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Consider a circuit consisting of a lightbulb and an inductor,as shown in Conceptual Example 24-14. If the frequency of thegenerator is increased, does the intensity of the lightbulb increase,decrease, or stay the same? Explain

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Physics 102 Electricity and Magnetism

Chapter 24

Alternating-Current Circuits

Current, Resistance, and Electromotive Force

Direct-Current Circuits

Electromagnetic Induction

Alternating Current

Cornell University

Simon Fraser University

University of Sheffield

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The brightness of the light bulb is related to the average power that consumes, which is even by the R. M s current squared times. The resistance off that light bulb we are modeling a lightbulb essentially as a resistor. No, Our question is the following. By increasing their frequents off the surface, what happens to the R. M s current in that circuit? Towards for that question, we have to remember that the remans voltage is it cost to the R. M s current times the impudence. Now we can solve for the currents to get the following the current Is it close to the voltage divided by the impedance? And now remember that in this case we have any doctor and the resistance. So the impudence is given by the square it off the resistance a squared plus the reactor's off that in Leuchter squared under devoted and remember that the reactor's off. Any doctor is given by the frequency, multiply it by the victims. So by increasing the frequency, we are increasing, the reactors have been doctor. So we are increasing the denominator off these equation and because of that, we are decreasing the RMX currents that flow for that circuit, and as a consequence, we are also decreasing the average power consumed by the light bulb, which results in the light bulb that do not shine as motors use it too. So it's Deemer now, then the intensity of the light bulb decreases.

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