In a single phase transformer, the total iron loss is 2500 W at nominal voltage of 440 V and frequency 50 Hz. The total iron loss is 850 W at 220 V and 25 Hz. Then, at nominal voltage and frequency, the hysteresis loss and eddy current loss respectively are

The correct option is 900 W and 1600 W.

Step-by-step explanation:

The total iron loss ($W_i$) in a transformer is the sum of hysteresis loss ($W_h$) and eddy current loss ($W_e$):

$W_i=W_h+W_e$

The relationships of these losses with voltage ($V$) and frequency ($f$) depend on the ratio $\frac{V}{f}$, which is proportional to the maximum flux density ($B_m$):

$B_m\propto \frac{V}{f}$

Let us check the ratio of voltage to frequency for both given conditions:
Case 1: Nominal voltage $V_1=440\text{V}$ and frequency $f_1=50\text{Hz}$.
Ratio: $\frac{V_1}{f_1}=\frac{440}{50}=8.8$

Case 2: Voltage $V_2=220\text{V}$ and frequency $f_2=25\text{Hz}$.
Ratio: $\frac{V_2}{f_2}=\frac{220}{25}=8.8$

Since the ratio $\frac{V}{f}$ is constant in both cases, the maximum flux density $B_m$ remains constant. Under this condition, the hysteresis loss is directly proportional to frequency, and the eddy current loss is proportional to the square of the frequency:
Hysteresis Loss: $W_h=A·f$
Eddy Current Loss: $W_e=B·f^2$
where $A$ and $B$ are constants.

Therefore, the total iron loss can be written as:

$W_i=Af+B{f}^2$

Dividing both sides by $f$ yields:

$\frac{W_i}{f}=A+Bf$

Now, we substitute the values of both cases into this equation to solve for the constants $A$ and $B$:

For Case 1 ($W_{i1}=2500\text{W}$ and $f_1=50\text{Hz}$):
$\frac{2500}{50}=A+50B$
$\Rightarrow 50=A+50B$ (Equation 1)

For Case 2 ($W_{i2}=850\text{W}$ and $f_2=25\text{Hz}$):
$\frac{850}{25}=A+25B$
$\Rightarrow 34=A+25B$ (Equation 2)

Subtracting Equation 2 from Equation 1:

$(A+50B)-(A+25B)=50-34$

$\Rightarrow 25B=16$

$\Rightarrow B=\frac{16}{25}=0.64$

Substituting $B=0.64$ back into Equation 2:

$34=A+25\left(0.64\right)$

$\Rightarrow 34=A+16$

$\Rightarrow A=34-16=18$

Finally, we calculate the individual losses at the nominal voltage and frequency ($f=50\text{Hz}$):

Hysteresis loss ($W_h$):
$W_h=A·f=18·50=900\text{W}$

Eddy current loss ($W_e$):
$W_e=B·f^2=0.64·{50}^2=0.64·2500=1600\text{W}$

Therefore, the hysteresis loss and eddy current loss are 900 W and 1600 W respectively.

Note: The attached image displays an AC circuit with a source frequency parameter of ω = 1 k rad/sec, a 4 mH inductor, a 0.5 mF capacitor, a 2 Ω resistor, and a load resistor R_L. This diagram is unrelated to the single-phase transformer losses problem, which has been solved directly using the parameters specified in the question text.