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An \(L-R-C\) series circuit consists of a \(50.0-\Omega\) resistor, a \(10.0-\mu \mathrm{F}\) capacitor, a \(3.50-\mathrm{mH}\) inductor, and an ac voltage source of voltage amplitude 60.0 \(\mathrm{V}\) operating at 1250 \(\mathrm{Hz}\) . (a) Find the current amplitude and the voltage amplitudes across the inductor, the resistor, and the capacitor. Why can the voltage amplitudes add up to more than 60.0 \(\mathrm{V}\) (b) If the frequency is now doubled, but nothing else is changed, which of the quantities in part (a) will change? Find the new values for those that do change.

Short Answer

Expert verified
Initial current amplitude is 1.167 A. When frequency doubles, current amplitude changes to 0.778 A. The other voltage amplitudes also change as recalculated. Voltage amplitudes add due to phase differences.

Step by step solution

01

Calculate the Angular Frequency

The angular frequency is given by \( \omega = 2\pi f \), where \( f \) is the frequency of the AC source. Substituting \( f = 1250 \text{ Hz} \), we get \( \omega = 2\pi \times 1250 \approx 7854 \text{ rad/s} \).
02

Calculate Reactances

Calculate the reactance of the inductor and capacitor. For the inductor, \( X_L = \omega L = 7854 \times 3.50 \times 10^{-3} = 27.49 \Omega\).For the capacitor, \( X_C = \frac{1}{\omega C} = \frac{1}{7854 \times 10 \times 10^{-6}} = 12.74 \Omega \).
03

Calculate Impedance

The impedance of the circuit is given by:\[ Z = \sqrt{R^2 + (X_L - X_C)^2} \]Substituting the values, we get:\[ Z = \sqrt{50^2 + (27.49 - 12.74)^2} = 51.46 \Omega \]
04

Calculate Current Amplitude

The current amplitude is given by:\[ I = \frac{V}{Z} = \frac{60}{51.46} = 1.167 \text{ A} \]
05

Calculate Voltage Amplitudes across Components

The voltage across each component is calculated as:1. Across the resistor: \( V_R = I \times R = 1.167 \times 50 = 58.35 \text{ V} \).2. Across the inductor: \( V_L = I \times X_L = 1.167 \times 27.49 = 32.08 \text{ V} \).3. Across the capacitor: \( V_C = I \times X_C = 1.167 \times 12.74 = 14.88 \text{ V} \).
06

Why Voltage Amplitudes Can Exceed the Source Voltage

The voltages across the inductor and capacitor are out of phase with the voltage across the resistor. Therefore, they can combine in such a way that their vector sum exceeds the source voltage amplitude.
07

Doubling the Frequency

If the frequency is doubled to \( 2500 \text{ Hz} \), the new angular frequency \( \omega = 2\pi \times 2500 \approx 15708 \text{ rad/s} \). This changes the reactances:1. \( X_L = 15708 \times 3.50 \times 10^{-3} = 54.95 \Omega \).2. \( X_C = \frac{1}{15708 \times 10 \times 10^{-6}} = 6.37 \Omega \).
08

Recalculate Impedance and Current at New Frequency

Recalculate the impedance:\[ Z = \sqrt{50^2 + (54.95 - 6.37)^2} = 77.11 \Omega \] The new current amplitude is:\[ I = \frac{60}{77.11} = 0.778 \text{ A} \]
09

Recalculate Voltage Amplitudes at New Frequency

The new voltage amplitudes are:1. Across the resistor: \( V_R = 0.778 \times 50 = 38.9 \text{ V} \).2. Across the inductor: \( V_L = 0.778 \times 54.95 = 42.73 \text{ V} \).3. Across the capacitor: \( V_C = 0.778 \times 6.37 = 4.95 \text{ V} \).

