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In an \(L-R-C\) series circuit, R = 400 \(\Omega\), \(L\) = 0.350 H, and \(C\) = 0.0120 \(\mu\)F. (a) What is the resonance angular frequency of the circuit? (b) The capacitor can withstand a peak voltage of 670 V. If the voltage source operates at the resonance frequency, what maximum voltage amplitude can it have if the maximum capacitor voltage is not exceeded?

Short Answer

Expert verified
(a) Resonance frequency is approximately \(1.53 \times 10^4\) rad/s. (b) Maximum source voltage is approximately 50.1 V.

Step by step solution

01

Understanding Resonance Frequency

In an L-R-C circuit, the resonance frequency occurs when the inductive reactance equals the capacitive reactance. The angular resonance frequency \( \omega_0 \) is given by the formula \( \omega_0 = \frac{1}{\sqrt{LC}} \). We need to substitute the given values for \(L\) and \(C\) into this formula.
02

Calculating Resonance Frequency

Substitute \(L = 0.350\, \text{H}\) and \(C = 0.0120 \times 10^{-6}\, \text{F}\) into the formula for resonance angular frequency:\[\omega_0 = \frac{1}{\sqrt{0.350 \times 0.0120 \times 10^{-6}}}\]Calculate to find \(\omega_0\).
03

Compute Resonance Frequency

Calculate the expression:\[\omega_0 = \frac{1}{\sqrt{0.350 \times 0.0120 \times 10^{-6}}} \approx 1.53 \times 10^4 \text{ rad/s}.\]
04

Understanding Maximum Voltage Condition

The peak voltage across the capacitor at resonance is given by \(V_C = Q V_s\), where \(Q\) is the quality factor \(Q = \frac{\omega_0 L}{R}\). We need to ensure that the peak voltage \(V_C\) does not exceed 670 V.
05

Calculate Quality Factor

Using the formula \(Q = \frac{\omega_0 L}{R}\), substitute \(\omega_0 = 1.53 \times 10^4\, \text{rad/s}, L = 0.350\, \text{H}, R = 400 \Omega\):\[Q = \frac{1.53 \times 10^4 \times 0.350}{400}\]Calculate \(Q\).
06

Compute Quality Factor

Calculate the expression:\[\Q \approx \frac{1.53 \times 10^4 \times 0.350}{400} \approx 13.375.\]
07

Find Maximum Source Voltage Amplitude

We have \(V_C = Q V_s = 670 \text{ V}\). Solve for \(V_s\):\[V_s = \frac{670}{Q} = \frac{670}{13.375}\].
08

Calculate Maximum Source Voltage Amplitude

Calculate:\[V_s \approx \frac{670}{13.375} \approx 50.1 \text{ V}.\]

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Key Concepts

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

L-R-C Circuit
An L-R-C circuit is an electronic circuit consisting of three main components: a resistor (R), an inductor (L), and a capacitor (C) connected in series. This type of circuit is used to analyze the behavior of AC (alternating current) signals. Each component plays its role in determining how the circuit responds to different frequencies.

The resistor offers resistance, measured in Ohms (\( \Omega \)), that limits the flow of current. The inductor stores energy in a magnetic field when current passes through it, and it’s measured in henrys (H). Finally, the capacitor stores energy in an electric field, measured in farads (F).

In an L-R-C circuit, the interplay of these components can create interesting phenomena, such as resonance. At resonance, the impedance of the circuit is minimized because the inductive and capacitive reactances cancel each other out. This happens at the resonance frequency, which is a pivotal characteristic of these circuits and dictates the condition for maximum current flow.
Inductive Reactance
Inductive reactance is a measure of how much an inductor opposes the change in current. It’s an important concept in alternating current (AC) circuits. When AC voltage is applied, the current through the inductor doesn’t immediately rise but is delayed due to the energy storage in its magnetic field.

The inductive reactance (\( X_L \)) of an inductor can be calculated using the formula:\[X_L = \omega L\]where \( \omega \) is the angular frequency of the AC source, and \( L \) is the inductance of the coil. The unit of inductive reactance is Ohms (\( \Omega \)).

The higher the frequency or the inductance, the greater the inductive reactance becomes, meaning more opposition to current changes. In an L-R-C circuit, it is crucial to balance inductive and capacitive reactance to achieve resonance, which leads to minimum impedance and maximum current flow.
Capacitive Reactance
Capacitive reactance is similar to inductive reactance but applies to capacitors in AC circuits. It measures how much a capacitor resists changes in voltage. When a voltage is applied, the capacitor charges and discharges, causing a phase difference between voltage and current.

The capacitive reactance (\( X_C \)) is given by the formula:\[X_C = \frac{1}{\omega C}\]where \( \omega \) is the angular frequency, and \( C \) is the capacitance. Like inductive reactance, it is measured in Ohms (\( \Omega \)).

For a lower frequency or higher capacitance, the capacitive reactance is high, meaning more resistance to current flow variations. In the context of L-R-C circuits, capacitive reactance plays a central role in determining whether the circuit is at resonance. At resonance, the inductive and capacitive reactances equal each other, thus canceling out, enabling maximum power transfer.

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Most popular questions from this chapter

An inductor with \(L\) = 9.50 mH is connected across an ac source that has voltage amplitude 45.0 V. (a) What is the phase angle \(\phi\) for the source voltage relative to the current? Does the source voltage lag or lead the current? (b) What value for the frequency of the source results in a current amplitude of 3.90 A?

In an \(L-R-C\) series circuit, the components have the following values: \(L = 20.0\space mH\), \(C = 140\space nF\), and R = 350 \(\Omega\). The generator has an rms voltage of 120 V and a frequency of 1.25 kHz. Determine (a) the power supplied by the generator and (b) the power dissipated in the resistor.

A 150-\(\Omega\) resistor is connected in series with a 0.250-H inductor and an ac source. The voltage across the resistor is \(v_R = (3.80 V)\)cos[720 rad/s)2t] . (a) Derive an expression for the circuit current. (b) Determine the inductive reactance of the inductor. (c) Derive an expression for the voltage \(v_L\) across the inductor.

You want the current amplitude through a 0.450-mH inductor (part of the circuitry for a radio receiver) to be 1.80 mA when a sinusoidal voltage with amplitude 12.0 V is applied across the inductor. What frequency is required?

A transformer connected to a 120-V (rms) ac line is to supply 13,000 V (rms) for a neon sign. To reduce shock hazard, a fuse is to be inserted in the primary circuit; the fuse is to blow when the rms current in the secondary circuit exceeds 8.50 mA. (a) What is the ratio of secondary to primary turns of the transformer? (b) What power must be supplied to the transformer when the rms secondary current is 8.50 mA? (c) What current rating should the fuse in the primary circuit have?

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