consider-a-circuit-consisting-of-a-lightbulb-and-a-capacitor-as-shown-in-circuit-2-of-conceptual-exa

Question

Consider a circuit consisting of a lightbulb
and a capacitor, as shown in circuit 2 of Conceptual Example 24-
10. (a) If the frequency of the generator is increased, does the
intensity of the lightbulb increase, decrease, or stay the same?
(b) Choose the best explanation from among the following:
I. As the frequency increases it becomes harder to force current through the capacitor, and therefore the intensity of the
lightbulb decreases.
II. The intensity of the lightbulb increases because as the frequency becomes higher the capacitor acts more like a short
circuit, allowing more current to flow.
III. The intensity of the lightbulb is independent of frequency
because the circuit contains a capacitor but not an inductor.

João Gabriel Alencar Caribé · Accepted Answer

in the first time thing, we have to consider the reactive capacitance off the capacitor, which is given by one divided by the frequency times the capacitance now no silly following. If we increase this frequency, their reactors will decrease. And by decreasing the reactions ah, the capacity behaves more and more like an open circuit. So what happens is that more current who flew through the capacitor and the ball will shine brighter. Because, remember, the bright news off the boat is related to the power that consumes and the power that is consumed by the boat. It&#x27;s proportional to the square off the current, then more current means more brightness in that light bulb. So again, the intensity of the light both increases as we increase the frequency off the source. Now, for the second item, we have to choose among the options. The better explanation for what you have just explained on the wound. That switch it better is the 2nd 1

João Gabriel Alencar Caribé · Answer

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&#x27;s off that in Leuchter squared under devoted and remember that the reactor&#x27;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&#x27;s Deemer now, then the intensity of the light bulb decreases.

João Gabriel Alencar Caribé · Answer

suppose that we are at the low frequency limit in the situation at that limit capacity for behaves just like unopened switch. Then no current will flow in these range and therefore the resistance of the circuit will be equals. Two are no supposed that In the second situation we are at the high frequency limit. In that limit, the capacitor acts just like on idea wire. So the current will flow both through the offer for the middle range. And there is no capacitor here sings. It behaves like an idea wire. Then the current faces resistance that is equivalent to our over, too. And therefore, as can be seen, the resistance in the high frequent situation is the lowest one, and then the current will be the highest in the high frequency situation. For the second item, the option that better sweets or explanation is the option number three

Ze-Han Lee · Answer

right. So, uh, in this problem, So basically we have Ah, we have a lighting moment here. So there&#x27;s the battery if we want to increase the intensity. So that means we want to increase the power. Okay, then we need to lower the resistance off this lighting bolt. Okay, Because this is Avi. This is our If art is lowered to say the power on the lining Mom is ah, it becomes larger, right becomes larger, so that intensity becomes larger. Only equipment all equivalent me. You can say that to the current through the lining moment snatcher. Eso this expression, you can also re express it as, uh, a&#x27;ight times V. Right. So Ah, because Ari, smaller than the current eyes, larger eyes larger and B&#x27;s constant. So peace lodger. Okay.

João Gabriel Alencar Caribé · Answer

in the first item. We are working in the low frequency limit. So we have low frequencies. And in that limit, a capacitor behaves like an open switch. Why you on in? Doctor behaves like an idea wire. So in that limit, the given circuit is equivalent to the following one. So we have this source and here we have a resistance and open circ. Wait. Now we have an idea wire on a resistance. So the current on that circle, it will flow through the following path. Then it will only meet a resistance are here. Therefore the current in that situation will be given by V. I remember. Is it closed? Why are a mass times the resistance So the current is equals to the voltage divided by the resistance? On the other hand, when we are at the high frequency limits, a capacitor behaves just like your wire on the doctor behaves just like unopened switch. Then in the high frequency limit. The given circuit is equivalent to the following one. We have the source. We have our existence. Idea why you&#x27;re I need your help. Your brain. JJ, we have Loken switch in the resistance. Then in the second situation at the high frequency limit. The path that the current follows is the following. Therefore it only meets again. Resistance are so again the current will be given by the voltage divided by the resistance. So in both limits, off low three princes and off high frequencies the current will be the same No as we have just seen. The best explanation for the second item is the 1st 1

