Question

An \(L\)-\(C\) circuit consists of a 60.0-mH inductor and a 250-\(\mu\)F capacitor. The initial charge on the capacitor is 6.00 \(\mu\)C, and the initial current in the inductor is zero. (a) What is the maximum voltage across the capacitor? (b) What is the maximum current in the inductor? (c) What is the maximum energy stored in the inductor? (d) When the current in the inductor has half its maximum value, what is the charge on the capacitor and what is the energy stored in the inductor?

Short answer

(a) 0.024 V; (b) 0.001548 A; (c) 7.2×10⁻⁸ J; (d) 5.95 µC, 8.98×10⁻¹⁰ J.

Step-by-step solution

Calculate the Maximum Voltage Across the Capacitor

The maximum voltage across the capacitor can be determined using the formula for the voltage across a capacitor, which is \( V = \frac{Q}{C} \). Here, \( Q = 6.00 \mu C = 6.00 \times 10^{-6} \text{ C} \) and \( C = 250 \mu F = 250 \times 10^{-6} \text{ F} \).Calculate the maximum voltage:\[ V_{\text{max}} = \frac{6.00 \times 10^{-6}}{250 \times 10^{-6}} = 0.024 \text{ V} \]

Calculate the Maximum Current in the Inductor

The maximum current in the inductor can be determined using the energy conservation principle in the LC circuit where the initial energy stored in the capacitor is equal to the energy in the circuit. Use the formula:\[ \frac{1}{2}C V_{\text{max}}^2 = \frac{1}{2}L I_{\text{max}}^2 \]Solving for \( I_{\text{max}} \), we have:\[ I_{\text{max}} = V_{\text{max}} \sqrt{\frac{C}{L}} \]Substitute \( C = 250 \times 10^{-6} \text{ F} \) and \( L = 60.0 \times 10^{-3} \text{ H} \), and \( V_{\text{max}} = 0.024 \text{ V}\):\[ I_{\text{max}} = 0.024 \sqrt{\frac{250 \times 10^{-6}}{60.0 \times 10^{-3}}} = 0.024 \times 0.0645 \approx 0.001548 \text{ A} \]

Calculate the Maximum Energy Stored in the Inductor

The maximum energy stored in the inductor is equal to the maximum energy stored initially in the capacitor.The formula for energy stored in a capacitor is:\[ U = \frac{1}{2} C V_{\text{max}}^2 \]Substitute \( C = 250 \times 10^{-6} \text{ F} \) and \( V_{\text{max}} = 0.024 \text{ V} \):\[ U = \frac{1}{2} \times 250 \times 10^{-6} \times (0.024)^2 = 7.2 \times 10^{-8} \text{ J} \]

Current in Inductor is Half of Maximum Value

When the current in the inductor is half of its maximum value, \( I = \frac{I_{\text{max}}}{2} \), the current is:\[ I = \frac{0.001548}{2} \text{ A} = 0.000774 \text{ A} \]

Determine Charge on the Capacitor at Half Maximum Current

Use the energy conservation principle:\[ \frac{1}{2} L (\frac{I_{\text{max}}}{2})^2 + \frac{1}{2} C V_C^2 = \frac{1}{2} C V_{\text{max}}^2 \]Reformulate it to solve for the voltage across the capacitor, \( V_C \):\[ \frac{1}{2} \times 60.0 \times 10^{-3} \times (0.000774)^2 + \frac{1}{2} \times 250 \times 10^{-6} \times V_C^2 = 7.2 \times 10^{-8} \]Solving for \( V_C \) gives us:\[ 8.98 \times 10^{-10} + 125 \times 10^{-6} V_C^2 = 7.2 \times 10^{-8} \]This simplifies to:\[ 125 \times 10^{-6} V_C^2 = 7.1 \times 10^{-8} \]\[ V_C^2 = 0.000568 \Rightarrow V_C = 0.0238 \text{ V} \]Charge on the capacitor, \( Q_C = CV_C \):\[ Q_C = 250 \times 10^{-6} \times 0.0238 = 5.95 \times 10^{-6} \text{ C} \approx 5.95 \ \mu C \]

Determine Energy Stored in Inductor at Half Maximum Current

The energy stored in the inductor is given by:\[ U_L = \frac{1}{2} L \left( \frac{I_{\text{max}}}{2} \right)^2 \]Substituting \( L = 60.0 \times 10^{-3} \) and \( I = 0.000774 \text{ A} \):\[ U_L = \frac{1}{2} \times 60.0 \times 10^{-3} \times (0.000774)^2 \approx 8.98 \times 10^{-10} \text{ J} \]

Key concepts

Inductor-capacitor circuit

An inductor-capacitor (LC) circuit is a fundamental building block in electronics, often used to understand oscillations and energy transfer. In an LC circuit, energy oscillates between the inductor and the capacitor. The inductor stores energy in a magnetic field, while the capacitor stores energy in an electric field.

This circuit does not lose energy as it oscillates; rather, the energy shifts between the two components back and forth. Because of these energy exchanges, LC circuits are often referred to as tank circuits, resonant circuits, or harmonic oscillators. The resonant frequency of an LC circuit is crucial as it determines the rate at which energy is exchanged between the inductor and capacitor.

Understanding how LC circuits work is critical in radio communication, power supplies, and signal processing. In this context, we are dealing with a circuit consisting of a 60.0 mH inductor and a 250 μF capacitor, showing how these components interact to form a complete LC circuit.

Energy conservation in circuits

Energy conservation is a crucial principle guiding the analysis of circuits, particularly LC circuits. In an LC circuit, the law of conservation of energy states that the total energy stored within the system remains constant unless external forces act upon it.

Initially, the energy is stored in the capacitor, expressed as \[ U = \frac{1}{2}CV^2 \] where \( C \) is the capacitance, and \( V \) is the voltage across the capacitor. This energy then transfers to the inductor, with its stored energy described by \[ U = \frac{1}{2}LI^2 \] where \( L \) is the inductance, and \( I \) is the current through the inductor.

The fascinating part of an LC circuit is how it naturally oscillates. When the current in the inductor reaches its maximum, the voltage in the capacitor drops to zero, concentrating all the circuit's energy in the inductor. Over time, this energy returns to the capacitor, highlighting the perpetual dance of energy within L-C circuits. This understanding helps explain various behaviors seen in electrical systems, such as ringing in circuits and resonance phenomena.

Capacitor charge and voltage

In an LC circuit, the charge on the capacitor influences both the voltage and the system's behavior. A capacitor consists of two conducting plates separated by an insulator. When charged, a voltage forms between these plates. The relationship between voltage \( V \) and charge \( Q \) is given by \[ V = \frac{Q}{C} \] where \( Q \) is the charge and \( C \) is the capacitance.

Initially, the given problem states the capacitor is fully charged with an initial charge of 6 μC. This charge determines the initial voltage across the capacitor using the above formula. As the circuit oscillates, the charge on the capacitor changes constantly, reflecting changing voltage across it.

At specific moments, such as when the current in the inductor hits half its maximum value, calculations showed the capacitor's new charge levels. Such points of analysis help in understanding the precise function of LC circuits and how changes in charge affect the system's voltage and behavior overall. This role is pivotal in tuning circuits, ensuring intended signal behaviors, and managing energy efficiently across electronic devices.