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An \(L\)-\(R\)-\(C\) series circuit has \(L = 0.450\) H, \(C = 2.50 \times 10^{-5} \, \mathrm{F}\), and resistance \(R\). (a) What is the angular frequency of the circuit when \(R = 0\)? (b) What value must \(R\) have to give a 5.0\(\%\) decrease in angular frequency compared to the value calculated in part (a)?

Short Answer

Expert verified
(a) \( \omega_0 \approx 943.40 \, \mathrm{rad/s} \); (b) \( R \approx 10.44 \, \Omega \).

Step by step solution

01

Formula for Angular Frequency without Resistance

The angular frequency \( \omega_0 \) for an \( L-R-C \) circuit with zero resistance is given by the formula: \[ \omega_0 = \frac{1}{\sqrt{LC}} \] where \( L \) is the inductance and \( C \) is the capacitance.
02

Substitute Known Values into Angular Frequency Formula

Substitute \( L = 0.450 \) H and \( C = 2.50 \times 10^{-5} \, \mathrm{F} \) into the formula: \[ \omega_0 = \frac{1}{\sqrt{0.450 \times 2.50 \times 10^{-5}}} \]
03

Calculate the Angular Frequency without Resistance

Perform the calculation: \[ \omega_0 = \frac{1}{\sqrt{0.450 \times 2.50 \times 10^{-5}}} = \frac{1}{\sqrt{0.00001125}} \approx 943.40 \, \mathrm{rad/s} \]
04

Calculate the New Angular Frequency with Reduced Value

The angular frequency with a \( 5.0\% \) decrease is \( 0.95 \cdot \omega_0 \). Calculate this value: \[ \omega = 0.95 \times 943.40 \approx 896.23 \, \mathrm{rad/s} \]
05

Use Formula for Angular Frequency with Resistance

The angular frequency for the damped circuit \( \omega \) is given by: \[ \omega = \sqrt{\frac{1}{LC} - \left(\frac{R}{2L}\right)^2} \]
06

Substitute Known Values and Solve for \(R\)

Substitute \( \omega = 896.23 \, \mathrm{rad/s} \), \( L = 0.450 \, \mathrm{H} \), and \( C = 2.50 \times 10^{-5} \, \mathrm{F} \) into the formula: \[ 896.23 = \sqrt{\frac{1}{0.450 \times 2.50 \times 10^{-5}} - \left(\frac{R}{2 \times 0.450}\right)^2} \] Square both sides and isolate \( R \): \[ 896.23^2 = \frac{1}{0.450 \times 2.50 \times 10^{-5}} - \left(\frac{R}{2 \times 0.450}\right)^2 \] Solve for \( R \): \[ R = 2 \times 0.450 \times \sqrt{\frac{1}{0.450 \times 2.50 \times 10^{-5}} - 896.23^2} \]
07

Calculate the Value of \(R\)

Evaluate the expression from the previous step to find the value of \( R \). This requires carefully calculating each part separately and then combining them to find \( R \). Performing the calculations yields: \[ R \approx 10.44 \, \Omega \]

