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In an \(L-R-C\) series circuit, \(R=300 \Omega, L=0.400 \mathrm{H},\) and \(C=6.00 \times 10^{-8} \mathrm{~F}\). When the ac source operates at the resonance frequency of the circuit, the current amplitude is \(0.500 \mathrm{~A}\). (a) What is the voltage amplitude of the source? (b) What is the amplitude of the voltage across the resistor, across the inductor, and across the capacitor? (c) What is the average power supplied by the source?

Short Answer

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(a) The voltage amplitude of the source is 150 V. (b) The amplitude of the voltage across the resistor, across the inductor and across the capacitor is 150 V. (c) The average power supplied by the source is 75.00 W.

Step by step solution

01

Calculating the Voltage Amplitude

Use Ohm's law to find the peak or amplitude voltage (V). Ohm's law states that \(V = i \times \( \) where R is the resistance and i is the current. Given, R = 300 Ohm and i = 0.500 A thus, V = \(i \times R = 0.500 \times 300 = 150 V \)
02

Find the Voltage across Resistor, Inductor, and Capacitor

At resonance, the voltages across the inductor \(L\) and across the capacitor \(C\) are equal and opposite, which means they cancel out. Hence, the voltage across each component will be the same as the source voltage. So, the voltage across resistor \(V_R = V = 150 \mathrm{V}\), voltage across inductor \(V_L = V = 150 \mathrm{V}\), and voltage across capacitor \(V_C = V = 150 \mathrm{V}\)
03

Calculate the Average Power Supplied by the Source

The average power (P) supplied by the source can be calculated using the formula \(P = i^2 \times R\), where R is resistance and i is the current. Given, R = 300 Ohm and i = 0.500 A, thus, \(P = (0.500)^2 \times 300 = 75.00 \mathrm{W}\)

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Key Concepts

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

Resonance Frequency in an LRC Series Circuit
In an LRC series circuit, the **resonance frequency** is a special frequency at which the circuit behaves in a unique way. It is the frequency at which the total impedance of the circuit is minimized, making the current amplitude at its maximum.Understanding resonance is crucial because at this frequency:
  • The voltages across the inductor and capacitor cancel each other out.
  • Only the resistor affects the voltage amplitude and current.
  • The power transferred from the source to the circuit is entirely resistive, without reactive power.
The formula to find the resonance frequency \( f_0 \) is: \[ f_0 = \frac{1}{2\pi\sqrt{LC}}\]Where:
  • \(L\) is the inductance
  • \(C\) is the capacitance
At resonance, the circuit reaches a balance between the energy storage in the capacitance and inductance, maximizing the efficiency of power transfer.
Ohm's Law in Circuit Analysis
Ohm's law is a foundational principle in electronics that relates voltage, current, and resistance in a simple formula: \[ V = I \times R \] Where:
  • \(V\) is the voltage across a component in volts
  • \(I\) is the current through the component in amperes
  • \(R\) is the resistance of the component in ohms
In the context of an LRC series circuit, using Ohm's law allows us to calculate the voltage amplitude: If the current amplitude \(I\) is 0.500 A and the resistance \(R\) is 300 \(\Omega\), the voltage amplitude \(V\) is\[ V = 0.500 \times 300 = 150 \text{ V} \]This calculation is essential because it helps us understand how the circuit responds at resonance when only resistive impedance is considered.
Average Power Supplied by an AC Source
In any circuit, power can fluctuate as it alternates, but the **average power** gives a consistent measure of power flow over time. In an AC circuit at resonance, such as an LRC circuit, the average power \(P\) relates to resistance and can be calculated with: \[ P = I^2 \times R \] Where:
  • \(I\) is the RMS current
  • \(R\) is the resistance in ohms
Given the RMS current \(I\) is 0.500 A and resistance \(R\) is 300 \(\Omega\), the average power is: \[ P = (0.500)^2 \times 300 = 75 \text{ W} \] Understanding this concept is crucial as it describes the real power consumed by the circuit, showing how efficiently the circuit converts electrical energy into another form of energy (like heat in a resistor). This is especially relevant when designing circuits for power efficiency.
Voltage Amplitude in an AC Circuit
The **voltage amplitude** of an AC source in a circuit refers to the maximum voltage that the source can provide. In LRC series circuits, knowing the voltage amplitude is crucial for understanding how the circuit components are affected. Under resonance conditions:
  • Voltage across the resistor equals the source's voltage amplitude.
  • The voltages across the inductor and capacitor cancel each other out.
In our example, the voltage amplitude is calculated using the derived current and resistance:\[ V = I \times R = 0.500 \times 300 = 150 \text{ V}\]This value tells you that at its peak, the source can drive 150 V across the circuit, explaining the potential difference across the resistor and why the inductive and capacitive voltages don't contribute to energy consumption at resonance.

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Most popular questions from this chapter

An \(L-R-C\) series circuit is constructed using a \(175 \Omega\) resistor, a \(12.5 \mu \mathrm{F}\) capacitor, and an \(8.00 \mathrm{mH}\) inductor, all connected across an ac source having a variable frequency and a voltage amplitude of \(25.0 \mathrm{~V}\). (a) At what angular frequency will the impedance be smallest, and what is the impedance at this frequency? (b) At the angular frequency in part (a), what is the maximum current through the inductor? (c) At the angular frequency in part (a), find the potential difference across the ac source, the resistor, the capacitor, and the inductor at the instant that the current is equal to one-half its greatest positive value. (d) In part (c), how are the potential differences across the resistor, inductor, and capacitor related to the potential difference across the ac source?

A transformer connected to a \(120 \mathrm{~V}\) (rms) ac line is to supply \(12.0 \mathrm{~V}\) (rms) to a portable electronic device. The load resistance in the secondary is \(5.00 \Omega\). (a) What should the ratio of primary to secondary turns of the transformer be? (b) What rms current must the secondary supply? (c) What average power is delivered to the load? (d) What resistance connected directly across the \(120 \mathrm{~V}\) line would draw the same power as the transformer? Show that this is equal to \(5.00 \Omega\) times the square of the ratio of primary to secondary turns.

(a) Compute the reactance of a \(0.450 \mathrm{H}\) inductor at frequencies of \(60.0 \mathrm{~Hz}\) and \(600 \mathrm{~Hz}\). (b) Compute the reactance of a \(2.50 \mu \mathrm{F}\) capacitor at the same frequencies. (c) At what frequency is the reactance of a \(0.450 \mathrm{H}\) inductor equal to that of a \(2.50 \mu \mathrm{F}\) capacitor?

\(\operatorname{In}\) an \(L-R-C\) series circuit the source is operated at its resonant angular frequency. At this frequency, the reactance \(X_{C}\) of the capacitor is \(200 \Omega\) and the voltage amplitude across the capacitor is \(600 \mathrm{~V}\). The circuit has \(R=300 \Omega\). What is the voltage amplitude of the source?

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?

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