You want the current amplitude through a inductor with an inductance of 5.00 m H

ID: 3901638 • Letter: Y

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You want the current amplitude through a inductor with an inductance of 5.00mH (part of the circuitry for a radio receiver) to be 2.70mA when a sinusoidal voltage with an amplitude of 12.0V is applied across the inductor. Part A What frequency is required? f = Hz SubmitMy AnswersGive Up You want the current amplitude through a inductor with an inductance of 5.00mH (part of the circuitry for a radio receiver) to be 2.70mA when a sinusoidal voltage with an amplitude of 12.0V is applied across the inductor. You want the current amplitude through a inductor with an inductance of 5.00mH (part of the circuitry for a radio receiver) to be 2.70mA when a sinusoidal voltage with an amplitude of 12.0V is applied across the inductor. You want the current amplitude through a inductor with an inductance of 5.00mH (part of the circuitry for a radio receiver) to be 2.70mA when a sinusoidal voltage with an amplitude of 12.0V is applied across the inductor. You want the current amplitude through a inductor with an inductance of 5.00mH (part of the circuitry for a radio receiver) to be 2.70mA when a sinusoidal voltage with an amplitude of 12.0V is applied across the inductor. Part A What frequency is required? f = Hz SubmitMy AnswersGive Up Part A What frequency is required? f = Hz SubmitMy AnswersGive Up Part A What frequency is required? f = Hz SubmitMy AnswersGive Up Part A What frequency is required? f = Hz SubmitMy AnswersGive Up Part A What frequency is required? f = Hz SubmitMy AnswersGive Up Part A What frequency is required? f = Hz SubmitMy AnswersGive Up f = Hz f = Hz SubmitMy AnswersGive Up You want the current amplitude through a inductor with an inductance of 5.00mH (part of the circuitry for a radio receiver) to be 2.70mA when a sinusoidal voltage with an amplitude of 12.0V is applied across the inductor. Part A What frequency is required? f = Hz SubmitMy AnswersGive Up

Explanation / Answer

f = V/2pi*I*L = 12/2pi*2.7x10^-3*5x10^-3 = 1.4147x10^5 Hz

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You want the current amplitude through a inductor with an inductance of 5.00 m H

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