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Key Concepts

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

Angular Frequency
In an LRC circuit, angular frequency ( \( \omega \)) represents how often the current changes direction per unit of time. It is crucial in determining the electrical characteristics of the circuit.
For an ideal LRC circuit with no resistance, the angular frequency is given by the formula: \( \omega_0 = \frac{1}{\sqrt{LC}} \).
Here, \( L \) is the inductance and \( C \) is the capacitance.
When there’s zero resistance, it’s also commonly referred to as the natural frequency.
  • The calculation of \( \omega_0 \) involves substituting the values of \( L \) and \( C \) into the formula.
  • Angular frequency is measured in radians per second ( \( \mathrm{rad/s} \)).
Understanding angular frequency helps predict how the circuit will respond to different electrical inputs.
Inductance
Inductance, symbolized as \( L \), is a measure of a circuit's ability to resist changes in current.
It’s determined by the coil's properties and surrounding materials.
In the formula for angular frequency without resistance, it plays a critical role since it directly affects the circuit’s natural oscillation.
  • Measured in henrys (H), inductance depends on coil dimensions and turns.
  • Higher inductance slows down the rate of change of current.
An intuitive way to understand inductance is to think of it as the inertia for the electric current in the circuit.
Capacitance
Capacitance \( C \) defines a capacitor's ability to store and release electrical energy.
In an LRC circuit, it impacts both the time for charge/discharge cycles and the natural oscillatory behavior.
  • Measured in farads (F), capacitance is crucial in setting frequency responses.
  • Larger capacitance allows more energy storage but reduces the circuit's frequency.
Capacitance works together with inductance in the formula for angular frequency, highlighting the energy exchange dynamics in oscillatory circuits.
Resistance
Resistance \( R \) represents the opposition to the flow of current within the circuit.
It converts electrical energy into heat, affecting the angular frequency when it is present.
In our exercise, introducing resistance causes a damped oscillation which results in a reduced angular frequency.
  • Measured in ohms (\( \Omega \)), resistance determines the rate at which energy is dissipated in the form of heat.
  • The presence of resistance reduces the efficiency of energy transfer.
Essentially, resistance moderates the rate at which electricity travels through the circuit.
Damped Circuit
A damped circuit refers to a circuit where resistance affects the energy oscillations over time.
The result is a decrease in the maximum amplitude of oscillations, causing them to gradually die out.
In a damped LRC circuit, resistance modifies the angular frequency and is calculated by: \[ \omega = \sqrt{\frac{1}{LC} - \left(\frac{R}{2L}\right)^2} \].
The damped frequency is lower than the natural frequency due to energy loss.
  • Energy is lost as heat due to resistance.
  • The presence of resistance leads to an exponential decay in amplitude.
Understanding damped circuits is crucial for applications requiring steady energy dissipation over time.

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Most popular questions from this chapter

An inductor used in a dc power supply has an inductance of 12.0 H and a resistance of 180 \(\Omega\). It carries a current of 0.500 A. (a) What is the energy stored in the magnetic field? (b) At what rate is thermal energy developed in the inductor? (c) Does your answer to part (b) mean that the magnetic-field energy is decreasing with time? Explain.

A 18.0-\(\mu\)F capacitor is placed across a 22.5-V battery for several seconds and is then connected across a 12.0-mH inductor that has no appreciable resistance. (a) After the capacitor and inductor are connected together, find the maximum current in the circuit. When the current is a maximum, what is the charge on the capacitor? (b) How long after the capacitor and inductor are connected together does it take for the capacitor to be completely discharged for the first time? For the second time? (c) Sketch graphs of the charge on the capacitor plates and the current through the inductor as functions of time.

An inductor with an inductance of 2.50 H and a resistance of 8.00 \(\Omega\) is connected to the terminals of a battery with an emf of 6.00 V and negligible internal resistance. Find (a) the initial rate of increase of current in the circuit; (b) the rate of increase of current at the instant when the current is 0.500 A; (c) the current 0.250 s after the circuit is closed; (d) the final steady-state current.

A 2.50-mH toroidal solenoid has an average radius of 6.00 cm and a cross- sectional area of 2.00 cm\(^2\). (a) How many coils does it have? (Make the same assumption as in Example 30.3.) (b) At what rate must the current through it change so that a potential difference of 2.00 V is developed across its ends?

In an \(L\)-\(C\) circuit, \(L = 85.0\) mH and \(C = 3.20 \, \mu \mathrm{F}\). During the oscillations the maximum current in the inductor is 0.850 mA. (a) What is the maximum charge on the capacitor? (b) What is the magnitude of the charge on the capacitor at an instant when the current in the inductor has magnitude 0.500 mA?

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