Question B.13: Finding the parallel resonant frequency The parameters of th...

(a) From Eq. (B.53),

f_{ n }=\frac{1}{2 \pi \sqrt{L C}}                                  (B.53)

f_{ n }=\frac{1}{2 \pi \times \sqrt{5 \times 10^{-3} \times 50 \times 10^{-12}}}=318.3 kHz

From Eq. (B.63),

f_p=f_{ n }\left[1-\frac{C R_{ C 1}^{2}}{L}\right]^{1 / 2}                               (B.63)

f_{ p }=318.3 \times 10^{3} \times\left[1-50 \times 10^{-12} \times \frac{47^{2}}{5 \times 10^{-3}}\right]^{1 / 2} \approx 318.3 kHz

(b) We know that

X_{ L }=2 \pi f_{ p } L=2 \pi \times 318.3 \times 10^{3} \times 5 \times 10^{-3}=9999.7 \Omega

From Eq. (B.59), the effective resistance R_{ p } of the parallel circuit is

R_{ p }=\frac{R_{ C 1}^{2}+X_{ L }^{2}}{R_{ C 1}}                                       (B.59)

R_{ p }=\frac{47^{2}+9999.7^{2}}{47}=2127.6 k \Omega

Using the current divider rule, the rms current I_{ p } through the parallel circuit is

\begin{aligned}&I_{ p }=\frac{R_{ S }}{R_{ S }+R_{ p }} I_{ s }=\frac{20 k \Omega \times 6 mA }{20 k \Omega+2127.6 k \Omega}=55.876 \mu A \\&V_{ p }=I_{ p } R_{ p }=55.876 \mu A \times 2127.6 k \Omega=118.88 V\end{aligned}

(c) We have

Q_{ C 1}=\frac{X_{ L }}{R_{ C 1}}=\frac{9999.7}{47}=212.8

(d) From Eq. (B.60), the effective inductive reactance X_{ p } of the parallel circuit is

X_{ p }=\frac{R_{ C 1}^{2}+X_{ L }^{2}}{X_{ L }}                                    (B.60)

\begin{aligned}&X_{ p }=\frac{47^{2}+9999.7^{2}}{9999.7}=9999.92 \Omega \\&R_{ S } \| R_{ p }=\frac{20 \times 2127.6}{20+2127.6}=19.81 k \Omega\end{aligned}

From Eq. (B.64), Q_{ p }=19,810 / 9999.92=1.98.

Q_{ p }=\frac{V_{ p }^{2} / X_{ p }}{V_{ p }^{2} /\left(R_{S} \| R_{ p }\right)}=\frac{R_{ S } \| R_{ p }}{X_{ p }}                                      (B.64)