Question

When the winding \(c\!-\!d\) of the single-phase, 50 Hz, two-winding transformer is supplied from an AC current source of frequency 50 Hz, the rated voltage of 200 V (rms), 50 Hz is obtained at the open-circuited terminals \(a\!-\!b\). The cross-sectional area of the core is \(5000 \, \text{mm}^2\) and the average core length traversed by the mutual flux is 500 mm. The maximum allowable flux density in the core is \(B_{max} = 1 \, \text{Wb/m}^2\) and the relative permeability of the core material is 5000. The leakage impedance of the winding \(a\!-\!b\) and winding \(c\!-\!d\) at 50 Hz are \((5 + j100\pi \times 0.16) \, \Omega\) and \((11.25 + j100\pi \times 0.36) \, \Omega\), respectively. Considering the magnetizing characteristics to be linear and neglecting core loss, the self-inductance of the winding \(a\!-\!b\) in millihenry is .............. (Round off to 1 decimal place). \begin{center} \includegraphics[width=0.5\textwidth]{24.jpeg} \end{center}

Hint:

Always use the emf equation \(E = 4.44 f N \Phi_{max}\) to determine turns, then compute inductance using \(L = \mu N^2 A / l\).

Answer 1

Step 1: Voltage equation.
The rms voltage induced in winding \(a\!-\!b\): \[ V_{ab} = 4.44 f N_1 \Phi_{max} \] where \(N_1\) = turns of winding \(a\!-\!b\).

Step 2: Flux.
Given: \[ B_{max} = 1 \, \text{Wb/m}^2, A = 5000 \, \text{mm}^2 = 5 \times 10^{-3} \, \text{m}^2 \] \[ \Phi_{max} = B_{max} \cdot A = 1 \times 5 \times 10^{-3} = 0.005 \, \text{Wb} \]

Step 3: Turns calculation.
\[ 200 = 4.44 \times 50 \times N_1 \times 0.005 \] \[ 200 = 1.11 N_1 \Rightarrow N_1 = \frac{200}{1.11} \approx 180 \]

Step 4: Self-inductance.
Inductance: \[ L = \frac{\mu N^2 A}{l} \] where \(\mu = \mu_r \mu_0 = 5000 \times 4\pi \times 10^{-7} = 0.006283 \, \text{H/m}\). \[ L = \frac{0.006283 \times (180)^2 \times 5 \times 10^{-3}}{0.5} \] \[ L \approx 0.2037 \, H = 203.7 \, mH \]

Final Answer:
\[ \boxed{203.7 \, \text{mH}} \]