Question 5.26: A transformer with a primary having R p = 100 Ω and Lp = 0.1......
A transformer with a primary having R_p = 100 \ Ω and L_p = 0.1 H and a secondary having R_s = 40 \ Ω and L_s = 0.4 \ H is connected between source voltage of 200 V at 159.2 Hz and a load resistance of 500 Ω. Determine the load current if k = 0.1.
The transformer connected between the source and load can be represented by the circuit shown in Fig. 1.
\text { Let, } \bar{I}_L=\text { Load current. }Primary inductive reactance, X_p=2 \pi f L_p=2 \times \pi \times 159.2 \times 0.1=100.0283 \Omega
\approx 100 \Omega
Secondary inductive reactance, \begin{aligned} & X_s=2 \pi fL _{ s }=2 \times \pi \times 159.2 \times 0.4=400.1132 \Omega \\ & \approx 400 \Omega \\ & \end{aligned}
Mutual inductance , M=k \sqrt{L_s L_p}=0.1 \sqrt{0.1 \times 0.4}=0.02 H
Mutual reactance, X_\text{m}=2 \pi f M=2 \times \pi \times 159.2 \times 0.02=20.0057 \Omega \approx 20 \Omega
The frequency domain representation of the transformer connected between the source and load is shown in Fig. 2. Let \bar{I}_1 \text { and } \bar{I}_2 be the mesh currents. Now the current enters at the dotted end in one coil and leaves at the dotted end in the other coil, and so the fluxes are opposing. Also, the mesh currents are in the same orientation. Hence, it is group-1 coupled coil. The electrical equivalent of the circuit of Fig. 2 is shown in Fig. 3.
With reference to Fig. 3, the mesh basis matrix equation is,
\left[\begin{array}{rr} 100+ j 80+ j 20 & – j 20 \\ – j 20 & j 20+ j 380+40+500 \end{array}\right]\left[\begin{array}{l} \overline{I}_1 \\ \overline{I}_2 \end{array}\right]=\left[\begin{array}{r} 200 \angle 0^{\circ} \\ 0 \end{array}\right]
\left[\begin{array}{rr}100+ j 100 & – j 20 \\- j 20 & 540+ j 400\end{array}\right]\left[\begin{array}{l} \bar{I}_1 \\\overline{ I }_2\end{array}\right]=\left[\begin{array}{r} 200 \\0\end{array}\right]
\text { Now, } \Delta=\left|\begin{array}{rr}100+ j 100 & – j 20 \\- j 20 & 540+ j 400\end{array}\right|=(100+ j 100) \times(540+ j 400)-(- j 20)^2
= 14400 + j94000
\Delta_2=\left|\begin{array}{rr}100+j 100 & 200 \\-j 20 & 0\end{array}\right|=0-(-j 20) \times 200=j 4000
\text { Load current, } \bar{I}_{ L }=\bar{I}_2=\frac{\Delta_2}{\Delta}=\frac{ j 4000}{14400+ j 94000}=0.0416+ j 0.0064 A
=0.0421 \angle 8.7^{\circ} A=42.1 \times 10^{-3} \angle 8.7^{\circ} A=42.1 \angle 8.7^{\circ} mA
