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operatorname{An} L-R-C\( scrics circuit has \)L=0.450 \mathrm{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

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Insert your mathematical solutions for the angular frequencies and the resistance here

Step by step solution

01

Calculation of angular frequency in LC circuit

Using the formula for the resonant angular frequency for an LC circuit, \(Ó¬ = 1 / \sqrt{LC} \). Substituting the values, L = 0.450 H and C = 2.50 x \(10^{-5} F\), we find the angular frequency when R = 0.
02

Calculation of resistance for specific decrease in angular frequency

The frequency is expected to decrease by 5% when the resistance is introduced. Thus, the new resonance frequency can be calculated as 0.95 of the original angular frequency computed in Step 1. Using the formula for the resonant angular frequency for an RLC circuit, \(Ó¬ = \sqrt{1/LC - (R/2L)^2}\), we substitute the values L = 0.450 H, C = 2.50 x \(10^{-5} F\) and the new frequency and solve to find the resistance R that will give a 5% decrease in angular frequency.

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Key Concepts

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

Angular Frequency
Angular frequency is a concept that helps us understand how fast an electrical circuit oscillates. Think of it as the frequency of the swings of a pendulum, but in circuits. In the context of L-R-C circuits, angular frequency can be expressed in radians per second. A special type of circuit is the LC circuit, which is composed of only an inductor (L) and a capacitor (C). When no resistance (R) is present, this circuit resonates at its natural frequency, calculated using:\[Ó¬ = \frac{1}{\sqrt{LC}}\]Here:
  • Ó¬ is the angular frequency.
  • L is the inductance in henries (H).
  • C is the capacitance in farads (F).
With this formula, you can determine how fast the energy will oscillate between the inductor and the capacitor when there's no resistance to dampen the process. Each component plays a role in the overall speed or frequency of these oscillations.
Resonant Frequency
The resonant frequency is the frequency at which a circuit naturally prefers to oscillate. This would be the loudest note an instrument naturally plays. In circuits, resonance makes the energy exchange between the inductor and the capacitor most efficient. A key case for resonance is when resistance is zero. For an LC circuit without resistance, resonant frequency is defined as the angular frequency calculated by:\[Ó¬_0 = \frac{1}{\sqrt{LC}}\]When resistance, R, enters the picture, the resonant frequency shifts slightly due to the damping effect. So, for an RLC circuit, the introduction of R changes the effective resonant frequency:\[Ó¬' = \sqrt{\frac{1}{LC} - \left(\frac{R}{2L}\right)^2}\]Understanding these changes helps engineers design circuits with accurate behavior under varying conditions. Resonant frequency is crucial in tuning circuits for applications ranging from radio receivers to filters that need precise frequency selectivity.
Electrical Resistance
Electrical resistance is a measure of how much a material or component opposes the flow of electrical current through it. Higher resistance means it's harder for the current to pass through. Resistance is measured in ohms (Ω). In an RLC circuit, introducing resistance affects the overall behavior. Primarily:
  • Resistance affects the energy storage pattern, slowing the exchange between the inductor and the capacitor.
  • High resistance dampens oscillations, leading to a slower angular frequency.
  • Partial resistance can shift the resonant frequency slightly from its natural value.
Imagine trying to move a heavy object through mud. The mud represents resistance, making it harder to move at the same speed. Similarly, resistance moderates the passage and speed of energy oscillation in a circuit. Therefore, it's key to carefully balance resistance in applications where oscillatory behavior is needed.

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Most popular questions from this chapter

{A} 6.40 \mathrm{nF} \text { capacitor is charged to } 24.0 \mathrm{~V} \text { and then discon- } nected from the battery in the circuit and connected in series with a coil that has \(L=0.0660 \mathrm{H}\) and negligible resistance. After the circuit has been completed, there are current oscillations. (a) At an instant when the charge of the capacitor is \(0.0800 \mu \mathrm{C}\), how much cnergy is stored in the capacitor and in the inductor, and what is the current in the inductor? (b) At the instant when the charge on the capacitor is \(0.0800 \mu \mathrm{C},\) what are the voltages across the capacitor and across the inductor, and what is the rate at which current in the inductor is changing?

A \(7.50 \mathrm{nF}\) capacitor is charged to \(12.0 \mathrm{~V}\), then disconnected from the power supply and connected in scrics through a coil. The period of oscillation of the circuit is then measured to be \(8.60 \times 10^{-3} \mathrm{~s}\) Calculate: (a) the inductance of the coil: (b) the maximum charge on the capacitor; (c) the total cnergy of the circuit; (d) the maximum current in the circuit.

C Consider a coil of wire that has radius \(3.00 \mathrm{~cm}\) and carries a sinusoidal current given by \(i(t)=I_{0} \sin (2 \pi f t),\) where the frequency \(f=60.0 \mathrm{IIz}\) and the initial current \(I_{0}=1.20 \mathrm{~A}\). (a) Estimate the magnetic flux through this coil as the product of the magnetic field at the center of the coil and the area of the coil. Use this magnctic flux to estimate the sclf-inductance \(L\) of the coil. (b) Use the value of \(L\) that you cstimated in part (a) to calculate the magnitudc of the maximum emf induced in the coil.

A \(35.0 \mathrm{~V}\) battery with negligible intemal resistance, a \(50.0 \Omega\) resistor, and a \(1.25 \mathrm{mH}\) inductor with negligible resistance are all connected in series with an open switch. The switch is suddenly closed. (a) How long after closing the switch will the current through the inductor reach onehalf of its maximum value? (b) How long after closing the switch will the energy stored in the inductor reach one-half of its maximum value?

It is possible to make your own inductor by winding wire around a cylinder, such as a pcncil. Assume you have a spool of AWG 20 copper wire, which has a diameter of \(0.812 \mathrm{~mm}\). (a) Estimate the diameter of a pencil. (b) Estimate how many times can you tightly wrap AWG 20 copper wire around a pencil to form a solenoid with a length of \(4.0 \mathrm{~cm}\). (c) Estimate the inductance of this solcnoid by assuming the magnetic field inside is constant. (d) If a current of 1.0 A flows through this solenoid, how much magnetic energy will be stored inside?

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