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An \(L-R-C\) series circuit is constructed using a 175-\(\Omega\) resistor, a 12.5-\(\mu\)F capacitor, and an 8.00-mH inductor, all connected across an ac source having a variable frequency and a voltage amplitude of 25.0 V. (a) At what angular frequency will the impedance be smallest, and what is the impedance at this frequency? (b) At the angular frequency in part (a), what is the maximum current through the inductor? (c) At the angular frequency in part (a), find the potential difference across the ac source, the resistor, the capacitor, and the inductor at the instant that the current is equal to one-half its greatest positive value. (d) In part (c), how are the potential differences across the resistor, inductor, and capacitor related to the potential difference across the ac source?

Short Answer

Expert verified
(a) 3162 rad/s, 175 Ω. (b) 0.143 A. (c) Resistor: 12.51 V. Capacitor: -12.51 V. Inductor: +12.51 V. (d) V_ac = V_R.

Step by step solution

01

Identify Resonance Frequency

The impedance of an L-R-C circuit is minimal at its resonance angular frequency. The formula for the resonance frequency \( \omega_0 \) in a series L-R-C circuit is given by:\[\omega_0 = \frac{1}{\sqrt{LC}}\]Where:- \( L = 8.00 \times 10^{-3} \text{ H} \)- \( C = 12.5 \times 10^{-6} \text{ F} \)Substitute in the values:\[\omega_0 = \frac{1}{\sqrt{8.00 \times 10^{-3} \cdot 12.5 \times 10^{-6}}} \approx 3162 \text{ rad/s} \]
02

Calculate Minimum Impedance

At resonance, the impedance \( Z \) of the circuit is smallest and equals the resistance \( R \), because the inductive and capacitive reactances cancel each other out.Thus, the minimum impedance \( Z_{min} \) is:\[ Z_{min} = R = 175 \text{ } \Omega \]
03

Compute Maximum Current

The maximum current \( I_{max} \) through the inductor at resonance is obtained using Ohm's law:\[I_{max} = \frac{V_0}{Z}\]Given \( V_0 = 25.0 \text{ V} \) and \( Z = 175 \text{ } \Omega \):\[ I_{max} = \frac{25.0}{175} \approx 0.143 \text{ A} \]
04

Find Voltage Relationships at Specific Current

The potential differences at the instant when the current is half its maximum, \( I = 0.5 I_{max} \), are analyzed. Since:- \( I = 0.5 \times 0.143 \approx 0.0715 \text{ A} \)The voltage across each component is:- Resistor: \( V_R = I \times R = 0.0715 \times 175 \approx 12.51 \text{ V} \)- Capacitor: At resonance, voltage across the capacitor \( V_C \) is \( -V_R \), and hence is \(-12.51 \) V.- Inductor: At resonance, voltage across the inductor \( V_L \) is \( +12.51 \) V.
05

Relate Potential Differences Across Elements

At resonance, potential differences across inductor and capacitor directly oppose each other, and their magnitudes are equal, thus canceling each other out. Consequently, the potential difference across the ac source equals the potential across the resistor:\[ V_{ac} = V_R \approx 12.51 \text{ V} \]

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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, the resonance frequency holds incredible importance due to its capability to minimize impedance. It is the frequency at which the inductive and capacitive reactances cancel each other out, making the net reactive part of the impedance zero. At this resonance frequency, only the resistive component contributes to the impedance, meaning it is at its minimum.The formula for finding the resonance angular frequency \( \omega_0 \) is:\[\omega_0 = \frac{1}{\sqrt{LC}}\]where \( L \) is the inductance and \( C \) is the capacitance. For our circuit with an inductor of 8.00 mH and a capacitor of 12.5 µF, substituting these values into the formula gives a resonance angular frequency of approximately 3162 rad/s. This frequency is crucial for achieving minimal impedance and optimal functioning of the RLC circuit.
Impedance
In an RLC circuit, impedance is the total opposition to the flow of alternating current and is symbolized as \( Z \). This impedance is frequency-dependent, composed of resistance \( R \), inductive reactance \( X_L = \omega L \), and capacitive reactance \( X_C = \frac{1}{\omega C} \).At resonance frequency, the impedance in a series RLC circuit is simplified to purely resistive, as the inductive and capacitive reactances cancel each other out, making the reactive component zero. Therefore, the impedance \( Z \) at resonance is equal to the resistance \( R \) alone. In our specific example, this results in a minimum impedance value of 175 Ω. Recognizing this resonating condition in circuits can significantly impact the design and functionality in real-world applications, such as tuning radio frequencies.
Maximum Current
Maximum current in an RLC circuit occurs when the impedance is at its minimum, which is at the resonance frequency. At this point, the circuit behaves almost like a resistive circuit, allowing for maximum current flow. The relationship can be defined by Ohm's Law:\[I_{max} = \frac{V_0}{Z}\]where \( I_{max} \) is the maximum current, \( V_0 \) is the voltage amplitude, and \( Z \) is the circuit's impedance. For our circuit with a voltage amplitude of 25.0 V and impedance of 175 Ω, the maximum current is approximately 0.143 A. Understanding this peak current is integral for safe and efficient circuit design, preventing component overload and maximizing performance.
Voltage Relationship
Inside an RLC circuit, the voltage relationships vary significantly among different components due to the presence of reactance. When the current is at half its maximum value, the potential differences can be uniquely distributed.- **Resistor:** The voltage across a resistor \( V_R \) follows Ohm's Law, calculated as \( I \times R \). For a half maximum current, this results in approximately 12.51 V.- **Inductor and Capacitor:** At resonance, the voltages across the inductor \( V_L \) and capacitor \( V_C \) are equal in magnitude but opposite in phase, thus directly counteracting each other. Therefore, \( V_C = -V_R = -12.51 V \) and \( V_L = +12.51 V \).This equality results in their contributions canceling each other out. The primary voltage across the AC source directly matches the resistor's voltage at resonance, simplifying analysis by equating \( V_{ac} \) to \( V_R \). This fundamental concept helps engineers design effective filtering and energy resonance systems.

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Most popular questions from this chapter

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 \(\omega3\)= \(\omega_1\)/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 an \(L-R-C\) series circuit, what will be the resonance angular frequency of the circuit?

A sinusoidal current \(i = I\space cos\space \omega\)t has an rms value \(I_{rms} =\)2.10 A. (a) What is the current amplitude? (b) The current is passed through a full-wave rectifier circuit. What is the rectified average current? (c) Which is larger: \(I_{rms}\) or \(I_{rav}\)? Explain, using graphs of \(i^2\) and of the rectified current.

In an \(L-R-C\) series circuit, R = 300 \(\Omega\), \(L\) = 0.400 H, and \(C\) = 6.00 \(\times\) 10\(^{-8}\) F. When the ac source operates at the resonance frequency of the circuit, the current amplitude is 0.500 A. (a) What is the voltage amplitude of the source? (b) What is the amplitude of the voltage across the resistor, across the inductor, and across the capacitor? (c) What is the average power supplied by the source?

A 0.180-H inductor is connected in series with a 90.0-\(\Omega\) resistor and an ac source. The voltage across the inductor is \(v_L\) = -(12.0 V)sin[1480 rad/s)t]. (a) Derive an expression for the voltage \(v_R\) across the resistor. (b) What is \(v_R\) at t = 2.00 ms?

You want the current amplitude through a 0.450-mH inductor (part of the circuitry for a radio receiver) to be 1.80 mA when a sinusoidal voltage with amplitude 12.0 V is applied across the inductor. What frequency is required?

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