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Chapter 30: Problem 34

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 inductor with an inductance of \(2.50 \mathrm{II}\) and a resistance of \(8.00 \Omega\) is connected to the terminals of a battery with an emf of \(6.00 \mathrm{~V}\) and negligible internal resistance. Find (a) the initial rate of increase of current in the circuit; (b) the rate of increase of current at the instant when the current is \(0.500 \mathrm{~A}\) : (c) the current \(0.250 \mathrm{~s}\) after the circuit is closed; (d) the final steady-state current.

Solar Magnetic Energy. Magnetic fields within a sunspot can be as strong as 0.4 T. (By comparison, the earth's magnetic field is about \(1 / 10,000\) as strong. ) Sunspots can be as large as \(25,000 \mathrm{~km}\) in radius. The material in a sunspot has a density of about \(3 \times 10^{-4} \mathrm{~kg} / \mathrm{m}^{3}\). Assume \(\mu\) for the sunspot material is \(\mu_{0}\). If \(100 \%\) of the magneticfield energy stored in a sunspot could be used to eject the sunspot's material away from the sun's surface, at what speed would that material be cjected? Compare to the sun's escape speed, which is about \(6 \times 10^{5} \mathrm{~m} / \mathrm{s} .\) (Hint: Calculate the kinetic energy the magnetic field could supply to \(1 \mathrm{~m}^{3}\) of sunspot material.)

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?

CP An altemating-current clectric motor includes a thin. hollow, cylindrical spool (similar to a ring) with mass \(M=1.11 \mathrm{~kg}\) and radius \(a=5.00 \mathrm{~cm}\) wrapped \(N=500\) times with a copper wire with resistance \(R=5.00 \Omega\) and inductance \(L=77.0 \mathrm{mH}\). Within the spool is a battery that supplics current \(I=1.00 \mathrm{~A}\). which makes the spool a magnetic dipole with dipole moment \(\overrightarrow{\boldsymbol{\mu}}\) parallel to the cylinder axis. A constant magnetic field with magnitude \(B=2.00 \mathrm{~T}\) is supplied by an extemal stator magnet, while the spool turns freely on an axis perpendicular to its own axis. At a certain time, a bar is inserted, stopping the spool's motion (Fig. \(\mathrm{P} 30.51\) ). At that instant the angle between the spool axis and the magnetic field is \(\theta=45^{\circ}\). (a) What is the magnitude of the downward force \(\vec{F}\) applicd by the bar onto the spool immediatcly after the bar is inserted? (b) Later, at time \(t=0\) with spool still at rest, the cril is short-circuited and a constant counter- toryue \(\tau=0.500 \mathrm{~N} \cdot \mathrm{m}\) is applied. The current subsides, and the magnetic torque decreases exponentially. At what time \(t\) does the force applied by the bar vanish'? (Hint: Determine when the magnetic torque balances the counter-torque.) (c) After the spool rotatcs \(180^{\circ}\) it hecomes stuck on the top side of the bar. The counter-torque is no longer applied, and the switch is returned to its original position. After a long time, the bar is removed. What is the angular acceleration of the spool immediately after the bar is removed? The moment of incrtia of the spool for an axis along its diamcter is \(I=\frac{1}{2} M a^{2}\).

CP CALC A Coaxial Cable. A small solid conductor with radius \(a\) is supported by insulating, nonmagnetic disks on the axis of a thin-walled tube with inner radius \(b\). The inner and outer conductors carry equal currents \(i\) in opposite directions. (a) Use Ampere's law to find the magnetic field at any point in the volume between the conductors. (b) Write the expression for the flux \(d \Phi_{B}\) through a narrow strip of length \(I\) parallel to the axis, of width \(d r\), at a distance \(r\) from the axis of the cable and lying in a planc containing the axis. (c) Intcgrate your expression from part (b) over the volume between the two conductors to find the total flux produced by a current \(i\) in the central conductor. (d) Show that the inductance of a length \(/\) of the cable is $$ I_{.}=l \frac{\mu_{0}}{2 \pi} \ln \left(\frac{b}{a}\right) $$ (e) Use Eq. (30.9) to calculate the energy stored in the magnetic field for a length \(l\) of the cable.
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