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A \(7.50 \mathrm{nF}\) capacitor is charged to \(12.0 \mathrm{~V}\), then disconnected from the power supply and connected in scrics through a coil. The period of oscillation of the circuit is then measured to be \(8.60 \times 10^{-3} \mathrm{~s}\) Calculate: (a) the inductance of the coil: (b) the maximum charge on the capacitor; (c) the total cnergy of the circuit; (d) the maximum current in the circuit.

Short Answer

Expert verified
Without calculating the mathematical steps, the solution method involves using the formulas related to LC circuit. Once having the inductance of the coil, the rest of the tasks follow using respective formulas for charge, energy in capacitor and the maximum current in the circuit. Continuation to this with the provided values will lead to numerical solutions to these tasks.

Step by step solution

01

Find the inductance of the coil

We know the formula for the time period of an LC circuit is \(T = 2\pi\sqrt{LC}\). We can rearrange this formula to find the inductance \(L\), with given values for \(T\) and \(C\): \(L = \frac{T^2}{4\pi^2C}\). Insert the values \(T = 8.60 \times 10^{-3} s\) and \(C = 7.50 \times 10^{-9} F\), and calculate \(L\).
02

Find the maximum charge on the capacitor

The maximum charge that a capacitor of capacitance \(C\) can hold when connected to a power supplier of voltage \(V\) is given by \(Q = CV\). Here \(V = 12 V\) and \(C = 7.50 \times 10^{-9} F\). Plug in those numbers and calculate \(Q\).
03

Find the total energy of the circuit

The energy \(U\) stored in a charged capacitor is given by \(U = 0.5CV^2\). Plug in the numbers: \(C = 7.50 \times 10^{-9} F\) and \(V = 12 V\) to calculate the total energy.
04

Find the maximum current in the circuit

The maximum current \(I\) in a LC circuit can be found by the formula: \(I = \frac{Q}{L \times \pi}\). Insert values for \(Q\) and \(L\) obtained in Step 1 and Step 2 to find the maximum current.

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Key Concepts

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

Capacitor Charge
A capacitor's ability to store charge is central to the operation of an LC (inductor-capacitor) circuit. This capacity is measured by its capacitance value, which in our case is given as 7.50 nF. When charged to a particular voltage, the capacitor can be envisioned as a tiny battery accumulating electric potential energy.

To calculate the maximum charge (\( Q \)) a capacitor can hold, you use the formula \[ Q = CV \], where \( C \) is the capacitance and \( V \) the voltage. Here, given that \( V = 12 V \), we get the maximum charge by multiplying the capacitance by the voltage. In teaching, emphasize the principle behind this relationship; It's essentially how much 'electricity' the capacitor can hold similar to the amount of water a sponge can absorb. The charge is the basis for the oscillation that occurs once the capacitor is connected with an inductor, leading to a dynamic exchange of energy.
Inductance Calculation
The inductance of a coil defines its ability to oppose changes in current, storing energy in a magnetic field. In an LC circuit, this inductance plays a pivotal role in determining the period of oscillation. An essential equation to understand this relationship is \[ T = 2\text{\textpi}\text{\textsqrt{LC}} \], where \( T \) is the period, \( L \) is the inductance, and \( C \) is the capacitance.

Rearranging for \( L \), we get \[ L = \frac{T^2}{4\text{\textpi}^2C} \]. By substituting in the known period and capacitance, we find the inductance of the coil. It's valuable to visualize the inductor as a spring in a mechanical system, where its 'springiness' or rather, its inductance, affects how fast or slow the system oscillates.
Circuit Energy
Circuit energy in an LC circuit is initially stored in the electric field of the capacitor when it's charged. This energy oscillates back and forth between the capacitor and the inductor's magnetic field. The total energy (\( U \)) stored can be found using \[ U = 0.5CV^2 \], which shows that it's directly proportional to the square of the voltage and the capacitance.

In effect, the greater the voltage and capacitance, the more energy can be stored. It is crucial to point out that no energy is lost in an ideal LC circuit; it is constantly transformed from electric to magnetic and back, akin to a frictionless pendulum swinging between its potential and kinetic energy phases.
Maximum Current in LC Circuit
The maximum current that flows in an LC circuit can be thought of as the peak flow rate of charge during the oscillations. It occurs precisely when all the stored energy in the electric field of the capacitor moves to the magnetic field of the inductor. For a simple LC circuit, the maximum current (\( I \)) is given by \[ I = \frac{Q}{L \text{\textpi}} \], where \( Q \) is the maximum charge and \( L \) the inductance.

Keep in mind that this maximum current occurs at the midpoint of the oscillation cycle, where the energy transfer is most efficient. It's helpful to compare this instance to the moment when a pendulum passes through its lowest point - the instant of maximum kinetic energy. It's also significant to denote that this current would decrease in a real-world circuit due to resistance not accounted for in this ideal scenario.

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Most popular questions from this chapter

An Wectromagnetic Car Alarm. Your latest invention is a car alarm that produces sound at a particularly annoying frequency of \(3500 \mathrm{~Hz}\). To do this, the car-alarm circuitry must produce an alternating electric current of the same frequency. That's why your design includes an inductor and a capacitor in series. The maximum voltage across the capacitor is to be \(12.0 \mathrm{~V}\). To produce a sufticiently loud sound, the capacitor must store \(0.0160 \mathrm{~J}\) of energy. What values of capacitance and inductance should you choose for your car-alarm circuit?

It has been proposed to use large inductors as energy storage devices. (a) Ilow much electrical energy is converied to light and thermal energy by a \(150 \mathrm{~W}\) light bulb in one day? (b) If the amount of energy calculated in part (a) is stored in an inductor in which the current is \(80.0 \mathrm{~A},\) what is the inductance?

A solenoidal coil with 25 turns of wire is wound tightly around another coil with 300 turns (see Example 30.1 ). The inner solenoid is \(25.0 \mathrm{~cm}\) long and has a diameter of \(2.00 \mathrm{~cm}\). At a certain time, the current in the inner solenoid is \(0.120 \mathrm{~A}\) and is increasing at a rate of \(1.75 \times 10^{3} \mathrm{~A} / \mathrm{s}\). For this time, calculate: (a) the average magnetic flux through each turn of the inner solenoid; (b) the mutual inductance of the two solenoids; (c) the emf induced in the outer solenoid by the changing current in the inner solenoid.

A resistor with \(R=30.0 \mathrm{~S}\) and an inductor with \(L=0.600 \mathrm{H}\) are connected in series to a hattery that has emf \(50.0 \mathrm{~V}\) and negligible internal resistance. At time \(t\) after the circuit is completed, the energy stored in the inductor is \(0.400 \mathrm{~J}\). At this instant, what is the voltage across the inductor?

A Radio Tuning Circuit. The minimum capacitance of a variable capacitor in a radio is \(4.18 \mathrm{pF}\). (a) What is the inductance of a coil connected to this capacitor if the oscillation frequency of the \(I\) - \(C\) circuit is \(1600 \times 10^{3} \mathrm{IIz}\), corresponding to one end of the AM radio broadcast band, when the capacitor is set to its minimum capacitance? (b) The frequency at the other end of the broadcast band is \(540 \times 10^{3} \mathrm{~Hz}\). What is the maximum capacitance of the capacitor if the oscillation frequency is adjustable over the range of the broadcast band?

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