An industrial load is modeled as a series combination of a capacitance and a resistance as shown in Fig. 9.89. Calculate the value of an inductance L across the series combination so that the net impedance is resistive at a frequency of 50 kHz.

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[latex]\begin{array}{l}{Z}_{\text {in }}=\mathrm{j} \omega \mathrm{L} \|\left(\mathrm{R}+\frac{1}{\mathrm{j} \omega \mathrm{C}}\right) \\{Z}_{\text {in }}=\frac{\mathrm{j} \omega \mathrm{L}\left(\mathrm{R}+\frac{1}{\mathrm{j} \omega \mathrm{C}}\right)}{\mathrm{R}+\mathrm{j} \omega \mathrm{L}+\frac{1}{\mathrm{j} \omega \mathrm{C}}}=\frac{\frac{\mathrm{L}}{\mathrm{C}}+\mathrm{j} \omega \mathrm{L} \mathrm{R}}{\mathrm{R}+\mathrm{j}\left(\omega \mathrm{L}-\frac{1}{\omega \mathrm{C}}\right)} \\{Z}_{\text {in }}=\frac{\left(\frac{\mathrm{L}}{\mathrm{C}}+\mathrm{j} \omega \mathrm{L} \mathrm{R}\right)\left(\mathrm{R}-\mathrm{j}\left(\omega \mathrm{L}-\frac{1}{\omega \mathrm{C}}\right)\right)}{\mathrm{R}^{2}+\left(\omega \mathrm{L}-\frac{1}{\omega \mathrm{C}}\right)^{2}}\\end{array}[/latex]

To have a resistive impedance, [latex]\operatorname{Im}\left({Z}_{\text {in }}\right)=0[/latex]. Hence,

[latex]\begin{array}{l}\omega \mathrm{L} \mathrm{R}^{2}-\left(\frac{\mathrm{L}}{\mathrm{C}}\right)\left(\omega \mathrm{L}-\frac{1}{\omega \mathrm{C}}\right)=0 \\\omega \mathrm{R}^{2} \mathrm{C}=\omega \mathrm{L}-\frac{1}{\omega \mathrm{C}} \\\omega^{2} \mathrm{R}^{2} \mathrm{C}^{2}=\omega^{2} \mathrm{LC}-1 \\\mathrm{L}=\frac{\omega^{2} \mathrm{R}^{2} \mathrm{C}^{2}+1}{\omega^{2} \mathrm{C}}\\end{array}[/latex]

Now we can solve for L.

[latex]\mathrm{L}=\mathrm{R}^{2} \mathrm{C}+1 /\left(\omega^{2} \mathrm{C}\right)\\[/latex]

[latex]\begin{array}{l}=\left(200^{2}\right)\left(50 \times 10^{-9}\right)+1 /\left((2 \pi \times 50,000)^{2}\left(50 \times 10^{-9}\right)\right. \\=2 \times 10^{-3}+0.2026 \times 10^{-3}={2.203 \mathrm{mH}}\\end{array}[/latex]

Checking, converting the series resistor and capacitor into a parallel combination, gives 220.3Ω in parallel with -j691.9Ω. The value of the parallel inductance is ωL =2πx50,000x2.203x[latex]10^{–3}[/latex] = 692.1Ω which we need to have if we are to cancel the effect of the capacitance. The answer checks.

Question 9.89: An industrial load is modeled as a series combination of a c...

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