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31.35. A series circuit consists of an ac source of variable frequency, a \(115-\Omega\) resistor, a \(1.25-\mu F\) capacitor, and a \(4.50-\mathrm{mH}\) inductor. Find the impedance of this circuit when the angular frequency of the ac source is adjusted to (a) the resonance angular frequency; (b) twice the resonance angular frequency; (c) half the resonance angular frequency.

Short Answer

Expert verified
(a) 115 Ω, (b) 292 Ω, (c) 1080 Ω.

Step by step solution

01

Calculate Resonance Angular Frequency

The resonance angular frequency \( \omega_0 \) for an RLC circuit is given by the formula:\[\omega_0 = \frac{1}{\sqrt{LC}}\]Given:\( L = 4.50 \times 10^{-3} \text{ H} \) and\( C = 1.25 \times 10^{-6} \text{ F} \).Substitute these values:\[\omega_0 = \frac{1}{\sqrt{(4.50 \times 10^{-3})(1.25 \times 10^{-6})}} \approx 1333.33 \text{ rad/s}\]
02

Calculate Impedance at Resonance Frequency

At resonance, the inductive reactance \( X_L = \omega_0 L \) and the capacitive reactance \( X_C = \frac{1}{\omega_0 C} \) cancel each other out, making the total impedance equal to the resistance:\[Z = R = 115 \text{ } \Omega\]
03

Calculate Impedance at Twice the Resonance Frequency

The angular frequency is now \( 2\omega_0 = 2 \times 1333.33 = 2666.66 \text{ rad/s} \).Calculate \( X_L \) and \( X_C \):\[X_L = (2666.66)(4.50 \times 10^{-3}) \approx 12 \text{ } \Omega\]\[X_C = \frac{1}{(2666.66)(1.25 \times 10^{-6})} \approx 300 \text{ } \Omega\]The impedance \( Z \) is given by:\[Z = \sqrt{R^2 + (X_L - X_C)^2} = \sqrt{115^2 + (12 - 300)^2} \approx 292 \text{ } \Omega\]
04

Calculate Impedance at Half the Resonance Frequency

Now the angular frequency is \( \frac{1}{2}\omega_0 = \frac{1}{2} \times 1333.33 = 666.67 \text{ rad/s} \).Calculate \( X_L \) and \( X_C \):\[X_L = (666.67)(4.50 \times 10^{-3}) \approx 3 \text{ } \Omega\]\[X_C = \frac{1}{(666.67)(1.25 \times 10^{-6})} \approx 1200 \text{ } \Omega\]The impedance \( Z \) is given by:\[Z = \sqrt{R^2 + (X_L - X_C)^2} = \sqrt{115^2 + (3 - 1200)^2} \approx 1080 \text{ } \Omega\]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Resonance Frequency
Resonance frequency is a special property of RLC circuits, where the inductive reactance and capacitive reactance are equal, allowing the circuit to oscillate naturally. At this frequency, these reactances cancel each other out. To find the resonance angular frequency, use the formula:
\[ \omega_0 = \frac{1}{\sqrt{LC}} \]
Here, \(L\) is the inductance, and \(C\) is the capacitance. For example, if you have an inductor of \(4.50\, \text{mH}\) and a capacitor of \(1.25\, \mu \text{F}\), the resonance angular frequency would be approximately \(1333.33\, \text{rad/s}\).

This concept is crucial because, at resonance frequency, the impedance of the circuit is at its minimum value, equaling the resistance \(R\). This leads to a maximum current flow through the circuit, which is particularly useful in applications like radio tuning or filtering specific signal frequencies.
Inductive Reactance
Inductive reactance occurs when a changing current flows through an inductor, causing it to resist changes in the current. It is symbolized by \(X_L\) and calculated using:
\[ X_L = \omega L \]
where \(\omega\) is the angular frequency and \(L\) is the inductance.

