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A damped \(L C\) circuit loses \(3.5 \%\) of its electromagnetic energy per cycle to thermal energy. If \(L=65 \mathrm{mH}\) and \(C=1.00 \mu \mathrm{F}\), what is the value of \(R ?\)

Short Answer

Expert verified
The value of \( R \) is approximately 1.42 Ohms.

Step by step solution

01

Calculate the Angular Frequency

First, find the angular frequency of the undamped LC circuit using the formula \( \omega_0 = \frac{1}{\sqrt{LC}} \). Plug in the given values: \( L = 65 \times 10^{-3} \) H and \( C = 1.00 \times 10^{-6} \) F. Thus, \( \omega_0 = \frac{1}{\sqrt{(65 \times 10^{-3})(1.00 \times 10^{-6})}} \). Simplifying, we get \( \omega_0 \approx 3.92 \times 10^3 \) rad/s.
02

Express Energy Loss in Terms of Quality Factor

The percentage energy loss per cycle is related to the quality factor \( Q \) by \( \text{energy loss per cycle} = \frac{2\pi}{Q} \). Given the circuit loses 3.5% of its energy per cycle, this translates to \( 0.035 = \frac{2\pi}{Q} \). Solving for \( Q \) gives \( Q = \frac{2\pi}{0.035} \approx 179.1 \).
03

Relate Quality Factor to Resistance

The quality factor \( Q \) for an RLC circuit is also given by \( Q = \frac{\omega_0 L}{R} \). Rearrange to find \( R \): \( R = \frac{\omega_0 L}{Q} \). Substitute \( \omega_0 = 3.92 \times 10^3 \) rad/s, \( L = 65 \times 10^{-3} \) H, and \( Q = 179.1 \) to find \( R \): \( R = \frac{(3.92 \times 10^3)(65 \times 10^{-3})}{179.1} \approx 1.42 \) Ohms.

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Key Concepts

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

Energy Loss
When you have a damped LC circuit, energy doesn't stay constant as it would in an ideal, undamped circuit. Instead, some of this energy is transformed into thermal energy due to resistance, manifesting as energy loss. In this specific scenario, the circuit loses 3.5% of its energy per cycle.

This percentage directly tells us how much energy is converted into heat during each oscillation of the circuit. In practical terms, if you started with 100 units of energy, you would have only 96.5 units left after one cycle.
  • This loss impacts how efficiently the circuit oscillates, dampening the amplitude over time.
  • Less efficient circuits will lose energy faster, while more efficient circuits maintain their oscillations longer.
The ability to quantify this loss through percentage helps us determine other parameters in the circuit, like the quality factor and the total resistance.
Angular Frequency
Angular frequency in an LC circuit tells us how quickly the circuit oscillates. It is denoted as \( \omega_0 \) and calculated using the formula \( \omega_0 = \frac{1}{\sqrt{LC}} \).

In this scenario, we plug in the given values for inductance \( L = 65 \times 10^{-3} \) H and capacitance \( C = 1.00 \times 10^{-6} \) F into the formula. The resulting angular frequency is approximately \( 3.92 \times 10^3 \) rad/s.

  • This value tells us how the circuit behaves in the absence of damping.
  • It's key for predicting the natural oscillation period and velocity of the circuit.
  • The higher the angular frequency, the faster the circuit will oscillate.
Remember, angular frequency is the fundamental backdrop against which the circuit's other characteristics are gauged.
Quality Factor
The quality factor, or \( Q \), of an LC circuit is a measure of its oscillatory behavior and its efficiency in retaining energy. It indicates how "sharp" or "damped" the resonance peak is.

In the calculation, the quality factor is linked to energy loss per cycle by the relation \( \text{energy loss per cycle} = \frac{2\pi}{Q} \). Given an energy loss of 3.5%, or \( 0.035 \), you can compute \( Q = \frac{2\pi}{0.035} \). This yields \( Q \approx 179.1 \).
  • A high \( Q \) value indicates a low rate of energy loss relative to the energy stored, meaning better efficiency.
  • A lower \( Q \) results in higher damping and more energy loss during each cycle.
The quality factor is integral to understanding the damping characteristics and enabling accurate predictions of the circuit's performance.
Resistance Calculation
Resistance in a damped LC circuit plays a crucial role in the damping process and energy dispersion into heat. The resistance \( R \) is directly connected to the quality factor and angular frequency.

Using the formula \( Q = \frac{\omega_0 L}{R} \), one can determine the circuit's resistance. In this setup, the quality factor \( Q = 179.1 \), angular frequency \( \omega_0 = 3.92 \times 10^3 \) rad/s, and inductance \( L = 65 \times 10^{-3} \) H are utilized to calculate \( R \).

Rearranging for \( R \), we get \( R = \frac{\omega_0 L}{Q} \).
  • By substituting the values, we find \( R \approx 1.42 \) Ohms.
  • This calculated resistance accounts for the damping in the circuit, handling the energy loss from oscillations.
Understanding resistance within this framework allows us to more precisely control how the circuit operates and responds to energy fluctuations.

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Most popular questions from this chapter

(I) \(\mathrm { A } 3800\) -pF capacitor is connected in series to a \(26.0 - \mu \mathrm { H }\) coll of resistance 2.00\(\Omega .\) What is the resonant frequency of this circuit?

A 2.44 -m-long coil containing 225 loops is wound on an iron core (average \(\mu=1850 \mu_{0}\) ) along with a second coil of 115 loops. The loops of each coil have a radius of \(2.00 \mathrm{~cm} .\) If the current in the first coil drops uniformly from \(12.0 \mathrm{~A}\) to zero in \(98.0 \mathrm{~ms},\) determine: \((a)\) the mutual inductance \(M ;(b)\) the emf induced in the second coil.

A resistor \(R\) is in parallel with a capacitor \(C,\) and this parallel combination is in series with a resistor \(R^{\prime}\). If connected to an ac voltage source of frequency \(\omega\), what is the equivalent impedance of this circuit at the two extremes in frequency \((a) \omega=0,\) and \((b) \omega=\infty ?\)

(II) A capacitor is placed in parallel with some device, B as in Fig. \(18 b ,\) to filter out stray high frequency signals, but to allow ordinary \(60 - H z\) ac to pass through with little loss. Suppose that circuit \(B\) in Fig. 18\(b\) is a resistance \(R = 490 \Omega\) connected to ground, and that \(C = 0.35 \mu \mathrm { F }\) . What percent of the incoming current will pass through \(C\) rather than \(R\) if it is \(( a )\) it is 60\(\mathrm { Hz }\) ; (b) \(60,000 \mathrm { Hz }\) ?

A \(35-\mathrm{mH}\) inductor with \(2.0-\Omega\) resistance is connected in series to a \(26-\mu \mathrm{F}\) capacitor and a \(60-\mathrm{Hz}, 45-\mathrm{V}(\mathrm{rms})\) source. Calculate \((a)\) the rms current, \((b)\) the phase angle, and \((c)\) the power dissipated in this circuit.

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