Figure 9.91 shows a parallel combination of an inductance and a resistance. If it is desired to connect a capacitor in series with the parallel combination such that the net impedance is resistive at 10 MHz, what is the required value of C?

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Accepted Answer

[latex]\begin{aligned}{Z}_{\text {in }} &=\frac{1}{j \omega C}+R \| \mathrm{j} \omega \mathrm{L} \\{Z}_{\text {in }} &=\frac{-\mathrm{j}}{\omega \mathrm{C}}+\frac{\mathrm{j} \omega \mathrm{LR}}{\mathrm{R}+\mathrm{j} \omega \mathrm{L}} \\&=\frac{-\mathrm{j}}{\omega \mathrm{C}}+\frac{\omega^{2} \mathrm{L}^{2} \mathrm{R}+\mathrm{j} \omega \mathrm{L} \mathrm{R}^{2}}{\mathrm{R}^{2}+\omega^{2} \mathrm{L}^{2}}\\end{aligned}[/latex]

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

[latex]\frac{-1}{\omega C}+\frac{\omega \mathrm{LR}^{2}}{\mathrm{R}^{2}+\omega^{2} \mathrm{L}^{2}}=0 \[/latex]

[latex]\frac{1}{\omega \mathrm{C}}=\frac{\omega \mathrm{LR}^{2}}{\mathrm{R}^{2}+\omega^{2} \mathrm{L}^{2}} \[/latex]

[latex]\mathrm{C}=\frac{\mathrm{R}^{2}+\omega^{2} \mathrm{L}^{2}}{\omega^{2} \mathrm{LR}^{2}} \[/latex]

[latex]\text { where } \omega=2 \pi \mathrm{f}=2 \pi \times 10^{7}\[/latex]

[latex]\mathrm{C}=\frac{9 \times 10^{4}+\left(4 \pi^{2} \times 10^{14}\right)\left(400 \times 10^{-12}\right)}{\left(4 \pi^{2} \times 10^{14}\right)\left(20 \times 10^{-6}\right)\left(9 \times 10^{4}\right)} \\[/latex]

[latex]\mathrm{C}=\frac{9+16 \pi^{2}}{72 \pi^{2}} \mathrm{nF} \\[/latex]

[latex]\mathrm{C}={235 \mathrm{p} \mathrm{F}}[/latex]

Question 9.91: Figure 9.91 shows a parallel combination of an inductance an...

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