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An air conditioner connected to 120 rms ac line is equivalent to 04 9 resistance and 1.72 $ inductive reactance in series_ Calculate (a) the impedance of the air co...

Question

An air conditioner connected to 120 rms ac line is equivalent to 04 9 resistance and 1.72 $ inductive reactance in series_ Calculate (a) the impedance of the air conditioner and (b) the average rate at which energy is supplied to the appliance_(a) NumberUnits(b) NumberUnits

An air conditioner connected to 120 rms ac line is equivalent to 04 9 resistance and 1.72 $ inductive reactance in series_ Calculate (a) the impedance of the air conditioner and (b) the average rate at which energy is supplied to the appliance_ (a) Number Units (b) Number Units



Answers

An air conditioner connected to a 120 $\mathrm{V} \mathrm{rms}$ ac line is equivalent to a 12.0$\Omega$ resistance and a 1.30$\Omega$ inductive reactance
in series. Calculate (a) the impedance of the air conditioner and
(b) the average rate at which energy is supplied to the appliance.

In the first item. We have to complete the impedance off that circuit. Then Peters is given by the square root off the resistance squared. Plus, they took the reactant squared and it Z goes to the square root off seven squared plus 15 squared. This is approximately 16.6 rooms, more 17 homes, making a further approximation to complete the arguments current that flows from that circulate. In the second item, we use a form off on law, which is a following. The voltage is equals to the current. Multiply it by it, impede ums. Then the current Zico stood involved, fish divided by the impudence, plugging the given value from the problem. On what we have just calculated, we get a current off 240 divided by 16.6, which is approximately 14.46 M peers, which can defer their approximated to 14 amperes in the last item. We have to complete the average power consumers by the air conditioning system, and it's given by the IRA man's voltage times. They are a mass current times the power factor, which is the resistance divided by an impedance, then, like in the values that were given by the problem and that we have calculated we get that. The average power as equals to 240 times 14.46 times seven divided by 16.6, which is approximately 1463.2 What's which can be further approximated to 1.5 kilowatts.

For this problem on the topic of a. C. Circuits. We have a 100 ohm resistor being connected across 100 and 20 volts RmS 60 hertz A. C. Power line. Using the information, we want to find the RmS current, The peak current as well as the average power that is dissipated in the resistor. Now for part A we know that the RMS voltage is equal to the RmS current times the resistance are which is owned law. And so the RMS current therefore is the RMS voltage divided by the resistance. And since we have both these values, if we put them in this is 120 balls Divided by the resistance, 100 homes which gives us the RMS current To be 1.2 and piers. Now we want to find the peak current and we know the peak current. I max is simply I not which is the RMS current Multiplied by the square root of two and this is 1.2 M. P. S. Times the square root of two which gives an RMS current of 17 and piers. And finally for for part C we have the average power dissipated to the resistor to be the RMS current times the RMS voltage. And this is 1.2 amperes times the RMS voltage 120 volts Which gives the average power to be 140 watts

In the first item. We have to evaluate the power factor of the circuit. So we have to evaluate the co sign other fees and the music goes to the resistance divided by BNP. This and in this case, the impudence has two contributions, one coming from the resistance and another one coming from they looked at us. So the feelings is it goes to the square it off squares plus the inductive reactors. Where'd they looked? Reactant is given by two quiet times the frequency times, the induct ins. No, we can plug in the values given by the problem in this equation and get the power factor. We choose the square root off two times, stand to the turns. There is a kilo there squared plus to pi times 1.55 Again there is a kilo times 292 merely squared. This is old under the resistance reaches two times 10 to return and these things approximately 0.5 seven 52 which we can for the approximate to 0.575 in the next item we have to complete. What is the average power consumers by that stir Great the average power is given by the are a mess voltage times the arguments current times the power factor. Now remember that there is a relation between the aromas voltage on the Aramis current and that relation is the following. The remans voltage is of course, to ari mass current. Multiply it biting Peters, then solving for the R. M s current, we get that the Aramis current is equals two The aromatase voltage divided by eating peanuts. No, the average power is then the remans voltage squared divided by they impede us and multiply it by the power factor and disease equals truth. A square root off the resistance is squared. Plus to buy f l squared under a square root under the aromatics voltage squared times the power factor This is just impedance which we have already calculated here. So we know just have to put in their values given by the problem on the power factor that we have completed before To get the following square root off two times 10 to the third squared plus to five times 1.55 time stand to deterred times 292 times 10 to the minus tree squared This is all under 20 points. Eight squared times the power factor, which is 0.575 and disease approximately 0.715 five, which we can write as 71.6. Really, What?


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