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(a) For an underdamped \(L R C\) circuit, determine a formula for the energy \(U = U _ { E } + U _ { B }\) stored in the electric and magnetic fields as a function of time. Give answer in terms of the initial charge \(Q _ { 0 }\) on the capacitor. (b) Show how \(d U / d t\) is related to the rate energy is transformed in the resistor, \(I ^ { 2 } R .\)

Short Answer

Expert verified
The total energy in an underdamped LRC circuit decreases over time at a rate equal to the power dissipated in the resistor, confirming \( \frac{dU}{dt} = -I^2 R \).

Step by step solution

01

Understanding Energy in an LRC Circuit

In an LRC circuit, energy is stored in two forms: electrical energy in the capacitor and magnetic energy in the inductor. The total energy is given by the sum of these energies. The electrical energy in the capacitor is given by \( U_E = \frac{1}{2} C V^2 \), where \( V \) is the voltage across the capacitor. The magnetic energy in the inductor is \( U_B = \frac{1}{2} L I^2 \), where \( I \) is the current through the inductor.
02

Express Initial Conditions

The initial charge on the capacitor is \( Q_0 \), which implies the initial voltage across the capacitor is \( V_0 = \frac{Q_0}{C} \). Initially, all the energy \( U_0 = \frac{1}{2} C V_0^2 = \frac{1}{2} \frac{Q_0^2}{C} \) is stored in the capacitor and released into the circuit as it oscillates.
03

Energy as a Function of Time

To find the energy as a function of time, we first use the equation for underdamped oscillation, where the charge \( q(t) = Q_0 e^{-\frac{R}{2L}t} \cos(\omega_d t) \) and current \( i(t) = -\frac{d}{dt}q(t) \). The angular frequency is \( \omega_d = \sqrt{\frac{1}{LC} - \left(\frac{R}{2L}\right)^2} \). Thus, the energy stored is \( U = \frac{1}{2}L (i(t)^2) + \frac{1}{2C} (q(t)^2) \).
04

Calculating \( \frac{dU}{dt} \)

The rate of energy change in the system \( \frac{dU}{dt} \) can be found by differentiating the expression for \( U \) found in the previous step. Given \( q(t) \) and \( i(t) \), use the product and chain rules to differentiate \( U \) with respect to time.
05

Relate to Power Loss in Resistor

In an LRC circuit, energy is dissipated as thermal energy in the resistor. The power loss due to the resistor is \( I^2 R \). By equating this with \(-\frac{dU}{dt} \), we show the rate of energy loss in the field equals the energy dissipated by the resistor, confirming \( \frac{dU}{dt} = -I^2 R \).

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Key Concepts

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

Energy in Capacitor
In an LRC circuit, capacitors play a crucial role in storing electrical energy. This energy is stored as electric potential energy and can be expressed mathematically using the formula \( U_E = \frac{1}{2} C V^2 \), where \( C \) is the capacitance and \( V \) is the voltage across the capacitor.
The interesting part about capacitors is how they initially hold all the energy when the circuit is charged. With the given initial charge \( Q_0 \), the initial voltage \( V_0 \) is calculated as \( V_0 = \frac{Q_0}{C} \). This means the initial energy stored solely in the capacitor is \( U_0 = \frac{1}{2} \frac{Q_0^2}{C} \).
As time progresses, this energy starts to oscillate between the capacitor and the inductor, demonstrating the dynamic nature of energy in LRC circuits. In an underdamped system, this oscillation is characterized by time functions of charge and current, showing how energy originally in the capacitor flows through the circuit.
Energy in Inductor
Inductors are equally significant in LRC circuits, as they store energy in the form of magnetic fields. The energy stored in an inductor is defined by the formula \( U_B = \frac{1}{2} L I^2 \), where \( L \) represents the inductance and \( I \) is the current flowing through it.
When a circuit oscillates, energy moves between the capacitor and the inductor. Initially, when the capacitor has more stored energy, the inductor's energy is relatively low. As the capacitor discharges, the energy is transferred to the inductor, emphasizing the interdependence of these two components.
This exchange of energy is well illustrated in an underdamped circuit using charge and current functions. With the charge \( q(t) = Q_0 e^{-\frac{R}{2L}t} \cos(\omega_d t) \) and current \( i(t) = -\frac{d}{dt}q(t) \), one can analyze how energy shifts over time. This dynamic interaction helps maintain the circuit's oscillatory behavior.
Energy Dissipation in Resistor
While capacitors and inductors store energy temporarily, resistors act as energy dissipators in LRC circuits. They convert the electrical energy into heat through a process known as energy dissipation.
In mathematical terms, the power loss in a resistor is expressed as \( I^2 R \), where \( I \) is the current through the resistor and \( R \) is its resistance. This power loss is continuous, and it results in a gradual decrease in the total stored energy over time.
For an underdamped LRC circuit, the energy dissipation in the resistor is directly related to the rate of change of the total energy. As shown by the relation \( \frac{dU}{dt} \), the decrease in energy stored in the capacitor and inductor (expressed as \( -\frac{dU}{dt} \)) equals the energy dissipated as heat in the resistor, confirming the equation \( \frac{dU}{dt} = -I^2 R \).
This relationship emphasizes the resistor's role in transforming electrical energy into thermal energy, ultimately leading to a damping effect in the circuit.

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Most popular questions from this chapter

\((a)\) What is the rms current in a series \(R C\) circuit if \(R=3.8 \mathrm{k} \Omega, \quad C=0.80 \mu \mathrm{F},\) and the rms applied voltage is \(120 \mathrm{~V}\) at \(60.0 \mathrm{~Hz} ?\) (b) What is the phase angle between voltage and current? \((c)\) What is the power dissipated by the circuit? ( \(d\) ) What are the voltmeter readings across \(R\) and \(C ?\)

You want to turn on the current through a coil of selfinductance \(L\) in a controlled manner, so you place it in series with a resistor \(R=2200 \Omega,\) a switch, and a dc voltage source \(V_{0}=240 \mathrm{~V}\). After closing the switch, you find that the current through the coil builds up to its steady- state value with a time constant \(\tau\). You are pleased with the current's steady-state value, but want \(\tau\) to be half as long What new values should you use for \(R\) and \(V_{0} ?\)

(II) A 75-W lightbulb is designed to operate with an applied ac voltage of \(120 \mathrm{~V} \mathrm{rms}\). The bulb is placed in series with an inductor \(L,\) and this series combination is then connected to a \(60-\mathrm{Hz} 240-\mathrm{V}\) rms voltage source. For the bulb to operate properly, determine the required value for \(L\). Assume the bulb has resistance \(R\) and negligible inductance.

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 \(R ?\)

A pair of straight parallel thin wires, such as a lamp cord, each of radius \(r,\) are a distance \(\ell\) apart and carry current to a circuit some distance away. Ignoring the field within each wire, show that the inductance per unit length is \(\left(\mu_{0} / \pi\right) \ln [(\ell-r) / r]\).

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