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31.33. In an \(L . R-C\) series circuit, \(R=150 \Omega, L=0.750 \mathrm{H},\) and \(C=0.0180 \mu \mathrm{F} .\) The source has voltage amplitude \(V=150 \mathrm{V}\) and a frequency equal to the resonance frequency of the circuit. (a) What is the power factor? (b) What is the average power delivered by the source? (c) The capacitor is replaced by one with \(C=0.0360 \mu \mathrm{F}\) and the source frequency is adjusted to the new resonance value. Then what is the average power delivered by the source?

Short Answer

Expert verified
(a) The power factor is 1. (b) Average power is 150 W. (c) Average power is still 150 W.

Step by step solution

01

Calculate Resonance Frequency

The resonance frequency of an L-R-C circuit is given by the formula \( f = \frac{1}{2\pi \sqrt{LC}} \). First, we need to use the given values of \( L = 0.750 \text{ H} \) and \( C = 0.0180 \times 10^{-6} \text{ F} \) to calculate it. Plugging in the values, we get: \[ f = \frac{1}{2\pi \sqrt{(0.750 \times 0.0180 \times 10^{-6})}} \]. Solving this gives the resonance frequency \( f \approx 1.374 \times 10^3 \text{ Hz} \).
02

Determine Power Factor at Resonance

At resonance, the inductive reactance \( X_L = 2\pi f L \) equals the capacitive reactance \( X_C = \frac{1}{2\pi f C} \), meaning the power factor is 1 because the reactive components cancel each other out.
03

Calculate Average Power Delivered (Part a)

The average power delivered in an L-R-C circuit at resonance can be calculated by \( P = \frac{V^2}{R} \) where \( V = 150 \text{ V} \) and \( R = 150 \Omega \). Substituting the given values, we get \[ P = \frac{150^2}{150} = 150 \text{ W} \].
04

Change in Capacitance and Frequency for New Resonance

When the capacitor is replaced by one with \( C = 0.0360 \times 10^{-6} \text{ F} \) and the source frequency is adjusted to the new resonance frequency, calculate the new resonance frequency using \( f' = \frac{1}{2\pi \sqrt{0.750 \times 0.0360 \times 10^{-6}}} \), which gives \( f' \approx 0.950 \times 10^3 \text{ Hz} \).
05

Calculate New Average Power Delivered (Part b)

Using the same formula \( P = \frac{V^2}{R} \) since the circuit will again be at resonance, thus the average power delivered by the source remains \( P = \frac{150^2}{150} = 150 \text{ W} \).

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Key Concepts

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

Resonance Frequency
In an RLC circuit, resonance frequency is a particular frequency where the circuit responds most efficiently to an external voltage. At this frequency, the inductive reactance and capacitive reactance are equal and cancel each other out. This results in the impedance being purely resistive, calculated only with the resistor value, which maximizes current flow.
To calculate the resonance frequency \( f \) in an RLC circuit, you use the formula:
  • \( f = \frac{1}{2\pi \sqrt{LC}} \)
Where \( L \) is inductance in henries and \( C \) is capacitance in farads.Plug in the circuit values \( L = 0.750 \text{ H} \) and \( C = 0.0180 \times 10^{-6} \text{ F} \), then calculate to find \( f \). This results in \( f \approx 1.374 \times 10^3 \text{ Hz} \) for the original setup. A change in capacitance and thus the recalculated frequency can impact circuit performance by altering the point of resonance, as we see when the capacitance changes to \( C = 0.0360 \mu \text{F} \), giving a new frequency of \( f' \approx 0.950 \times 10^3 \text{ Hz} \).
Understanding resonance is crucial because it determines the conditions under which the circuit operates most efficiently.
Power Factor
Power factor in an AC circuit describes the efficiency of power usage. It is defined as the cosine of the phase angle \( \phi \) between voltage and current. The formula is:
  • Power Factor (PF) = \( \cos \phi \)
In an RLC circuit, when the power factor is 1, the circuit is said to be at resonance. This means all power supplied by the source is effectively used. This ideal situation happens when inductive and capacitive reactances cancel, leading to voltage and current being perfectly in phase.
In our given exercise, since the power factor at resonance is calculated as 1, the circuit is using energy most effectively. This implies there's minimal reactive power (the power that bounces between source and reactive components without doing actual work), achieving maximum real power usage.
This optimal condition not only saves energy costs but also includes other benefits, like reduced heating and prolonged lifespan for the components used.
Average Power
Average power in an AC circuit signifies the power actually consumed or converted into work. For a series RLC circuit at resonance, average power \( P \) can be represented by:
  • \( P = \frac{V^2}{R} \)
Here, \( V \) is the voltage amplitude supplied by the source, and \( R \) is the resistance. Substitute \( V = 150 \text{ V} \) and \( R = 150 \Omega \) into the formula to find \( P \). This gives an average power output of 150 W.
Even when the circuit's capacitance changes, resulting in a new resonance frequency, the average power calculation remains unchanged if the circuit remains at resonance. Therefore, with a new capactior or frequency adjustment as seen here, the average power still equals 150 W.
Understanding average power is essential in designing efficient electrical systems as it directly affects the cost of electricity and engineering requirements of electrical systems.

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Most popular questions from this chapter

31\. A8. At a frequency \(\omega_{1}\) the reactance of a certain capacitor equals that of a certain inductor. (a) If the frequency is changed to \(\omega_{2}=2 \omega_{1},\) what is the ratio of the reactance of the inductor to that of the capacitor? Which reactance is larger? (b) If the frequency is changed to \(\omega_{3}=\omega_{1} / 3,\) what is the ratio of the reactance of the inductor to that of the capacitor? Which reactance is larger? (c) If the capacitor and inductor are placed in series with a resistor of resistance \(R\) to form a series \(L-R-C\) circuit? what will be the resonance angular frequency of the circuit?

31.32. An \(L-R-C\) series circuit consists of a source with voltage amplitude 120 \(\mathrm{V}\) and angular frequency 50.0 \(\mathrm{rad} / \mathrm{s}\) , a resistor with \(R=400 \Omega\) an inductor with \(L=9.00 \mathrm{H}\) , and a capacitor with capacitance \(C .\) (a) For what value of \(C\) will the current amplitude in the circuit be a maximum? (b) When \(C\) has the value calculated in part (a), what is the amplitude of the voltage across the inductor?

31.12. A \(250-\Omega\) resistor is connected in series with a \(4.80-\mu \mathrm{F}\) capacitor. The voltage across the capacitor is \(v_{c}=\) \((7.60 \mathrm{V}) \sin [(120 \mathrm{rad} / \mathrm{s}) t] .\) (a) Determine the capacitive reactance of the capacitor. (b) Derive an expression for the voltage \(v_{R}\) across the resistor.

3.1. The plate on the back nf a certain computer scanner says that the unit draws 0.34 \(\mathrm{A}\) of current from a \(120-\mathrm{V}, 60-\mathrm{Hz}\) line. Find (a) the root-mean-square current, (b) the current amplitude, (c) the average current; (d) the average square of the current.

31.28. An L-R-C series circuit is connected to a \(120-\mathrm{Hz}\) ac source that has \(V_{\mathrm{rms}}=80.0 \mathrm{V}\) . The circuit has a resistance of 75.0\(\Omega\) and an impedance at this frequency nf 105\(\Omega\) . What average power is delivered to the circuit by the source?

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