Find the value of C in the circuit shown in Fig. P8.21 so that Z is purely resistive at a frequency of 60 Hz.
Figure P8.21
Refer to Figure \(\mathrm{P} 8.16\) in the textbook.
Determine the impedance of inductor.
$$ \begin{aligned} Z_{L} &=j \omega L \\ &=j(2 \pi f) L \end{aligned} $$
Substitute \(60 \mathrm{~Hz}\) for \(f, \quad 5 \times 10^{-3} \mathrm{H}\) for \(L\).
$$ \begin{aligned} Z_{L} &=j(2 \pi(60))\left(5 \times 10^{-3} \mathrm{H}\right) \\ &=j 1.885 \Omega \end{aligned} $$
Determine the impedance of capacitor.
$$ \begin{aligned} Z_{C} &=\frac{1}{j \omega C} \\ &=\frac{1}{j(2 \pi f) C} \end{aligned} $$
Substitute \(60 \mathrm{~Hz}\) for \(f\).
$$ \begin{aligned} Z_{C} &=\frac{1}{j(2 \pi(60)) C} \\ &=\frac{2.65 \times 10^{-3}}{j C} \Omega \end{aligned} $$
Consider that the impedance in the circuit is purely resistive. Therefore,
$$ \begin{array}{l} Z_{L}+Z_{C}=0 \\ j 1.885+\frac{2.65 \times 10^{-3}}{j C}=0 \\ \frac{2.65 \times 10^{-3}}{j C}=-j 1.885 \end{array} $$
Find the value of capacitance, \(C\).
$$ \begin{aligned} C &=\frac{2.65 \times 10^{-3}}{j(-j 1.885)} \\ &=\frac{2.65 \times 10^{-3}}{1.885} \\ &=1.4 \times 10^{-3} \mathrm{~F} \\ &=1.4 \mathrm{mF} \end{aligned} $$
Therefore, the value of capacitance, \(C\) is, \(1.4 \mathrm{mF}\).