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Key Concepts

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

Reactance
Reactance is a measure of how much an inductor or a capacitor resists the change of electric current in a circuit. It is part of what makes circuits behave differently at various frequencies. For inductors, the reactance is directly proportional to the frequency; thus, it increases when the frequency increases. For capacitors, reactance is inversely proportional to the frequency, meaning it decreases when frequency rises. The reactance of an inductor is given by the formula \( X_L = \omega L \), and for a capacitor, it is \( X_C = \frac{1}{\omega C} \), where \( \omega \) is the angular frequency, \( L \) is inductance, and \( C \) is capacitance. Both measurements are in ohms (\(\Omega\)). Understanding these relationships helps in calculating how resistive components in a circuit will react to changes in frequency.
- Inductor reactance increases with higher frequencies.
- Capacitor reactance decreases with higher frequencies.
Impedance
Impedance is a more comprehensive measure of resistance in AC circuits that includes not just resistance, but also reactance. It's a combination of a circuit's resistance (Ohm's Law) along with any additional opposition from reactance due to capacitors and inductors. In a series LRC circuit, the total impedance \( Z \) can be calculated with the formula:\[ Z = \sqrt{R^2 + (X_L - X_C)^2} \] where \( R \) is the resistance, \( X_L \) is the inductive reactance, and \( X_C \) is capacitive reactance. Impedance is measured in ohms (\( \Omega \)). By knowing impedance, we can determine how much current will flow for a given voltage—essential for predicting circuit behavior in AC systems. It illustrates the combined effect of different components in a simplified form.
- Impedance integrates resistance and reactance.
- It dictates how effectively a circuit can transmit AC power.
Angular Frequency
Angular frequency is an essential concept in understanding AC circuits and is denoted by \( \omega \). It represents how fast the electric current oscillates in radians per second. The formula relating angular frequency to the standard frequency is \( \omega = 2\pi f \), where \( f \) is the frequency in Hz. Angular frequency plays a pivotal role in calculations for reactance and impedance. A change in frequency will alter the angular frequency and consequently affect the circuit's behavior, particularly its reactance and impedance values.
- Angular frequency is proportional to the regular frequency.
- It's crucial for determining how various circuit properties change with frequency.
Current Amplitude
Current amplitude refers to the maximum flow rate of charge or peak current that occurs in a circuit. In AC circuits, this is influenced heavily by the impedance. The amplitude can be calculated using Ohm's Law adapted for AC circuits, written as:\[ I = \frac{V}{Z} \] where \( V \) is the voltage amplitude and \( Z \) is the impedance. Current amplitude provides insight into how robustly current flows for given conditions in a circuit and helps determine the efficiency and effectiveness in delivering energy. In situations where impedance changes due to varying frequencies, the current amplitude will adjust accordingly, affecting how much work the circuit can perform.
- Current amplitude shows peak current flow.
- It is affected by both voltage amplitude and impedance.
Voltage Amplitude
Voltage amplitude in a circuit represents the maximum voltage present across the entire circuit. In an LRC series circuit, the source voltage distributes across each component creating different voltage amplitudes—which may add up to more than the source voltage due to phase differences between reactance components. The calculation of each component's voltage is based on the formula:
  • For the resistor: \( V_R = I \times R \)
  • For the inductor: \( V_L = I \times X_L \)
  • For the capacitor: \( V_C = I \times X_C \)
Each voltage drop relates to its respective component's resistance or reactance when multiplied by the current amplitude \( I \). These distributions illustrate why the sum of voltage amplitudes across components can exceed the source voltage in AC systems.
- Voltage amplitude can exceed source voltage due to AC characteristics.
- It's vital for understanding power distribution in the circuit.

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Most popular questions from this chapter

A transformer consists of 275 primary windings and 834 secondary windings. If the potential difference across the primary coil is \(25.0 \mathrm{V},\) (a) what is the voltage across the secondary coil, and (b) what is the effective load resistance of the secondary coil if it is connected across a \(125-\Omega\) resistance of the secondary coil if it is

At a frequency \(\omega_{1}\) the reactance of a certain capacitor equals that of a certain inductor. (a) If the frequency is changed to \(\omega_{2}=2 \omega_{1},\) what is the ratio of the reactance of the inductor to that of the capacitor? Which reactance is larger? (b) If the frequency is changed to \(\omega_{3}=\omega_{1} / 3,\) what is the ratio of the reactance of the inductor to that of the capacitor? Which reactance is larger? (c) If the capacitor and inductor are placed in series with a resistor of resistance \(R\) to form an \(L-R-C\) series circuit, what will be the resonance angular frequency of the circuit?

A series circuit consists of an ac source of variable frequency, a \(115-\Omega\) resistor, a \(1.25-\mu F\) capacitor, and a \(4.50-\mathrm{mH}\) inductor. Find the impedance of this circuit when the angular frequency of the ac source is adjusted to (a) the resonance angular frequency; (b) twice the resonance angular frequency; (c) half the resonance angular frequency.

Consider an \(L-R-C\) series circuit with a \(1.80-\mathrm{H}\) inductor, a \(0.900-\mu F\) capacitor, and a \(300-\Omega\) resistor. The source has terminal rms voltage \(V_{\text { rms }}=60.0 \mathrm{V}\) and variable angular frequency \(\omega\) . (a) What is the resonance angular frequency \(\omega_{0}\) of the circuit? (b) What is the rms current through the circuit at resonance, \(I_{\mathrm{rms}-0} ?(\mathrm{c})\) For what\ two values of the angular frequency, \(\omega_{1}\) and \(\omega_{2},\) is the rms current half the resonance value? (d) The quantity \(\left|\omega_{1}-\omega_{2}\right|\) defines the resonance width. Calculate \(I_{\text { rms- } 0 \text { and the resonance width for }}\) \(R=300 \Omega, 30.0 \Omega,\) and 3.00\(\Omega .\) Describe how your results compare to the discussion in Section \(31.5 .\)

A Radio Inductor. You want the current amplitude through a \(0.450-\mathrm{mH}\) inductor (part of the circuitry for a radio receiver) to be 2.60 \(\mathrm{mA}\) when a sinusoidal voltage with amplitude 12.0 \(\mathrm{V}\) is applied across the inductor. What frequency is required?

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