Question 9.90: An industrial coil is modeled as a series combination of an ...

An industrial coil is modeled as a series combination of an inductance L and resistance R, as shown in Fig. 9.90. Since an ac voltmeter measures only the magnitude of a sinusoid, the following measurements are taken at 60 Hz when the circuit operates in the steady state:

|V_{s}| = 145 V, |V_{1}| = 50 V, |V_{o}| = 110 V

Use these measurements to determine the values of L and R.

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\begin{aligned}&\text { Let } \quad {V}_{\mathrm{s}}=145 \angle 0^{\circ}, \quad \mathrm{X}=\omega \mathrm{L}=(2 \pi)(60) \mathrm{L}=377 \mathrm{L}\\\\&{I}=\frac{{V}_{\mathrm{s}}}{80+\mathrm{R}+\mathrm{j} \mathrm{X}}=\frac{145 \angle 0^{\circ}}{80+\mathrm{R}+\mathrm{j} \mathrm{X}}\\\\&{V}_{1}=80 {I}=\frac{(80)(145)}{80+\mathrm{R}+\mathrm{j} \mathrm{X}}\\\\&50=\left|\frac{(80)(145)}{80+\mathrm{R}+\mathrm{j} \mathrm{X}}\right|         (1)\\\\&{V}_{o}=(R+j X) {I}=\frac{(R+j X)\left(145 \angle 0^{\circ}\right)}{80+R+j X}\\\\&110=\left|\frac{(\mathrm{R}+\mathrm{j} \mathrm{X})(145)}{80+\mathrm{R}+\mathrm{j} \mathrm{X}}\right|         (2)\\\end{aligned}

From (1) and (2)

\begin{array}{l}\frac{50}{110}=\frac{80}{|\mathrm{R}+\mathrm{j} \mathrm{X}|} \\\\|\mathrm{R}+\mathrm{j} \mathrm{X}|=(80)\left(\frac{11}{5}\right) \\\\\mathrm{R}^{2}+\mathrm{X}^{2}=30976\end{array}

From (1)

|80+\mathrm{R}+\mathrm{j} \mathrm{X}|=\frac{(80)(145)}{50}=232\\\\ \begin{array}{l}6400+160 R+R^{2}+X^{2}=53824 \\\\160 R+R^{2}+X^{2}=47424\end{array}

Subtracting (3) from (4)

160 \mathrm{R}=16448 \longrightarrow \mathrm{R}={102.8} \Omega\\\\

From (3)

\begin{array}{l}X^{2}=30976-10568=20408 \\\\X=142.86=377 \mathrm{L} \longrightarrow \mathrm{L}=0.3789 \mathrm{H}\end{array}

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