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What resistance \(R\) should be connected in series with an inductance \(L=220 \mathrm{mH}\) and capacitance \(C=12.0 \mu \mathrm{F}\) for the maximum charge on the capacitor to decay to \(99.0 \%\) of its initial value in 50.0 cycles? (Assume \(\left.\omega^{\prime} \approx \omega .\right)\)

Short Answer

Expert verified
Connect approximately 0.11 ohms resistance in series.

Step by step solution

01

Understanding the LC Circuit in Series

In a series LC circuit, the charge on the capacitor decays over time. The decay is influenced by the resistance in the circuit, which impacts the rate at which the charge decreases. This phenomenon can be described by the damping factor in the equation of a damped harmonic oscillator. Our goal is to find the resistance that allows the charge to decay to 99.0% of its initial value after 50 cycles.
02

Calculating Angular Frequency

First, calculate the angular frequency of the LC circuit without considering any resistance. The angular frequency \(\omega\) is given by the formula \(\omega = \frac{1}{\sqrt{LC}}\). Here, \(L = 220 \, \text{mH} = 220 \times 10^{-3} \, \text{H}\) and \(C = 12.0 \, \mu\text{F} = 12.0 \times 10^{-6} \, \text{F}\). Calculate \(\omega\):\[\omega = \frac{1}{\sqrt{(220 \times 10^{-3})(12.0 \times 10^{-6})}}\]
03

Understanding Damping Factor

The resistance causes the oscillation to be under-damped. We know that the charge decreases as:\[Q(t) = Q_0 e^{-\frac{Rt}{2L}}\]We are given that the charge decays to 99% after 50 cycles. Therefore:\[\frac{Q(t)}{Q_0} = 0.99 = e^{-\frac{R \times 50T}{2L}}\]where \(T\) is the period of one cycle, and it is related to \(\omega\) by \(T = \frac{2\pi}{\omega}\). Substituting \(T\) in the equation setups the formulation to solve for \(R\).
04

Solve for Resistance \(R\)

Using the expression from Step 3 and knowing the values of \(L\) and \(\omega\),Calculate \(T\):\[T = \frac{2\pi}{\omega}\]Substitute \(T\) into the expression:\[0.99 = e^{-\frac{R \times 50 \times \frac{2\pi}{\omega}}{2 \times 220 \times 10^{-3}}}\]Solve for \(R\):\[R = -\frac{2 \times 220 \times 10^{-3}}{50 \times \frac{2\pi}{\omega}} \ln(0.99)\]Calculate \(R\) using the computed \(\omega\) from Step 2.

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Key Concepts

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

Damping Factor
The damping factor is a critical component in understanding how oscillations in an RLC circuit decay over time. In simple terms, it helps describe how quickly the amplitudes of oscillations decrease because of energy losses in the system.

In a series RLC circuit, the damping factor is influenced by the resistance component. Higher resistance leads to greater damping, which means the circuit will lose energy faster, causing the oscillations (or the charge on the capacitor) to decrease more quickly. This damping is often characterized by an exponential decay formula, represented as
  • \(Q(t) = Q_0 e^{-\frac{Rt}{2L}}\)
Here, \(Q_0\) is the initial charge on the capacitor, \(t\) is time, and \(R\) and \(L\) are the resistance and inductance, respectively.

Understanding how the damping factor impacts an RLC circuit allows for precise control over the decay rate of the charge, which is especially useful in applications like signal processing or electronics that require stable performance over time.
Angular Frequency
Angular frequency is essential in oscillatory systems, including RLC circuits. It is denoted by \(\omega\) and helps describe the rapidity of oscillations or cycles within a circuit over time.

In a simple LC circuit (comprising only an inductor and a capacitor), angular frequency can be calculated using the formula
  • \(\omega = \frac{1}{\sqrt{LC}}\)
where \(L\) is inductance and \(C\) is capacitance.

The unit of angular frequency is radians per second, and it gives a clear picture of how many oscillations occur in a particular interval of time. In our situation, the angular frequency of the circuit is the foundation upon which we calculate other properties like the damping factor and time period. This is crucial in determining how different resistances can affect the circuit’s performance, particularly the decay of the capacitor’s charge.
Series Circuit
In a series circuit, components are connected end-to-end in a single path for current to flow.

This arrangement creates a unique interaction between inductors, capacitors, and resistors. Specifically, in an RLC series circuit,
  • the inductor, resistor, and capacitor are connected one after another.
Understanding series circuits is vital because the same current flows through each component, but the voltage across each component can be different.

Moreover, in a series RLC circuit, the characteristics like total impedance and phase angle depend heavily on the frequency of the input signal and the individual properties of each component. These individual components together impact how efficiently the circuit can store and release energy, a mechanism which is finely tuned by adjusting the resistance.
Capacitor Decay
Capacitor decay refers to the reduction of charge stored in the capacitor over time, especially when subjected to a resistive force as seen in an RLC circuit.

This decay can be analyzed through the lens of the exponential decay law, which helps predict how quickly the charge diminishes in a circuit with specific parameters. In mathematical terms, this decay behavior is expressed as
  • \(\frac{Q(t)}{Q_0} = e^{-\frac{Rt}{2L}}\)
where \(Q(t)\) is the charge at time \(t\), \(Q_0\) is the initial charge, \(R\) is resistance, and \(L\) is inductance.

The decay rate is influenced by the doping factor, where a higher resistance results in a faster decrease in charge. Capacitor decay is particularly important when designing circuits that require stable voltage levels, as understanding and controlling this decay ensures the functionality and efficiency of electronic devices over their operational lifespan.

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Most popular questions from this chapter

A transformer has 500 primary turns and 10 secondary turns. (a) If \(V_{p}\) is \(120 \mathrm{~V}(\mathrm{rms}),\) what is \(V_{s}\) with an open circuit? If the secondary now has a resistive load of \(15 \Omega,\) what is the current in the (b) primary and (c) secondary?

A series circuit containing inductance \(L_{1}\) and capacitance \(C_{1}\) oscillates at angular frequency \(\omega .\) A second series circuit, containing inductance \(L_{2}\) and capacitance \(C_{2},\) oscillates at the same angular frequency. In terms of \(\omega,\) what is the angular frequency of oscillation of a series circuit containing all four of these elements? Neglect resistance. (Hint: Use the formulas for equivalent capacitance and equivalent inductance; see Module \(25-3\) and Problem 47 in Chapter \(30 .\)

In a certain oscillating \(L C\) circuit, the total energy is converted from electrical energy in the capacitor to magnetic energy in the inductor in \(1.50 \mu\) s. What are (a) the period of oscillation and (b) the frequency of oscillation? (c) How long after the magnetic energy is a maximum will it be a maximum again?

A \(1.50 \mu \mathrm{F}\) capacitor has a capacitive reactance of \(12.0 \Omega .\) (a) What must be its operating frequency? (b) What will be the capacitive reactance if the frequency is doubled?

A variable capacitor with a range from 10 to \(365 \mathrm{pF}\) is used with a coil to form a variable-frequency \(L C\) circuit to tune the input to a radio. (a) What is the ratio of maximum frequency to minimum frequency that can be obtained with such a capacitor? If this circuit is to obtain frequencies from \(0.54 \mathrm{MHz}\) to \(1.60 \mathrm{MHz}\), the ratio computed in (a) is too large. By adding a capacitor in parallel to the variable capacitor, this range can be adjusted. To obtain the desired frequency range, (b) what capacitance should be added and (c) what inductance should the coil have?

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