/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 26 One Inductor, Two Capacitors You... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

One Inductor, Two Capacitors You are given a \(10 \mathrm{mH}\) inductor and two capacitors, of \(5.0 \mu \mathrm{F}\) and \(2.0 \mu \mathrm{F}\) capacitance. List the oscillation fre- quencies that can be generated by connecting these elements in various combinations.

Short Answer

Expert verified
The oscillation frequencies are approximately 712.25 Hz, 1125.94 Hz, 1331.91 Hz, and 600.48 Hz.

Step by step solution

01

- Formulas for LC Oscillations

The frequency of oscillation for an LC circuit is given by the formula: \[ f = \frac{1}{2\pi \sqrt{LC}} \] where \(L\) is the inductance and \(C\) is the capacitance.
02

- Calculate Frequency with 5.0 µF Capacitor

Given the inductance \(L = 10\, \text{mH} = 0.01\, \text{H}\) and capacitance \(C = 5.0\, \text{µF} = 5.0 \times 10^{-6}\, \text{F}\), calculate the frequency: \[ f = \frac{1}{2\pi \sqrt{0.01 \times 5.0 \times 10^{-6}}} \approx 712.25 \, \text{Hz} \]
03

- Calculate Frequency with 2.0 µF Capacitor

Using the same inductance with capacitance \(C = 2.0\, \text{µF} = 2.0 \times 10^{-6}\, \text{F}\), calculate the frequency: \[ f = \frac{1}{2\pi \sqrt{0.01 \times 2.0 \times 10^{-6}}} \approx 1125.94 \, \text{Hz} \]
04

- Calculate Frequency with Combined Capacitors in Series

When capacitors are connected in series, the total capacitance \(C_{total}\) is given by: \[ \frac{1}{C_{total}} = \frac{1}{C_1} + \frac{1}{C_2} \] For \(C_1 = 5.0 \text{µF}\) and \(C_2 = 2.0 \text{µF}\): \[ \frac{1}{C_{total}} = \frac{1}{5.0 \times 10^{-6}} + \frac{1}{2.0 \times 10^{-6}} \] \[ C_{total} \approx 1.43 \times 10^{-6} \text{F} \] Using this capacitance, calculate the frequency: \[ f = \frac{1}{2\pi \sqrt{0.01 \times 1.43 \times 10^{-6}}} \approx 1331.91 \, \text{Hz} \]
05

- Calculate Frequency with Combined Capacitors in Parallel

When capacitors are connected in parallel, the total capacitance \(C_{total}\) is given by: \[ C_{total} = C_1 + C_2 \] For \(C_1 = 5.0 \text{µF}\) and \(C_2 = 2.0 \text{µF}\): \[ C_{total} = 5.0 \times 10^{-6} + 2.0 \times 10^{-6} = 7.0 \times 10^{-6} \text{F} \] Using this capacitance, calculate the frequency: \[ f = \frac{1}{2\pi \sqrt{0.01 \times 7.0 \times 10^{-6}}} \approx 600.48 \, \text{Hz} \]

