A series audio circuit is shown in Fig. 9.88.
(a) What is the impedance of the circuit?
(b) If the frequency were halved, what would be the impedance of the circuit?
A series audio circuit is shown in Fig. 9.88.
(a) What is the impedance of the circuit?
(b) If the frequency were halved, what would be the impedance of the circuit?
(a) {Z}=-\mathrm{j} 20+\mathrm{j} 30+120-\mathrm{j} 20\\
\mathrm{Z}={120-\mathrm{j} 10 \Omega}\\(b) If the frequency were halved, \frac{1}{\omega C}=\frac{1}{2 \pi f C} would cause the capacitive
impedance to double, while \omega \mathrm{L}=2 \pi \mathrm{f} \mathrm{L} would cause the inductive impedance to halve. Thus,