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Here is the compound DC machine schematic:

enter image description here

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?

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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.

What is the direction of this equivalent exciting current in the schematic?

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.

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  • \$\begingroup\$ 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 \$\endgroup\$ Commented Mar 8, 2023 at 23:28
  • \$\begingroup\$ It's a compound motor and armature current excites the series field winding @user16266657 \$\endgroup\$
    – Andy aka
    Commented Mar 8, 2023 at 23:46

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