Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

In an \(L-R-C\) series circuit, \(L=0.280 \mathrm{H}\) and \(C=\) \(4.00 \mu \mathrm{F}\). The voltage amplitude of the source is \(120 \mathrm{~V}\). (a) What is the resonance angular frequency of the circuit? (b) When the source operates at the resonance angular frequency, the current amplitude in the circuit is 1.70 A. What is the resistance \(R\) of the resistor? (c) At the resonance angular frequency, what are the peak voltages across the inductor, the capacitor, and the resistor?

Short Answer

Expert verified
(a) \(2998 \; \mathrm{rad/s}\); (b) \(70.59 \; \Omega\); (c) \(V_L = 1427 \; \mathrm{V}, V_C = 1427 \; \mathrm{V}, V_R = 120 \; \mathrm{V}\).

Step by step solution

01

Identify Known Variables

We are given that the inductance \( L = 0.280 \; \mathrm{H} \), the capacitance \( C = 4.00 \; \mu \mathrm{F} = 4.00 \times 10^{-6} \; \mathrm{F} \), and the voltage amplitude \( V_0 = 120 \; \mathrm{V} \). Furthermore, the current amplitude at resonance is \( I_0 = 1.70 \; \mathrm{A} \).
02

Calculate Resonance Angular Frequency

Use the formula for the resonance angular frequency: \( \omega_0 = \frac{1}{\sqrt{LC}} \).Substitute \( L = 0.280 \; \mathrm{H} \) and \( C = 4.00 \times 10^{-6} \; \mathrm{F} \) into the formula:\[\omega_0 = \frac{1}{\sqrt{0.280 \times 4.00 \times 10^{-6}}} = \frac{1}{\sqrt{1.12 \times 10^{-6}}} \approx 2998 \; \mathrm{rad/s}.\]
03

Determine Resistance at Resonance

At resonance, the impedance is equal to the resistance, \( R \). Use the relation \( V_0 = I_0 R \) to find \( R \):\[R = \frac{V_0}{I_0} = \frac{120}{1.70} \approx 70.59 \; \Omega.\]
04

Calculate Peak Voltage Across Inductor and Capacitor

At resonance, the peak voltage across the inductor \( V_L \) is given by \( V_L = I_0 \cdot \omega_0 \cdot L \) and across capacitor \( V_C \) by \( V_C = \frac{I_0}{\omega_0 \cdot C} \).For inductor:\[V_L = 1.70 \times 2998 \times 0.280 \approx 1427 \; \mathrm{V}.\]For capacitor:\[V_C = \frac{1.70}{2998 \times 4.00 \times 10^{-6}} \approx 1427 \; \mathrm{V}.\]
05

Calculate Peak Voltage Across Resistor

At resonance, the peak voltage across the resistor \( V_R \) is simply \( V_R = I_0 \cdot R \):\[V_R = 1.70 \times 70.59 \approx 120 \; \mathrm{V}.\]This matches the source voltage, confirming our calculations.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

