Given the parallel network in Fig. 21.34 composed of “ideal” elements:

a. Determine the resonant frequency $f_p$ .

b. Find the total impedance at resonance.

c. Calculate the quality factor, bandwidth, and cutoff frequencies $f_1 and f_2$ of the system.

d. Find the voltage $V_C$ at resonance.

e. Determine the currents $I_L and I_C$ at resonance.

Question

Given the parallel network in Fig. 21.34 composed of “ideal” elements:

a. Determine the resonant frequency [latex]f_p[/latex].

b. Find the total impedance at resonance.

c. Calculate the quality factor, bandwidth, and cutoff frequencies [latex]f_1 and f_2[/latex] of the system.

d. Find the voltage [latex]V_C[/latex] at resonance.

e. Determine the currents [latex]I_L and I_C[/latex] at resonance.

Accepted Answer

a. The fact that [latex]R_l[/latex] is zero ohms results in a very high [latex]Q_l(= X_L/R_l )[/latex], permitting the use of the following equation for [latex]f_p[/latex]:

[latex]f_p=f_s=\frac{1}{2\pi\sqrt{LC} } =\frac{1}{2\pi\sqrt{(1mH)1\mu F)} }=5.03 kHz [/latex]

b. For the parallel reactive elements,

[latex]Z_L\parallel Z_C=\frac{(X_L ∠90°)(X_C ∠-90°)}{+j(X_L - X_C)} [/latex]

but [latex]X_L = X_C[/latex] at resonance, resulting in a zero in the denominator of the equation and a very high impedance that can be approximated by an open circuit. Therefore,

[latex]Z_{T_p}=R_s\parallel Z_L\parallel Z_C=R_s = 10 k\Omega[/latex]

c. [latex]Q_p=\frac{R_s}{X_{L_p}}=\frac{R_s}{2\pi f_pL}=\frac{10 kΩ}{2\pi(5.03 kHz)(1 mH)}=316.41\[0.5cm] BW=\frac{f_p}{Q_p}=\frac{5.03 kHz}{316.41 }=15.90 Hz [/latex]

Eq. (21.41):

[latex]f_1=\frac{1}{4\pi C}\left[-\frac{1}{R}+\sqrt{\frac{1}{R^2 }+\frac{4C}{L} } \right] \[0.5cm] =\frac{1}{4\pi (1\mu F)}\left[\frac{1}{10k\Omega}-\sqrt{\frac{1}{(10k\Omega)^2 }+\frac{4(1\mu F)}{1mH} } \right]=5.025 kHz [/latex]

Eq. (21.42):

[latex]f_2=\frac{1}{4\pi C}\left[\frac{1}{R}+\sqrt{\frac{1}{R^2}+\frac{4C}{L} } \right]=5.041 kHz [/latex]

d. [latex]V_C=IZ_{T_p}=(10 mA)(10 kΩ) = 100 V[/latex]

e. [latex]I_L=\frac{V_L}{X_L}=\frac{V_C}{2\pi f_pL} =\frac{100 V}{2\pi (5.03 kHz)(1 mH)}=\frac{100 V}{31.6 Ω}=3.16 A\[0.5cm] I_C=\frac{V_C}{X_C}=\frac{100 V}{31.6 Ω}=3.16 A (= Q_pI) [/latex]

Question 21.6: Given the parallel network in Fig. 21.34 composed of “ideal”...

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