Question 6.5: Series Resonant Circuit Consider the series resonant circuit...
Series Resonant Circuit
Consider the series resonant circuit shown in Figure 6.27. Compute the resonant frequency, the bandwidth, and the half-power frequencies. Assuming that the frequency of the source is the same as the resonant frequency, find the phasor voltages across the elements and draw a phasor diagram.
First, we use Equation 6.30 to compute the resonant frequency:
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\mathrm{~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\mathrm{~Hz}
f_L\cong f_0-\frac{B}{2}=1000-\frac{100}{2}=950\mathrm{~Hz}
At resonance, the impedance of the inductance and capacitance are
Z_L=j2\pi f_0 L=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
\mathrm{I}=\frac{\mathrm{V}_s}{Z_s}=frac{1\angle 0^\circ}{100}=0.01\angle 0^\circ
The voltages across the elements are
\mathrm{V}_R=R\mathrm{I}=100\times 0.01\angle 0^\circ=1\angle 0^\circ
\mathrm{V}_L=Z_L\mathrm{I}=j1000\times 0.01\angle 0^\circ=10\angle 90^\circ
\mathrm{V}_C=Z_C \mathrm{I}=-j1000\times 0.01\angle 0^\circ=10\angle -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 VL and VC are out of phase and cancel.
