Chapter 31, Problem 31.66P

DATA You are given this table of data recorded for a circuit that has a resistor, an inductor with negligible resistance, and a capacitor, all in series with an ac voltage source:

f(Hz)	80	160
z (Ω)	15	13
ϕ (°)	–71	67

Here f is the frequency of the voltage source, Z is the impedance of the circuit, and ϕ is the phase angle. (a) Use the data at both frequencies to calculate the resistance of the resistor. (Hint: Use the results of Exercise 31.21.) Calculate the average of these two values of the resistance, and use the result as the value of R in the rest of the analysis. (b) Use the data at 80 Hz and 160 Hz to calculate the inductance L and capacitance C of the circuit. (c) What is the resonance frequency for the circuit, and what are the impedance and phase angle at the resonance frequency?

31.66 IDENTITY and SET UP: For an L-R-C series circuit, $\text{tan}\varphi \text{=}\frac{\omega L-\frac{1}{\omega C}}{R}$ and the power factor is cosϕ = R/Z.

EXECUTE: (a) cosϕ = R/Z, so R = Z cosϕ.

At 80 Hz: R = (15 Ω) cos(−71°) = 4.88 Ω

At 160 Hz: R = (13 Ω) cos(67°) = 5.08 Ω

The average resistance is (4.88 Ω + 5.08 Ω)/2 = 5.0 Ω

(b) We use $\text{tan}\varphi \text{=}\frac{\omega L-\frac{1}{\omega C}}{R}$ with R = 5.0 Ω from part (a).

At 80 Hz: $\text{tan}\left(-{71}^{\circ }\right)\text{=}\frac{2\pi \left(80\text{ Hz}\right)L-\frac{1}{2\pi \left(80\text{ Hz}\right)C}}{5.0\Omega }$.

−14.52 = 160π Hz L − 1/[(160π Hz)C]. Eq. (1)

At 160 Hz: $\text{tan}\left({67}^{\circ }\right)\text{=}\frac{2\pi \left(160\text{ Hz}\right)L-\frac{1}{2\pi \left(160\text{ Hz}\right)C}}{5.0\Omega }$.

11.78 = 320π Hz L − 1/[(320π Hz)C]. Eq. (2)

Multiply Eq. (1) by −2 and add it to Eq. (2), giving 2(14.52) + 11.78 = (1/C)(1/80π − 1/320π).

C = 7.31 × 10⁻⁵ F, which rounds to C = 73 μF.

Substituting this result into either Eq. (1) or Eq. (2) gives L = 25.3 mH, which rounds to L = 25 mH.

(c) The resonance angular frequency is ${\omega }_0=\frac{1}{\sqrt{LC}}$, so the resonance frequency is

$f_0=\frac{{\omega }_0}{2\pi }=\frac{1}{2\pi \sqrt{LC}}=\frac{1}{2\pi \sqrt{\left(2.53\times {10}^{-2}\text{ H}\right)\left(73.1\times {10}^{-6}\text{ F}\right)}}=117\text{ Hz}$

At resonance, Z = R = 5.0 Ω and ϕ = 0.

EVALUATE: It is only at resonance that Z = R, not at the other frequencies.