb}$$

Answer: (a) $I_w = 3$ A, $I_\mu = 9.54$ A; (b) iron loss = 1200 W; (c) $\phi_m = 3.604$ mWb, $\phi_{rms} = 2.548$ mWb.