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

Resonance Angular Frequency
In an LRC circuit, the resonance angular frequency, represented as \( \omega_0 \), is a crucial concept. It determines the frequency at which the circuit naturally prefers to oscillate. Understanding this enables us to predict the circuit's behavior under various conditions.
At resonance, the impedance of the circuit is minimized, typically resulting in the maximum possible current flow. The formula to calculate the resonance angular frequency is:
  • \( \omega_0 = \frac{1}{\sqrt{LC}} \)
where \( L \) is the inductance in henries (H) and \( C \) is the capacitance in farads (F).
Using our exercise values, \( L = 0.280 \, \mathrm{H} \) and \( C = 4.00 \times 10^{-6} \, \mathrm{F} \), we find \( \omega_0 \) by substituting the known values into the formula. After calculation, \( \omega_0 \approx 2998 \, \mathrm{rad/s} \).
This frequency is where the inductor and capacitor store and release energy maximally, leading to a significant current with minor resistance influence.
Impedance
Impedance in an LRC circuit, denoted as \( Z \), is a measure of the circuit's opposition to the current when voltage is applied. It is more than simple resistance as it considers the effects of inductance and capacitance.
At resonance, however, the impedance simplifies greatly.
  • In a series LRC circuit at resonance, the inductive reactance \( \omega L \) and capacitive reactance \( \frac{1}{\omega C} \) cancel each other out.
  • Thus, the impedance \( Z \) equals the resistance \( R \).
This simplification allows us to calculate \( R \) directly from the peak current and voltage values using the equation:
\[ R = \frac{V_0}{I_0} \]
where \( V_0 \) is the source's voltage amplitude and \( I_0 \) is the current amplitude. From the given exercise, using \( V_0 = 120 \, \mathrm{V} \) and \( I_0 = 1.70 \, \mathrm{A} \), we find that \( R \approx 70.59 \, \Omega \). This resistance is crucial as it dictates the circuit's energy loss at resonance.
Peak Voltage Calculation
In an LRC circuit at resonance, understanding the distribution of voltage across each component is vital. The peak voltage across the inductor \( V_L \), the capacitor \( V_C \), and the resistor \( V_R \) can be calculated as follows:
For the inductor:
  • \( V_L = I_0 \cdot \omega_0 \cdot L \)
Substituting the known values leads to \( V_L \approx 1427 \, \mathrm{V} \).
For the capacitor:
  • \( V_C = \frac{I_0}{\omega_0 \cdot C} \)
This also results in \( V_C \approx 1427 \, \mathrm{V} \).
The equality of \( V_L \) and \( V_C \) at resonance implies energy is perfectly exchanged between the inductor and capacitor.
Finally, for the resistor:
  • \( V_R = I_0 \cdot R \)
From calculations, \( V_R \approx 120 \, \mathrm{V} \), matching the source, which confirms our understanding of the voltage distribution at resonance. This helps recognize the separate roles each component plays in the circuit.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

An \(L-R-C\) series circuit has \(C\) = 4.80 \(\mu\)F, \(L\) = 0.520 H, and source voltage amplitude \(V\) = 56.0 V. The source is operated at the resonance frequency of the circuit. If the voltage across the capacitor has amplitude 80.0 V, what is the value of \(R\) for the resistor in the circuit?

Consider an \(L-R-C\) series circuit with a 1.80-H inductor, a 0.900-\(\mu\)F capacitor, and a 300-\(\Omega\) resistor. The source has terminal rms voltage V\(_{rms}\) = 60.0 V and variable angular frequency \(\omega\). (a) What is the resonance angular frequency \(\omega_0\) of the circuit? (b) What is the rms current through the circuit at resonance, I\(_{rms}\)-0? (c) For what two values of angular frequency, \(\omega\)1 and \(\omega\)2, is the rms current half the resonance value? (d) The quantity \(\omega\)1 - \(omega\)2 defines the resonance \(width\). Calculate I\(_{rms}\)-0 and the resonance width for R = 300 \(\Omega\), 30.0 \(\Omega\), and 3.00 \(\Omega\). Describe how your results compare to the discussion in Section 31.5. the

The signal from the oscillating electrode is fed into an amplifier, which reports the measured voltage as an rms value, 1.5 nV. What is the potential difference between the two extremes? (a) 1.5 nV; (b) 3.0 nV; (c) 2.1 nV; (d) 4.2 nV.

A transformer connected to a 120-V (rms) ac line is to supply 13,000 V (rms) for a neon sign. To reduce shock hazard, a fuse is to be inserted in the primary circuit; the fuse is to blow when the rms current in the secondary circuit exceeds 8.50 mA. (a) What is the ratio of secondary to primary turns of the transformer? (b) What power must be supplied to the transformer when the rms secondary current is 8.50 mA? (c) What current rating should the fuse in the primary circuit have?

A series circuit has an impedance of 60.0 \(\Omega\) and a power factor of 0.720 at 50.0 Hz. The source voltage lags the current. (a) What circuit element, an inductor or a capacitor, should be placed in series with the circuit to raise its power factor? (b) What size element will raise the power factor to unity?

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free