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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

An \(L-R-C\) series circuit has R = 500 \(\Omega\), L = 2.00 H, \(C\) = 0.500 \(\mu\)F, and \(V\) = 100 V. (a) For \(\omega\) = 800 rad/s, calculate \(V_R , V_L, V_C\), and \(\phi\). Using a single set of axes, graph \(v\), \(v_R , v_L\), and \(v_C\) as functions of time. Include two cycles of \(v\) on your graph. (b) Repeat part (a) for \(\omega\) = 1000 rad/s. (c) Repeat part (a) for \(\omega = 1250\space rad/s\).

An \(L-R-C\) series circuit has \(R\) = 60.0 \(\Omega\), \(L\) = 0.800 H, and \(C\) = 3.00 \(\times\) 10\(^{-4}\) F. The ac source has voltage amplitude 90.0 V and angular frequency 120 rad/s. (a) What is the maximum energy stored in the inductor? (b) When the energy stored in the inductor is a maximum, how much energy is stored in the capacitor? (c) What is the maximum energy stored in the capacitor?

In an \(L-R-C\) series circuit, \(L\) = 0.280 H and C = 4.00 \(\mu\)F. The voltage amplitude of the source is 120 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?

In an \(L-R-C\) series circuit, the phase angle is 40.0\(^\circ\), with the source voltage leading the current. The reactance of the capacitor is 400 \(\Omega\), and the resistance of the resistor is 200 \(\Omega\). The average power delivered by the source is 150 W. Find (a) the reactance of the inductor, (b) the rms current, (c) the rms voltage of the source.

A capacitance \(C\) and an inductance \(L\) are operated at the same angular frequency. (a) At what angular frequency will they have the same reactance? (b) If \(L\) = 5.00 mH and \(C\) = 3.50 \(\mu\)F, what is the numerical value of the angular frequency in part (a), and what is the reactance of each element?

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