ω =120 rad/sT =5.26×10−2 sQ =6.00×10−4 C initial charge part E time 0.0250 s A capacitor with a capacitance 5.00 × 10-5 F ischarged by connecting it to a 12.0 V battery. Thecapacitor is then disconnected from the battery andconnected across an inductor with L = 1.40 H.For related problemsolving tips and strategies, youmay want to view a Video Tutor Solution of Anoscillating circuit.The minus sign means that the capacitor has discharged fully and then partially charged again by themaintained by the inductor; the plate that initially had positive charge now has negative charge and theinitially had negative charge now has positive charge.Part GAt the time given in part E, what is the current in the inductor?Express your answer with the appropriate units.I =μAValueSubmitAPrevious Answers Request Answer?

Given:Max charge on capacitor, Qo = 6.00×10−4 Cangular frequency, ω = 120 rad/sMax voltage on the…

Answered: A capacitor with a capacitance 5.00 ×…

A capacitor with a capacitance 5.00 × 10-5 F is charged by connecting it to a 12.0 V battery. The capacitor is then disconnected from the battery and connected across an inductor with L = 1.40 H.

Principles of Physics: A Calculus-Based Text

5th Edition

ISBN:9781133104261

Author:Raymond A. Serway, John W. Jewett

Publisher:Raymond A. Serway, John W. Jewett

Chapter23: Faraday’s Law And Inductance

Section: Chapter Questions

Problem 7OQ

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Question

ω =120 rad/s

T =	5.26×10⁻² s

Q =	6.00×10⁻⁴ C initial charge

part E time 0.0250 s

A capacitor with a capacitance 5.00 × 10-5 F is
charged by connecting it to a 12.0 V battery. The
capacitor is then disconnected from the battery and
connected across an inductor with L = 1.40 H.
For related problemsolving tips and strategies, you
may want to view a Video Tutor Solution of An
oscillating circuit.
The minus sign means that the capacitor has discharged fully and then partially charged again by the
maintained by the inductor; the plate that initially had positive charge now has negative charge and the
initially had negative charge now has positive charge.
Part G
At the time given in part E, what is the current in the inductor?
Express your answer with the appropriate units.
I =
μA
Value
Submit
A
Previous Answers Request Answer
?

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