Electronic – Synchronous Generator EMI

The easiest way of solving such problems is to use complex phasors. The total complex impedance is

$$Z_T=R_1+R_2||j\omega L=R_1+\frac{j\omega R_2L}{R_2+j\omega L}=877.18\cdot e^{30.67^{\circ}}\Omega$$

with \$\omega=2\pi\cdot 10,000 Hz\$, \$R_1=470\Omega\$, \$R_2=988\Omega\$, and \$L=10mH\$ (as shown in your circuit diagram).

This gives for the total current

$$I_s=\frac{V_0}{Z_T}=9.12\cdot e^{-30.67^{\circ}} mA$$ with \$V_0=8V\$.

The voltage across \$R_1\$ is

$$V_1=I_sR_1=4.28\cdot e^{-30.67^{\circ}} V$$

The voltage across \$R_2\$ and \$L\$ is

$$V_2=I_s(R_2||j\omega L)=I_s\frac{j\omega R_2L}{R_2+j\omega L}=4.83\cdot e^{26.88^{\circ}} V\quad(=V_0-V_1)$$

The current through \$R_2\$ is

$$I_2=\frac{V_2}{R_2}=4.89\cdot e^{26.88^{\circ}} mA$$

The current through the inductor is

$$I_L=\frac{V_2}{j\omega L}=7.70\cdot e^{-63.12^{\circ}} mA\quad(=I_s-I_2)$$

I realize that there are quite a few differences between these results and your calculations and measurements. I just used the values you gave in the circuit diagram, and I did my best to avoid any errors.

When the synchoronous generator loaded, an angle δ occurs between stator magnetic field and rotor magnetic field. It is called "load angle". Hence the IA represents the stator magnetic fied and the BS is now perpendicular to this surface. Now the terminal voltage VΦ is related to Bnet, Bnet=BS+Br. You can observe exact load angle δ between VΦ and EA in your figure (d). Also the load angle δ is shown on voltage vector diagram between Ea and VΦ. Sorry my poor English. I hope i can explain it.