Examples of Parallel Resonance

Electrical Circuit Analysis > Resonance

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Example 1: Given the parallel network of [Fig. 1] composed of ideal elements:
a. Determine the resonant frequency fp.
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 VC at resonance.
e. Determine the currents IL and IC at resonance.

Fig. 1: Example 1.

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

$$f_p = f_s = { 1 \over 2 \pi \sqrt{LC}}\\ = { 1 \over 2 \pi \sqrt{(1 mH)(1 \mu F)}} = 5.03 kHz$$

b. For the parallel reactive elements:

$$ Z_L || Z_C = {jX_L)(jX_C) \over j(X_L -X_C)}$$

but $X_L = X_C$ 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,

$$Z_{T_{p}} = R_s || Z_L || Z_C = R_s = 10 k Ω$$

$$ Q_p = { R_s \over X_{L_{p}}} = { R_s \over 2 \pi f_p L}\\ = { 10 k Ω \over 2 \pi (5.03 kHz) (1 mH)} = 316.41$$

$$ BW = {f_p \over Q_p} = {5.03 kHz \over 316.41} = 15.90 Hz$$

$$ f_1 = { 1 \over 4 \pi C}[ { 1\over R} - \sqrt{{1 \over R^2} + { 4C \over L}}]\\ = { 1 \over 4 \pi (1 \mu F)}[ { 1\over 10k} - \sqrt{{1 \over (10k)^2} + { 4(1 \mu F) \over (1 mH)}}]\\ = 5.025 kHz$$

$$ f_2 = { 1 \over 4 \pi C}[ { 1\over R} + \sqrt{{1 \over R^2} + { 4C \over L}}]\\ = { 1 \over 4 \pi (1 \mu F)}[ { 1\over 10k} + \sqrt{{1 \over (10k)^2} + { 4(1 \mu F) \over (1 mH)}}]\\ = 5.041 kHz$$

$$ V_C = I Z_{T_{p}} = ( 10mA)(10kΩ) = 100V$$

$$ I_L = { V_L \over X_L} = {V_C \over 2 \pi f_p L}\\ = {100 \over 2 \pi (5.03 kHz) (1 mH)} = { 100V \over 31.6Ω} = 3.16 A$$

$$ I_C = { V_C \over X_C} = { 100V \over 31.6 Ω} = 3.16A$$

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