The inductive reactance increases with frequency. This means a higher frequency results in a bigger reactance, causing more opposition to the current flow through the inductor. If the circuit is operating at twice its resonance frequency, the inductive reactance was calculated as about \(12 \Omega\), as per the example problem.

Understanding this concept helps in designing circuits that respond differently to various frequencies, emphasizing or blocking specific signals as needed.
Capacitive Reactance
Capacitive reactance is the opposition that a capacitor presents to the change of voltage across its plates. It is denoted as \(X_C\) and calculated by:
\[ X_C = \frac{1}{\omega C} \]
where \(\omega\) is the angular frequency and \(C\) is the capacitance.

Capacitive reactance is inversely proportional to frequency. This means as the frequency increases, \(X_C\) decreases, offering less resistance to the current. From the exercise, at twice the resonance frequency, the capacitive reactance was found to be about \(300 \Omega\).

This characteristic is leveraged in applications such as signal coupling and filtering. Capacitors can selectively allow certain frequencies to pass through while blocking others, depending on the design of the circuit and the reactance values needed for each frequency.

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Most popular questions from this chapter

31.11. Kitchen Capacitance. The wiring for a refrigerator contains a starter capacitor. A voltage of amplitude 170 \(\mathrm{V}\) and frequency 60.0 \(\mathrm{Hz}\) applied across the capacitor is to produce a current amplitude of 0.850 A through the capacitor. What capacitance \(C\) is required?

31.41. A coil has a resistance of 48.0\(\Omega\) . At a frequency of 80.0 \(\mathrm{Hz}\) the voltage across the coil leads the current in it by \(52.3^{\circ} .\) Determine the inductance of the coil.

31.73. In an \(L-R-C\) series circuit the current is given by \(i=I \cos \omega t .\) The voltage amplitudes for the resistor, inductor, and capacitor are \(V_{R}, V_{L},\) and \(V_{C}\) (a) Show that the instantaneous power into the resistor is \(p_{R}=V_{R} I \cos ^{2} \omega t=\frac{1}{2} V_{R} I(1+\cos 2 \omega t) .\) What does this expression give for the average power into the resistor? (b) Show that the instantaneous power into the inductor is \(p_{L}=\) \(-V_{L}\) Isin \(\omega t \cos \omega t=-\frac{1}{2} V_{L} I \sin 2 \omega t .\) What does this expression give for the average power into the inductor?(c) Show that the instantaneous power into the capacitor is \(p_{C}=V_{C}\) Isin\omegat cos \(\omega t=\) \(\frac{1}{2} V_{C}\) Isin \(2 \omega t .\) What does this expression give for the average power into the capacitor? (d) The instantancous power delivered by the source is shown in Section 31.4 to be \(p=V I \cos \omega t\) (cos \(\phi \cos \omega t-\) sin \(\phi \sin \omega t ) .\) Show that \(p_{R}+p_{L}+p_{C}\) equals \(p\) at each instant of time.

31.14. You have a \(200-\Omega\) resistor, a \(0.400-\mathrm{H}\) inductor, and a \(6.00-\mu \mathrm{F}\) capacitor. Suppose you take the resistor and inductor and make a series circuit with a voltage source that has voltage amplitude 30.0 \(\mathrm{V}\) and an angular frequency of 250 \(\mathrm{rad} / \mathrm{s}\) . (a) What is the impedance of the circuit? (b) What is the current amplitude? (c) What are the voltage amplitudes across the resistor and across the inductor? (d) What is the phase angle \(\phi\) of the source voltage with respect to the current? Does the source voltage lag or lead the conrent? (e) Construct the phasor diagram.

31\. 27. (a) Show that for an \(L-R-C\) series circuit the power factor is equal to \(R / Z\) . (Hint: Use the phasor diagram; see Fig. \(31.13 b . )\) (b) Show that for any ac circuit, not just one containing pure resistance only, the average power delivered by the voltage source is given by \(P_{\mathrm{av}}=I_{\mathrm{rms}}^{2} R\)

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