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

inductor in circuits
An inductor plays a crucial role in electrical circuits, especially in LC circuits used for oscillation. An inductor stores energy in a magnetic field when electric current flows through it. Inductance, measured in henrys (H), is the property of the inductor. In an LC circuit, the inductor combines with a capacitor to create oscillations, where energy shifts between the magnetic field of the inductor and the electric field of the capacitor.
If we consider an inductor of 10 mH (which is 0.01 H), it will determine the frequency of oscillations in the LC circuit, in conjunction with the capacitance of the capacitor(s) used.
capacitors in series
When capacitors are connected in series, the total capacitance (C_total) is less than the smallest individual capacitor in the series. The formula for the total capacitance is:
\(\frac{1}{C_{total}}= \frac{1}{C_{1}} + \frac{1}{C_{2}}\)
This happens because the potential difference gets divided among the capacitors while the charge remains constant along the series. For example, if we have a 5.0 µF and 2.0 µF capacitor connected in series, the calculation for \ C_{total} will be:
\(\frac{1}{C_{total}}= \frac{1}{5.0 \times 10^{-6}} + \frac{1}{2.0 \times 10^{-6}}\)
After solving, we get:
\(C_{total} \approx 1.43 \times 10^{-6} \text{F} (or 1.43 µF)\)
This smaller capacitance will result in a higher oscillation frequency when used in an LC circuit with a fixed inductance.
capacitors in parallel
In a parallel connection, capacitors sum up their capacitances to form the total capacitance. This setup is used where higher capacitance is required. The formula to calculate total capacitance \(C_{total}\) is given by:
\(C_{total} = C_{1} + C_{2}\)
Using our given values, we can find the combined capacitance of capacitors in parallel:
\(C_{total} = 5.0 \text{µF} + 2.0 \text{µF} = 7.0 \times 10^{-6} \text{F (or 7.0 µF)}\)
With a greater total capacitance in parallel configuration, the resulting oscillation frequency in an LC circuit will be lower compared to the series configuration.
oscillation frequency calculation
The oscillation frequency of an LC circuit depends on the inductance (L) and capacitance (C). The formula to find the frequency \(f\) is:
\[f = \frac{1}{2\text{π}\text{√}(LC)}\]
Let's calculate different scenarios:
- For \(L = 0.01 \text{H}\) and \(C = 5.0 \text{µF}\):
\[f = \frac{1}{2\text{π}\text{√}(0.01 \times 5.0 \times 10^{-6})} \approx 712.25 \text{Hz}\]
- For \(L = 0.01 \text{H}\) and \(C = 2.0 \text{µF}\):
\[f = \frac{1}{2\text{π}\text{√}(0.01 \times 2.0 \times 10^{-6})} \approx 1125.94 \text{Hz}\]
- For capacitors in series (\(C_{total} \approx 1.43 \text{µF})\):
\[f = \frac{1}{2\text{π}\text{√}(0.01 \times 1.43 \times 10^{-6})} \approx 1331.91 \text{Hz}\]
- For capacitors in parallel (\(C_{total} = 7.0 \text{µF})\):
\[f = \frac{1}{2\text{π}\text{√}(0.01 \times 7.0 \times 10^{-6})} \approx 600.48 \text{Hz}\]
Understanding these principles helps in optimizing LC circuits for desired oscillation frequencies.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Sinusoidal Voltage , \(R=15.0 \Omega, C=4.70 \quad \mu \mathrm{F}\), and \(L=\) \(25.0 \mathrm{mH}\). The generator provides a sinusoidal voltage of \(75.0 \mathrm{~V}(\mathrm{rms})\) and frequency \(f=550 \mathrm{~Hz}\). (a) Calculate the rms current. (b) Find the rms voltages \(\Delta V_{a b}, \Delta V_{b c}\), \(\Delta V_{c d}, \Delta V_{b d}, \Delta V_{a d} \cdot\) (c) At what average rate is energy dissipated by each of the three circuit elements?

Capacitor Is Charging In an oscillating \(L C\) circuit, \(L=25.0 \mathrm{mH}\) and \(C=7.80 \mu \mathrm{F}\). At time \(t=0 \mathrm{~s}\) the current is \(9.20 \mathrm{~m} \mathrm{~A}\), the charge on the capacitor is \(3.80 \mu \mathrm{C}\), and the capacitor is charging. (a) What is the total energy in the circuit? (b) What is the maximum charge on the capacitor? (c) What is the maximum current? (d) If the charge on the capacitor is given by \(q=|Q| \cos (\omega t+\phi)\), what is the phase angle \(\phi ?\) (e) Suppose the data are the same, except that the capacitor is discharging at \(t=0 \mathrm{~s}\). What then is \(\phi\) ?

Magnitude of \(E\) -Field What must be the magnitude of a uniform electric field if it is to have the same energy density as that possessed by a \(0.50\) T magnetic field?

Damped \(L C\) Consider a damped \(L C\) circuit. (a) Show that the damping term \(e^{-R t / 2 L}\) (which involves \(L\) but \(\operatorname{not} C\) ) can be rewritten in a more symmetric manner (involving \(L\) and \(C\) ) as \(e^{-\pi R(\sqrt{C / L}) t T}\). Here \(T\) is the period of oscillation (neglecting resistance). (b) Using (a), show that the SI unit of \(\sqrt{L / C}\) is the ohm. (c) Using (a), show that the condition that the fractional energy loss per cycle be small is \(R \ll \sqrt{L / C}\).

Maximum Potential Difference In an oscillating \(L C\) circuit in which \(C=4.00 \mu \mathrm{F}\), the maximum potential difference across the capacitor during the oscillations is \(1.50 \mathrm{~V}\) and the maximum current through the inductor is \(50.0 \mathrm{~mA}\). (a) What is the inductance \(L\) ? (b) What is the frequency of the oscillations? (c) How much time is required for the charge on the capacitor to rise from zero to its maximum value?

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.