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An array of solar panels produces 9.45 A of direct current at a potential difference of 192 $\mathrm{V}$ . The current flows into an inverter that produces a $60-\mathrm{Hz}$ alternating current with $V_{\max }=171 \mathrm{V}$ and $I_{\max }=19.5 \mathrm{A}$ (a) What rms power is produced by the inverter? (b) Use the rms values to find the power efficiency $P_{\text { out }} / P_{\text { lin }}$ of the inverter.

a) 1.67 $\mathrm{kW}$b) 0.923

Physics 102 Electricity and Magnetism

Chapter 24

Alternating-Current Circuits

Current, Resistance, and Electromotive Force

Direct-Current Circuits

Electromagnetic Induction

Alternating Current

Cornell University

Rutgers, The State University of New Jersey

University of Washington

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the first item off this question asked us to cooperate. What is the average are amass power. Produce it by the inverter. The average power producer by the inverter is equals to our times. I are a mess squared. We begin by calculating the resistance off that circuit. So by using arms law V is equals toe are times I solving for the resistance. We get that our is the coast movie over I then plugging the values for the maximum current and the maximum voltage we get that the resistance is equals to 171 or her 19.5. And no, we get that the average power is equals to 107 to 1, divided by 19.5 times. I are a mess to get that I wear a mask current. We have to do the following. So remember that the i r. A. Mast is it close to the maximum current divided by the square root of two. Then week unplugging these equations in our equation for the power to get that these easy goes to 107 to 1 divided by 19.5. Thanks. I'm Max squared divided bite you and finally bringing their value off the maximum current we get 107 21 times 19.5 squared, divided by two times 19.5. Finally we get that. The average power is he goes to 100.6 to 7 kilo walks. The second item. We have to evaluate the efficiency of the inverter. So the efficiency is given by the power output divided by the power input off the inverter. We already know the power output. It Z equals to 1.627 kilo. What's so the efficiency is equals to 1.6 to 7, divided by the power input. The power input is very simple to copulate in his ego's, too. The input current times, the input voltage And these are these values that were given in the question. So by plugging these values in the equation for the input power, we get that these these equals to 9.45 times 192 and these is equals true, one 0.8 to 1. What off it With power pointing, then you put power in our equation for the efficiency we get 1.67 divided by 1.8 to 1, and this is approximately 92.3% off efficiency.

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