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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?

Short Answer

Expert verified
(a) 150 V; (b) 150 V, large, large; (c) 75 W.

Step by step solution

01

Find the Resonance Frequency

In an L-R-C series circuit, the resonance frequency \( f_0 \) is calculated using the formula \( f_0 = \frac{1}{2\pi\sqrt{LC}} \). Substitute \( L = 0.400 \text{ H} \) and \( C = 6.00 \times 10^{-8} \text{ F} \) into the formula:\[ f_0 = \frac{1}{2\pi\sqrt{0.400 \times 6.00 \times 10^{-8}}} \] Calculating this gives us \( f_0 \approx 325 \text{ Hz} \).
02

Calculate the Voltage Amplitude of the Source

At resonance, the impedance \( Z \) is equal to the resistance \( R \). Therefore, the voltage amplitude \( V \) can be calculated as \( V = IR \), where \( I = 0.500 \text{ A} \) and \( R = 300 \Omega \).\[ V = 0.500 \times 300 = 150 \text{ V} \]
03

Calculate the Voltage Across Each Component

In a resonance situation, the voltage across the resistor \( V_R \) is also \( IR \), so:\[ V_R = 0.500 \times 300 = 150 \text{ V} \] The voltages across the inductor \( V_L \) and the capacitor \( V_C \) are equal because the reactances are equal and opposite, given by \( V_L = V_C = I \times \omega_0 \times L = I \times \frac{1}{\omega_0 \times C} \), which resolves to: \[ V_L = V_C = \frac{150}{R} \times 300 = \text{Large}\] We replace plug this condition in for one complete cycle.
04

Calculate the Average Power Supplied

The average power \( P \) supplied by the source is given by \( P = I^2 R \). Plugging in \( I = 0.500 \text{ A} \) and \( R = 300 \Omega \):\[ P = 0.500^2 \times 300 = 75 \text{ W} \]

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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 key concept in LCR circuits. It is the frequency at which the circuit naturally oscillates when not driven by an external source. At this frequency, the energy alternates perfectly between the magnetic field of the inductor and the electric field of the capacitor. In an LCR series circuit, it is defined by the formula:
  • \( f_0 = \frac{1}{2\pi\sqrt{LC}} \)
For instance, if we have an inductor \( L = 0.400 \) H and a capacitor \( C = 6.00 \times 10^{-8} \) F, the resonance frequency \( f_0 \) can be calculated to be approximately 325 Hz.
This means that at this frequency, the circuit will exhibit maximum current throughput because the impedance of the circuit is at its minimum, consisting solely of the resistive component.
Impedance
Impedance in an LCR series circuit is the total opposition that the circuit presents to the flow of alternating current (AC). It is a combination of resistance and reactance, which includes both inductive and capacitive effects. Mathematically, impedance \( Z \) is given by:
  • \( Z = \sqrt{R^2 + (X_L - X_C)^2} \)
Where \( X_L \) is the inductive reactance and \( X_C \) is the capacitive reactance. At resonance frequency, the reactances cancel each other out \((X_L = X_C)\), and thus the impedance equals the resistance \( R \) only.
In our example, with a resistance \( R = 300 \Omega \), the impedance at resonance is also \( 300 \Omega \). This is why the current reaches its maximum amplitude, and phase angles align optimally for minimal resistance beyond the inherent resistor's.
Average Power
Average power in an LCR circuit refers to the power being supplied to and dissipated by the circuit on average over a complete cycle of AC. At resonance, the average power is at its maximum due to the minimal impedance (only resistive). It is calculated with the formula:
  • \( P = I^2 R \)
Where \( I \) is the current amplitude, and \( R \) is the resistance.
In the exercise provided, the average power supplied by the AC source with a current amplitude of 0.500 A through a resistor of 300 \( \Omega \) is:
  • \( P = 0.500^2 \times 300 = 75 \text{ W}\)
Hence, 75 watts is the average thermal power dissipated by the resistor, demonstrating energy conversion at its peak efficiency.
Voltage Amplitude
Voltage amplitude in an LCR circuit refers to the maximum voltage supplied by the source. At resonance, due to the minimum impedance, we achieve the greatest current but using the relationship \( V = IR \), the source voltage can be found directly.
For a given exercise, the source voltage amplitude \( V_s \) is:
  • \( V_s = I \times R = 0.500 \times 300 = 150 \text{ V} \)
This value ensures that the power supply can drive the conditions of resonance. Similarly, the voltage across the resistor at resonance is also given as:
  • \( V_R = V_s = 150 \text{ V} \)
In resonance, voltage drops across the inductor and capacitor can be quite large, owing to their reactance canceling each other out. This scenario leads to a redistribution of voltage values along the circuit but with zero net voltage across the LC combination.

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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 \(C\) = 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?

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?

(a) Compute the reactance of a 0.450-H inductor at frequencies of 60.0 Hz and 600 Hz. (b) Compute the reactance of a 2.50-\(\mu\)F capacitor at the same frequencies. (c) At what frequency is the reactance of a 0.450-H inductor equal to that of a 2.50-\(\mu\)F capacitor?

(a) What is the reactance of a 3.00-H inductor at a frequency of 80.0 Hz? (b) What is the inductance of an inductor whose reactance is 120\(\Omega\) at 80.0 Hz? (c) What is the reactance of a 4.00-\(\mu\)F capacitor at a frequency of 80.0 Hz? (d) What is the capacitance of a capacitor whose reactance is 120 \(\Omega\) at 80.0 Hz?

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