Question 6.6: Series Resonant Circuit , Consider the series resonant circu...

First, we use Equation 6.30 to compute the resonant frequency:

f_0=\frac{1}{2 \pi \sqrt{LC}}=\frac{1}{2 \pi \sqrt{0.1592 \times 0.1592 \times 10^{-6}}} =1000 \text{ Hz}

The quality factor is given by Equation 6.31

Q_s=\frac{2 \pi f_0 L}{R}=\frac{2 \pi \times 1000 \times 0.1592}{100} =10

The bandwidth is given by Equation 6.35

B=\frac{f_0}{Q_s}=\frac{1000}{10}=100 \text{ Hz}

Next, we use Equations 6.36 and 6.37 to find the approximate half-power frequencies:

f_H\cong f_0+\frac{B}{2}=1000+\frac{100}{2}=1050 \text{ Hz} \\ f_L\cong f_0-\frac{B}{2}=1000-\frac{100}{2}=950 \text{ Hz}

At resonance, the impedance of the inductance and capacitance are

Z_L=j 2 \pi f_0L=j2 \pi \times 1000 \times 0.1592=j1000 \ \Omega \\ Z_C=-j\frac{1}{2 \pi f_0 C}=-j\frac{1}{2 \pi \times 1000 \times 0.1592 \times 10^{-6}}=-j1000 \ \Omega

As expected, the reactances are equal in magnitude at the resonant frequency. The total impedance of the circuit is

Z_s=R+Z_L+Z_C=100+j1000-j1000=100 \ \Omega

The phasor current is given by

\pmb{\text{I}}=\frac{\pmb{\text{V}}_s}{Z_s} =\frac{1 \underline{/0^\circ}}{100}=0.01 \underline{/0^\circ}

The voltages across the elements are

\pmb{\text{V}}_R=R \pmb{\text{I}}=100 \times 0.01 \underline{/0^\circ}=1 \underline{/0^\circ} \\ \pmb{\text{V}}_L=Z_L \pmb{\text{I}}=j1000 \times 0.01 \underline{/0^\circ}=10 \underline{/90^\circ} \\ \pmb{\text{V}}_C=Z_C \pmb{\text{I}}=-j1000 \times 0.01 \underline{/0^\circ}=10 \underline{/-90^\circ}

The phasor diagram is shown in Figure 6.28. Notice that the voltages across the inductance and capacitance are much larger than the source voltage in magnitude.
Nevertheless, Kirchhoff’s voltage law is satisfied because \pmb{\text{V}}_L \text{ and } \pmb{\text{V}}_C are out of phase and cancel.