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An \(L-R-C\) series circuit has \(R\) = 60.0 \(\Omega\), \(L\) = 0.800 H, and \(C\) = 3.00 \(\times\) 10\(^{-4}\) F. The ac source has voltage amplitude 90.0 V and angular frequency 120 rad/s. (a) What is the maximum energy stored in the inductor? (b) When the energy stored in the inductor is a maximum, how much energy is stored in the capacitor? (c) What is the maximum energy stored in the capacitor?

Short Answer

Expert verified
(a) Maximum energy in inductor: calculate using \( E_L = \frac{1}{2} L I^2 \). (b) When inductor's energy is maximum, capacitor's energy is zero. (c) Maximum energy in capacitor: use \( E_C = \frac{1}{2} C V^2 \).

Step by step solution

01

Energy in Inductor Formula

The energy stored in the inductor can be expressed as \( E_L = \frac{1}{2} L I^2 \). To determine the maximum energy, we first need to calculate the maximum current (\( I \)) using the RMS voltage provided and the impedance of the circuit.
02

Calculate Impedance of Circuit

The impedance \( Z \) in an \( L-R-C \) series circuit is given by \( Z = \sqrt{R^2 + (\omega L - 1/\omega C)^2} \). Substitute \( R = 60.0 \Omega \), \( L = 0.800 \text{ H} \), \( C = 3.00 \times 10^{-4} \text{ F} \), and \( \omega = 120 \text{ rad/s} \) to find \( Z \).
03

Compute Maximum Current

The maximum current is given by \( I = \frac{V_0}{Z} \), where \( V_0 = 90.0 \text{ V} \). Substitute the impedance calculated in the previous step into this formula to find \( I \).
04

Calculate Maximum Energy in Inductor

Substitute \( L = 0.800 \text{ H} \) and the newly calculated \( I \) into \( E_L = \frac{1}{2} L I^2 \) to find the maximum energy stored in the inductor.
05

Find Energy in Capacitor when Inductor Energy is Maximum

When the energy stored in the inductor is maximum, the energy stored in the capacitor is zero, because these energies interchange completely in an ideal LC combination.
06

Determine Maximum Energy in Capacitor

The maximum energy stored in the capacitor is given by \( E_C = \frac{1}{2} C V^2 \), where \( V = V_0 = 90.0 \text{ V} \). Substitute this value to find the energy stored in the capacitor.

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Key Concepts

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

Impedance
Impedance in an LRC series circuit is a key factor that helps us understand how much opposition the circuit presents to alternating current flow. It is a combination of resistance, inductance, and capacitance, creating a total effective opposition known as impedance, denoted as \( Z \).
In a circuit with resistance \( R \), inductance \( L \), and capacitance \( C \), the impedance is calculated using the formula:
  • \( Z = \sqrt{R^2 + (\omega L - \frac{1}{\omega C})^2} \)
Here, \( \omega \) is the angular frequency of the AC source. For the given exercise values: \( R = 60.0 \, \Omega \), \( L = 0.800 \, \text{H} \), \( C = 3.00 \times 10^{-4} \, \text{F} \), and \( \omega = 120 \, \text{rad/s} \), we calculate the impedance.
Replacing these into the formula helps find \( Z \), which is necessary to determine how much current flows through the circuit under a given voltage.
Energy Stored in Inductor
Energy in an inductor is all about magnetic fields. When current runs through an inductor, it creates a magnetic field storing energy. The maximum energy stored in an inductor is crucial for understanding how the circuit functions during alternating current flow.
The energy stored in an inductor \( L \) with current \( I \) is expressed as:
  • \( E_L = \frac{1}{2} L I^2 \)
To calculate the maximum energy, we first need to discover the maximum current, which is found using the impedance and the voltage amplitude of the AC source.
  • Maximum current, \( I = \frac{V_0}{Z} \)
Here, \( V_0 \) is the voltage amplitude (90.0 V). Calculating \( I \) using the impedance we found, allows us to substitute into \( E_L = \frac{1}{2} L I^2 \) to find the maximum energy stored in the inductor. This shows the peak energy storage capability in a magnetic field for our given circuit.
Energy Stored in Capacitor
Capacitors store energy electrically, unlike inductors that store it magnetically. When a capacitor is fully charged, it holds maximum energy, which is key in transitioning areas between energy states.
The energy stored in a capacitor \( C \) with voltage \( V \) is given by the formula:
  • \( E_C = \frac{1}{2} C V^2 \)
In the context of an LRC circuit, when the energy in the inductor is at its maximum, the energy in the capacitor is zero, highlighting their complementary role.
To find the maximum energy stored in the capacitor, we use the same voltage amplitude \( V = V_0 = 90.0 \, \text{V} \). Plugging this into the energy formula gives the peak energy storage capability of the capacitor. This shows how energy transitions between components in the oscillatory process of the AC circuit.

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Most popular questions from this chapter

An \(L-R-C\) series circuit is connected to a 120-Hz ac source that has \(V_{rms} = 80.0 V\). The circuit has a resistance of 75.0 \(\Omega\) and an impedance at this frequency of 105 \(\Omega\). What average power is delivered to the circuit by the source?

A resistance \(R\), capacitance \(C\), and inductance \(L\) are connected in series to a voltage source with amplitude \(V\) and variable angular frequency \(\omega\). If \(\omega\) = \(\omega$$_0\) , the resonance angular frequency, find (a) the maximum current in the resistor; (b) the maximum voltage across the capacitor; (c) the maximum voltage across the inductor; (d) the maximum energy stored in the capacitor; (e) the maximum energy stored in the inductor. Give your answers in terms of \(R\), \(C\), \(L\), and \(V\).

In an \(L-R-C\) series circuit, \(R\) = 300 \(\Omega\), X\(_C\) = 300 \(\Omega\), and X\(_L\) = 500 \(\Omega\). The average electrical power consumed in the resistor is 60.0 W. (a) What is the power factor of the circuit? (b) What is the rms voltage of the source?

An \(L-R-C\) series circuit has \(C\) = 4.80 \(\mu\)F, \(L\) = 0.520 H, and source voltage amplitude \(V\) = 56.0 V. The source is operated at the resonance frequency of the circuit. If the voltage across the capacitor has amplitude 80.0 V, what is the value of \(R\) for the resistor in the circuit?

In an \(L-R-C\) series circuit, \(L=0.280 \mathrm{H}\) and \(C=\) \(4.00 \mu \mathrm{F}\). The voltage amplitude of the source is \(120 \mathrm{~V}\). (a) What is the resonance angular frequency of the circuit? (b) When the source operates at the resonance angular frequency, the current amplitude in the circuit is 1.70 A. What is the resistance \(R\) of the resistor? (c) At the resonance angular frequency, what are the peak voltages across the inductor, the capacitor, and the resistor?

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