Induction Motor Model: Equivalent Circuit Breakdown

Equivalent Circuit of an IM

2-winding Transformer:
- \(I_0\) is 1% of the F.L Current
- Reason: magnetic flux is confined completely in the steel core of low reluctance, hence \(I_\mu\) is small, as a result \(I_0\) is small
Induction Motor:
- presence of air-gap (high reluctance) necessitates a large \(I_\mu\), hence \(I_0\) is very large (40-50 % of F.L Current)
Although, stator and rotor operates at different frequency, but represented in the same vector diagram because magnetic field relative to them are synchronous with each other

\[I_{2}=s\dfrac{E_{2}}{\sqrt{R_{2}^{2}+\left(sX_{2}\right)^{2}}}=\dfrac{E_{2}}{\sqrt{\left(R_{2}/s\right)^{2}+X_{2}^{2}}}\]
When motor is loaded, rotor current is given by
From above equation, rotor circuit consists of
- a fixed \(R_2\) and variable \(sX_2\) (proportional to slip) connected across \(E_r = sE_2\), or
- Fixed reactance \(X_2\) connected in series with a variable \(R_2/s\) (inversely proportional to slip) and supplied with constant \(E_2\)

\[\begin{aligned} \dfrac{R_{2}}{s} & =R_{2}+R_{2}\left(\dfrac{1}{s}-1\right)\\ \Rightarrow R_{2} & =\mbox{represents the rotor Cu loss}\\ \Rightarrow R_{L}&=R_{2}\left(\dfrac{1}{s}-1\right) \\ & =\mbox{electrical equivalent of the mechanical load} \end{aligned}\]

The exciting current may be transferred to the left because inaccuracy involved is negligible and calculations is very much simplified

Induction Motor Model: Equivalent Circuit Breakdown

Demonstrative Video

Equivalent Circuit of an IM