A single-phase, 4 kVA, 200 V/100 V, 50 Hz transformer with laminated CRGO steel core has rated noload loss of 450 W. When the high-voltage winding is excited with 160 V, 40 Hz sinusoidal ac supply, the no-load losses are found to be 320 W. When the high-voltage winding of the same transformer is supplied from a 100 V, 25 Hz sinusoidal ac source, the no-load losses will be __________ W (rounded off to 2 decimal places).

The correct answer is 162.5

Step 1: Understand the core loss equations for a transformer
The total no-load loss ($P_{\text{nl}}$) in a transformer core consists of hysteresis loss ($P_{\text{h}}$) and eddy current loss ($P_{\text{e}}$):

$P_{\text{nl}}=P_{\text{h}}+P_{\text{e}}$

The losses depend on the operating voltage ($V$) and frequency ($f$). The maximum magnetic flux density $B_{\text{m}}$ is proportional to $\frac{V}{f}$.

Notice the ratio of voltage to frequency for the given cases:

Case 1 (Rated):
$\frac{V_1}{f_1}=\frac{200}{50}=4$ V/Hz

Case 2:
$\frac{V_2}{f_2}=\frac{160}{40}=4$ V/Hz

Case 3:
$\frac{V_3}{f_3}=\frac{100}{25}=4$ V/Hz

Since $\frac{V}{f}$ is constant in all cases, the maximum flux density $B_{\text{m}}$ remains constant. When $\frac{V}{f}$ is constant, the loss components simplify to:

$P_{\text{h}}=A»f$

$P_{\text{e}}=B»f^2$

Thus, the total no-load loss can be expressed as:

$P_{\text{nl}}=Af+B{f}^2$

or, dividing both sides by $f$:

$\frac{P_{\text{nl}}}{f}=A+Bf$

Step 2: Set up equations using Case 1 and Case 2

From Case 1 ($V_1=200$ V, $f_1=50$ Hz, $P_{\text{nl}1}=450$ W):

$\frac{450}{50}=A+50B$

$9=A+50B$   —  (Equation 1)

From Case 2 ($V_2=160$ V, $f_2=40$ Hz, $P_{\text{nl}2}=320$ W):

$\frac{320}{40}=A+40B$

$8=A+40B$   —  (Equation 2)

Step 3: Solve for constants $A$ and $B$

Subtract Equation 2 from Equation 1:

$(9-8)=(A+50B)-(A+40B)$

$1=10B\Rightarrow B=0.1$

Substitute $B=0.1$ back into Equation 1:

$9=A+50\left(0.1\right)$

$9=A+5\Rightarrow A=4$

Step 4: Calculate the no-load loss for Case 3

For Case 3, $V_3=100$ V and $f_3=25$ Hz:

$P_{\text{nl}3}=A{f}_3+B{f}_3^2$

$P_{\text{nl}3}=4\left(25\right)+0.1\left({25}^2\right)$

$P_{\text{nl}3}=100+0.1\left(625\right)$

$P_{\text{nl}3}=100+62.5=162.5\text{W}$

Therefore, the no-load loss when supplied from a 100 V, 25 Hz source will be 162.5 W.