Question 6.6: Parallel Resonant Circuit Find the Land C values for a paral...

First, we compute the quality factor of the circuit. Rearranging Equation 6.46 and substituting values, we have

B=\frac{f_0}{Q_p}         (6.46)

Q_p=\frac{f_0}{B}=\frac{10^6}{10^5}=10

Solving Equation 6.41 for the inductance and substituting values, we get

Q_p=\frac{R}{2\pi f_0 L}        (6.41)

L=\frac{R}{2\pi f_0 Q_p}=\frac{10^4}{2\pi \times 10^6 \times 10 }=159.2~\mu \mathrm{H}

Similarly, using Equation 6.42, we find that

Q_p=2\pi f_0 CR        (6.42)

C=\frac{Q_p}{2 \pi f_0 R}=\frac{10}{2\pi \times 10^6 \times 10^4}=159.2\mathrm{~pF}

At resonance, the voltage is given by

\mathrm{V_{out}}=\mathrm{I}R=(10^{-3}\angle 0^\circ)\times 10^4 =10\angle 0^\circ

and the currents are given by

\mathrm{I}_R=\frac{\mathrm{V_{out}}}{R}=\frac{10\angle 0^\circ}{10^4}=10^{-3}\angle 0^\circ

\mathrm{I}_L=\frac{\mathrm{V_{out}}}{j2\pi f_0 L}=\frac{10\angle 0^\circ}{j10^3}=10^{-2}\angle -90^\circ

\mathrm{I}_C=\frac{\mathrm{V_{out}}}{-j/2\pi f_0 C}=\frac{10\angle 0^\circ}{-10^3}=10^{-2}\angle 90^\circ

The phasor diagram is shown in Figure 6.31. Notice that the currents through the inductance and capacitance are larger in magnitude than the applied source current. However, since I_C and I_L are out of phase, they cancel.