Why do we still need to calculate the equivalent exciting current $I_{eq}$ when we have the exciting current $I_F$ already?

Here is the compound DC machine schematic:

The equivalent exciting current is $$\I_{eq}=I_F+\frac{N_{SE}}{N_{SH}}I_A-\frac{F_{AR}}{N_{SH}}\$$

• $$\F_{AR}\$$: the equivalent magnetomotive force of armature-reaction
• $$\N_{SE}\$$: the number of turns of $$\L_S\$$
• $$\N_{SH}\$$: the number of turns of $$\L_F\$$
1. How can I prove this equation? Apparently, I can't prove it by using KVL or KCL.
2. Why do we still need to calculate the equivalent exciting current $$\I_{eq}\$$ when we have the exciting current $$\I_F\$$ already and don't have other equivalent schematic here? Where is the $$\I_{eq}\$$ in this schematic?

The book just tells me the formula of equivalent exciting current $$\I_{eq}\$$, but it doesn't tell me why should we need this. What is the direction of this equivalent exciting current in the schematic?

Why do we still need to calculate the equivalent exciting current $$\I_{eq}\$$ when we have the exciting current $$\I_F\$$
It's a compound wound DC machine and, the field is produced by two windings; one in series ($$\L_S\$$) and one in parallel ($$\L_F\$$). Hence, $$\I_F\$$ is only part of the excitation current story.
There are two currents; $$\I_A\$$ and $$\I_F\$$. They are indicated on your diagram. Both these currents excite the machine. Given the dots on the excitation inductors, you can see that they will be in-phase with each other.
• so the exciting current is actually the current who go through that two windings? I thought the $I_A$ is the armature current,so it won't be part of the equivalent exciting current Commented Mar 8, 2023 at 23:28