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A 18.0-\(\mu\)F capacitor is placed across a 22.5-V battery for several seconds and is then connected across a 12.0-mH inductor that has no appreciable resistance. (a) After the capacitor and inductor are connected together, find the maximum current in the circuit. When the current is a maximum, what is the charge on the capacitor? (b) How long after the capacitor and inductor are connected together does it take for the capacitor to be completely discharged for the first time? For the second time? (c) Sketch graphs of the charge on the capacitor plates and the current through the inductor as functions of time.

Short Answer

Expert verified
(a) Maximum current is 0.385 A, charge is 0 C at this time. (b) Capacitor fully discharges in 0.0231 s and 0.0693 s. (c) Graphs are sinusoidal.

Step by step solution

01

Calculate Initial Charge on Capacitor

First, we need to find the initial charge \( Q_0 \) on the capacitor. We use the formula \( Q = C \cdot V \), where \( C \) is the capacitance (18.0 \( \mu \)F) and \( V \) is the voltage (22.5 V).\[Q_0 = (18.0 \times 10^{-6} \, \text{F}) \cdot (22.5 \, \text{V}) = 4.05 \times 10^{-4} \, \text{C}\]
02

Determine Maximum Current in the LC Circuit

To find the maximum current \( I_{max} \), use conservation of energy. The initial energy stored in the capacitor \( \frac{1}{2} Q_0^2/C \) is equal to the maximum energy in the inductor \( \frac{1}{2} L I_{max}^2 \). Set up the equation and solve for \( I_{max} \):\[I_{max} = \sqrt{\frac{Q_0^2}{LC}} = \sqrt{\frac{(4.05 \times 10^{-4} \, \text{C})^2}{12.0 \times 10^{-3} \, \text{H} \cdot 18.0 \times 10^{-6} \, \text{F}}}\]Solving this gives \(I_{max} = 0.385 \, \text{A}\).
03

Calculate Charge on Capacitor at Maximum Current

At maximum current, all energy is stored in the inductor and the charge on the capacitor is zero. Thus, \( Q = 0 \) for maximum current.
04

Calculate Time for Capacitor to First Fully Discharge

The time to discharge the capacitor completely for the first time is a quarter of the period of oscillation of the LC circuit. The period \( T \) is given by \( T = 2\pi\sqrt{LC} \):\[T = 2\pi\sqrt{(12.0 \times 10^{-3} \, \text{H})(18.0 \times 10^{-6} \, \text{F})} = 0.0925 \, \text{s}\]Therefore, the time to discharge completely is \( \frac{T}{4} = 0.0231 \, \text{s} \).
05

Calculate Time for Capacitor to Fully Discharge a Second Time

The second time the capacitor is completely discharged is at \( \frac{3T}{4} \). This is because the charge will be zero twice in each full oscillation (once at \( \frac{T}{4} \) and the other at \( \frac{3T}{4} \)): \[\frac{3T}{4} = 3 \times 0.0231 \, \text{s} = 0.0693 \, \text{s}\]
06

Sketch Graphs of Charge and Current Over Time

The charge \( Q(t) \) on the capacitor and the current \( I(t) \) through the inductor both oscillate sinusoidally with time. \( Q(t) \) is maximized when \( I(t) \) is zero, and vice versa. These functions are described as:\[Q(t) = Q_0 \cos\left(\frac{2\pi}{T} t \right)\]\[I(t) = I_{max} \sin\left(\frac{2\pi}{T} t \right)\]These expressions show periodic behavior with period \( T = 0.0925 \, \text{s} \). Plot these two equations to visualize charge and current fluctuations over time.

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Key Concepts

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

Capacitor Discharge
In an LC circuit, the capacitor stores energy in the form of an electric field when it is charged. After connecting a charged capacitor with an inductor, the process of capacitor discharge begins. Capacitor discharge in an LC circuit is a cyclic event where stored electrical energy is transferred to magnetic energy and vice versa. Initially, when connected, the charge on the capacitor decreases as it starts discharging through the inductor.
  • The charge at any time, during oscillation, follows a sinusoidal pattern.
  • The maximum charge the capacitor can hold is defined initially.
This discharge process continues until all energy is transferred into the inductor, reaching a point where the capacitor is momentarily without charge. The discharge sequence is periodic, causing the capacitor to continually charge and discharge over time.
Maximum Current Calculation
The point of maximum current in an LC circuit is crucial as it indicates the moment when all stored energy in the circuit is transferred to the inductor. This maximum current arises due to conservation of energy principles.Conservation of energy in the LC circuit means the initial energy from the capacitor is equal to the energy at that time in the inductor: \[\frac{1}{2} C V^2 = \frac{1}{2} L I_{max}^2\]Where:
  • \(C\) is the capacitance of the capacitor.
  • \(V\) is the voltage across the capacitor.
  • \(L\) is the inductance of the inductor.
  • \(I_{max}\) is the maximum current through the inductor.
Solving this, we find the maximum current using:\[I_{max} = \sqrt{\frac{Q_0^2}{LC}}\]Where \(Q_0\) is the initial charge on the capacitor.
Energy Conservation in Circuits
Energy conservation is an essential principle in analyzing LC circuits. It ensures that the total energy within the circuit is always conserved, merely converting from one form to another.In a closed LC circuit:
  • The initial energy is stored in the capacitor as electrical energy: \(\frac{1}{2} C V^2\).
  • Energy is alternated between the capacitor and inductor, turning into magnetic energy in the inductor: \(\frac{1}{2} L I^2\).
When the capacitor is fully charged, all energy resides electrically. On discharging fully into the inductor, the energy is magnetic.As the charge and current oscillate, so too does the distribution of energy, demonstrating a beautiful conservation mechanism in LC circuits.
Oscillation Period Calculation
In an LC circuit, the charge and current oscillate. Therefore, it is important to measure how quickly these oscillations occur through the period calculation.The period of oscillation, \(T\), is determined by both the capacitance and inductance:\[T = 2\pi\sqrt{LC}\]
  • C is the capacitance.
  • L is the inductance.
The formula signifies that the oscillation period depends on the values of both capacitor and inductor. A larger inductance or capacitance will yield a slower oscillation rate.By calculating the period, one can ascertain the precise timing for various significant events within the oscillation cycle, such as when the capacitor is fully discharged.

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Most popular questions from this chapter

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