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Exercise E1 Component Parameters
(written test, approx. 10 % of a 120-minute written test, SS2021)

The equivalent circuit diagram of an electric motor represents a resistive-inductive load.
The values of the series resistance $R_{\rm M}$ and the inductance $L_{\rm M}$ are to be determined below. Both result in the impedance of the motor.

a) Derive in general the equation for the absolute value of the impedance of the motor.

Path

The complex impedance $\underline{Z}$ for a resistive-inductive load (=$R$-$L$ series circuit) is given as \begin{align*} \underline{Z} &= {\rm j} \cdot X_L + R_{\rm M} \\ &= {\rm j} \cdot 2\pi \cdot f \cdot L_{\rm M} + R_{\rm M} \\ \end{align*}

The Pythagorean theorem can derive the absolute value: \begin{align*} |\underline{Z}|&= \sqrt{ (2\pi \cdot f \cdot L_{\rm M})^2 + R_{\rm M}^2 }\\ \end{align*}

Result

\begin{align*} Z = \sqrt{ (2\pi \cdot f \cdot L_{\rm M})^2 + R_{\rm M}^2 } \end{align*}


For the next exercises consider the following:
In two measurements, an AC voltage with a constant RMS value of $U = U_1 = U_2 = 50 ~\rm V$ but two different frequencies, $f_1$ and $f_2$ was applied.
This resulted in the recorded current of

  • for $f_1 = 50 ~\rm Hz$: $I_1 = 8~\rm A$
  • for $f_2 = 100~\rm Hz$: $I_2 = 5~\rm A$

b) Determine the absolute values of the impedances from the specified RMS values at $f_1$ and $f_2$ (independent).

Path

The absolute value of the impedance is given as $Z = {{U}\over{I}}$.
This leads to:

  • $Z_1 = {{U_1}\over{I_1}} = {{50 ~\rm V}\over{8~\rm A}} $
  • $Z_2 = {{U_2}\over{I_2}} = {{50 ~\rm V}\over{5~\rm A}}$

Result

  • $Z_1 = 6.25~\rm \Omega$
  • $Z_2 = 10~\rm \Omega$

c) Determine the component parameters $R_{\rm M}$ and $L_{\rm M}$ from a) and b)! (hard).

Path

Since we have $Z_1$ and $Z_2$ from b) we can subtract two of the formulas from a).
This has the advantage that $R_{\rm M}$ will cancel out: \begin{align*} Z_2^2 - Z_1^2 &= (2\pi \cdot f_2 \cdot L_{\rm M})^2 + R_{\rm M}^2 - \left( (2\pi \cdot f_1 \cdot L_{\rm M})^2 + R_{\rm M}^2 \right) \\ &= (2\pi \cdot f_2 )^2 \cdot L_{\rm M}^2 - (2\pi \cdot f_1)^2 \cdot L_{\rm M}^2 \\ \end{align*}

Now we can rearrange to $L_{\rm M}^2$:

\begin{align*} Z_2^2 - Z_1^2 &= L_{\rm M}^2 \cdot \left( (2\pi \cdot f_2 )^2 - (2\pi \cdot f_1)^2 \right) \\ L_{\rm M}^2 &= {{Z_2^2 - Z_1^2} \over { (2\pi \cdot f_2 )^2 - (2\pi \cdot f_1)^2 }} \\ L_{\rm M}^2 &= {{Z_2^2 - Z_1^2} \over { (2\pi)^2 \cdot ( f_2^2 - f_1^2 ) }} \\ \end{align*}

And then to $L_{\rm M}$:

\begin{align*} L_{\rm M} &={{1}\over{2\pi}} \sqrt{{{Z_2^2 - Z_1^2} \over { f_2^2 - f_1^2 }} }\\ \end{align*}

With the values:

\begin{align*} L_{\rm M} &={{1}\over{2\pi}} \sqrt{{{(10~\Omega)^2 - (6.25~\Omega)^2} \over { (100 {{1}\over{s}})^2 - (50 {{1}\over{s}})^2 }} }\\ &=14.346... ~\rm mH\\ \end{align*}

The resistance value $R_{\rm M}$ can be derived from \begin{align*} Z_2^2 &= (2\pi \cdot f_2 \cdot L_{\rm M})^2 + R_{\rm M}^2 \\ R_{\rm M}^2 &= Z_2^2 - (2\pi \cdot f_2 \cdot L_{\rm M})^2 \\ R_{\rm M} &=\sqrt{ Z_2^2 - (2\pi \cdot f_2 \cdot L_{\rm M})^2}\\ \end{align*}

The values have to be inserted also for $R_{\rm M}$: \begin{align*} R_{\rm M} &=\sqrt{ (10~\rm \Omega)^2 - (2\pi \cdot 100 {{1}\over{s}} \cdot 0.014346... ~\rm H)^2}\\ &= 4.3301...~\Omega \end{align*}

Result

  • $R_{\rm M} = 4.33 ~\Omega$
  • $L_{\rm M} = 14.35 ~\rm mH$