{{tag>complex_voltage_divider RMS inductor exam_ee2_SS2021}}{{include_n>1250}}
 
#@TaskTitle_HTML@##@Lvl_HTML@#~~#@ee1_taskctr#~~ Component Parameters \\
<fs medium>(written test, approx. 10 % of a 120-minute written test, SS2021)</fs> #@TaskText_HTML@#

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. 

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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*}

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#@HiddenBegin_HTML~WJTtvMydRSkZhcIM_12,Result~@#
\begin{align*}
Z = \sqrt{ (2\pi \cdot f \cdot L_{\rm M})^2 + R_{\rm M}^2 }
\end{align*}
#@HiddenEnd_HTML~WJTtvMydRSkZhcIM_12,Result~@#

\\
__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). 

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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}} $
  * $Z_2 = {{U_2}\over{I_2}} = {{50 ~\rm V}\over{5~\rm A}}$

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#@HiddenBegin_HTML~WJTtvMydRSkZhcIM_22,Result~@#
  * $Z_1 = 6.25~\rm \Omega$
  * $Z_2 = 10~\rm \Omega$
#@HiddenEnd_HTML~WJTtvMydRSkZhcIM_22,Result~@#

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

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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*}
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#@HiddenBegin_HTML~WJTtvMydRSkZhcIM_32,Result~@#
  * $R_{\rm M}  = 4.33  ~\Omega$
  * $L_{\rm M}  = 14.35 ~\rm mH$
#@HiddenEnd_HTML~WJTtvMydRSkZhcIM_32,Result~@#

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