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A parallel-plate capacitor having square plates 4.50 $\mathrm{cm}$ on each side and 8.00 $\mathrm{mm}$ apart is placed in series with an ac source of angular frequency 650 $\mathrm{rad} / \mathrm{s}$ and voltage amplitude $22.5 \mathrm{V},$ a 75.0$\Omega$ resistor, and an ideal solenoid that is 9.00 $\mathrm{cm}$ long, has a circular cross section 0.500 $\mathrm{cm}$ in diameter, and carries 125 coils per centimeter. What is the resonance angular frequency of this circuit? (See problem 36 in Chapter $21 . )$

$w_{o}=3.95 .10^{7} \mathrm{rad} / \mathrm{s}$

Physics 102 Electricity and Magnetism

Chapter 22

Alternating Current

Current, Resistance, and Electromotive Force

Direct-Current Circuits

Electromagnetic Induction

Tom C.

April 19, 2021

A transformer connected to a 120-V (rms) ac line is to supply 12.0 V (rms) to a portable electronic device. The load resistance in the secondary is 5.00 LaTeX: \Omega

Cornell University

University of Michigan - Ann Arbor

University of Washington

McMaster University

Lectures

04:44

Alternating current (AC) is an electric current which periodically reverses direction, in contrast to direct current (DC) which flows only in one direction. Alternating current is the form in which electric power is delivered to businesses and residences, and it is the form of electrical energy that consumers typically use when they plug electrical appliances into a wall socket. A common source of DC power is a battery cell in a flashlight. The abbreviations AC and DC are often used to mean simply alternating and direct, as when they modify current or voltage.

11:31

In electrical engineering, a direct current (DC) circuit is an electrical circuit operating with a constant voltage (or current), as opposed to alternating current (AC) circuits. Direct current may flow in a conductor such as a wire, but can also flow through semiconductors, insulators, or even through a vacuum as in electron or ion beams.

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A parallel-plate capacitor…

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31.41 -- CP A parallel-pla…

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The circuit shown has a so…

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A parallel plate capacitor…

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The circuit shown has a

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A series circuit consists …

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31.35. A series circuit co…

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For the circuit in Figure …

you see, here we have the self induct INTs of a solenoid. This but actually be apologies. This would be actually l so L equals the Premier Bill. The magnetic permeability of free space multiplied by the area of the still annoyed that the solenoid is encompassing rather multiplied by the total number of turns squared, divided by l the length we have a which would be the length of each side of the parallel plate of a capacitor. And we have to convert this two meters. Now we can also say d would be the separation between the parallel plates. This would be eight millimeters or again 80.8 meters. We have the radius of the coil, are equal in the diameter of the coil, divided by 2/2 a centimeter, divided by two equaling 20.25 meters. And we have the length of the entire solenoid at nine centimeters, Alice equaling points their own 90 meters. Additionally, we have the number of Terrence per unit length, so we have 125 recoils per centimeter. We can convert this into 12,500 coils per meter and we have the total number of turns equal in the number of turns per unit length multiplied by the entire length of the soul annoyed. L. We have the area that the coil is encompassing pi r squared that is simply the area of a circle from this week and then plug it into the huh equation for the self inducting ts of a still annoyed and big and solved. So we have the self inducting ts of so annoyed equaling again new, not a and squared, divided by l. This would be equal in two new not multiplied by pi r squared, multiplied by and most part by Elle's quantity squared, divided by l and this is simply equaling knew not pie r squared and squared l and so we can solve Ella's equaling You knocked pie these air simply Constance, you not equaling, of course, four pi times 10 to the negative seventh Tesla's meters per amps and we kit will be multiplying. This rather by 0.25 meters multiplied by another quantity squared multiplied by 12,500 coils per meter. Quantity squared multiplied by 0.9 meters, and we find that the self induct in CE of the solenoid is 3.46 times 10 to the negative fourth Henry's. So this would be your answer for the self induct its of the solenoid, the capacitance of the parallel plate capacitor. In this case, see would be equaling two. Absolutely not. The electric primitive e multiplied by a the area divided by D and in this case we can say that the areas is not pi r squared, but rather a squared for the square plates. And so this would be equaling two Absolutely not a squared divided by D and this is equaling the electric primitive ity, which is 8.8 to 5 times 10 to the negative 12th. This is Newton's meters squared per Coolum squared multiplied by 0.45 meters. My apologies for it being cramped and this would be divided by D of 0.8 meters and we find that the capacitance see is equaling 2.23 times 10 to the negative 12th ferrets. This would be your answer for the capacitance. See, And finally we can solve for the and the residents angular frequency. Given that we have the induct its and the capacitance now, so this would be one over the square root of L. C. This is equaling one over the square root of, uh, 3.46 times 10 to the negative. Fourth Henry's multiplied by 2.233 Go around at the very end times 10 to the negative 12 Fareed's and we find that the resident's frequency is equaling 3.6 times, 10 to the Seventh radiance for a second. This is our final answer for the residents. Angular frequency. That is the end of the solution. Thank you for watching.

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