Question 6.7: Parallel Resonant Circuit , Find the L and C values for a pa...

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

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

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

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

Similarly, using Equation 6.42, we find that

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

At resonance, the voltage is given by

\pmb{\text{V}}_{\text{out}}=\pmb{\text{I}}R=(10^{-3} \underline{/0^\circ}) \times 10^4=10 \underline{/0^\circ} \text{ V}

and the currents are given by

\pmb{\text{I}}_R=\frac{\pmb{\text{V}}_{\text{out}}}{R} =\frac{10 \underline{/0^\circ}}{10^4}=10^{-3} \underline{/0^\circ}\text{ A}\\ \pmb{\text{I}}_L=\frac{\pmb{\text{V}}_{\text{out}}}{j2 \pi f_0L} =\frac{10 \underline{/0^\circ}}{j10^3} =10^{-2} \underline{/-90^\circ} \text{ A} \\ \pmb{\text{I}}_C=\frac{\pmb{\text{V}}_{\text{out}}}{-j/2 \pi f_0C} =\frac{10 \underline{/0^\circ}}{-j10^3} =10^{-2}\underline{/90^\circ} \text{ A}

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 \pmb{\text{I}}_C \text{ and }\pmb{\text{I}}_L are out of phase, they cancel.