Ben Nicholson · Answer

31.56 were considering one possible way of doing light. This is evidently used in theaters for dimming stage lights, the, uh, more typical light dimmer circuits for things in your house or more complicated in some ways. Then, uh, then this. Although they are cheaper to produce, it uses a solid state, uh, component that we, you know, don&#x27;t worry about right now because that&#x27;s not what this problem is, Anyway, So are voltage supply. Here is 100 and 20 volts are a mess at a frequency of 60 hertz, and our light bulb is rated for 120 volts. So that&#x27;s good. We aren&#x27;t going to make it explode. Uh, and its power rating is one kilowatt. So first we want to find what the maximum inductive sze of this in doctor needs to be in order for the minimum power in this light bulb to be 2200 watts, 1/5 of its maximum value. We want to know if you could, uh, use a resistor instead of a doctor, and if so, what is its maximum value? And then we want to discuss why you wouldn&#x27;t actually use a resistor. So the power in like the bowl be is that its current squared times its resistance that you&#x27;re told doesn&#x27;t change with temperatures. We don&#x27;t need to worry about that. That would make the problem of not not that much more complicated, but it would be enough to be obnoxious and scared the point of what we&#x27;re trying to do here. Um, this is the r. M s power. So this is the r. M s current. This is the R. M s power. Um, if you wanted to use the peak for one of them or both of them and you get a factor of the square root of two but that&#x27;s going to cancel out because we want he macks overpay men, which is going to be I over. I&#x27;m in squared to be equal to five. And so the you know, whatever factors we have to that account for whether you&#x27;re using the maximum or the R M s voltage ends up the end of cancelling out, it doesn&#x27;t matter. So in each case, the current it&#x27;s going to be the e m f divided by the students. So you have the maximum power with the minimum impedance and the minimum power with the maximum Eaton&#x27;s And then these cancel out. So we get C max over Zemun Squared. So just rewriting this in this new page so that we don&#x27;t lose track of what we&#x27;re doing. The max over a seamen quantity squared. So the light bulb has resistance of our. So the maximum impedance is going to be r squared us omega el max, which is what we want to find squared square root of that because their penises ad in this way or the magnitude do, at least and then the minimum impedance is just the resistance of the light Old this needs to be fine, like we said, and so we so for L. Max, this is too are over omega. Now we know that the hour the light bulb will dissipate is 1000 watts. And with our equals, I squared R implies that, um or no, actually, the more useful one is, uh, our equals. He squared over r. So are is going to be our voltage squared. Divided by the power. This will be 100 and 20 volts squared, divided by 1000 watts and then two pi times. Our frequency gives us the angular frequency. 60 hertz. So we&#x27;re probably in North America. A lot of the rest of the world uses 50 hertz. Yeah, so that&#x27;s 7.64 times 10 to the negative to Henry&#x27;s. So are in. Doctor needs to be variable from zero Henry&#x27;s to this. So for part B, um, can you use a resistor? The answer, of course, is yes. There&#x27;s, uh the same logic works for a resistor. You increase the total resistance of the circuit or its impedance, which you could do with a resistor, and that reduces the current and so reduces the power being dissipated by delightful. So let&#x27;s figure out what value of the what? Well, what we need for the maximum value of this resistor. So using what we already did, we can write this as, uh, are Max. Plus, are the resistance of the bulb divided by the resistance of the bulb squared equals five. Now what we want for now, we saw for our mats we get square root of five minus one times the resistance of the bulb tree already knows our 120 fold squared, divided by 1000 locks if you want. You could work out with that, Actually, is it, homes? But you could also just played those numbers indirectly, like we did before. We find that it needs to have a resistance of 17.8 homes. Now, why don&#x27;t you do this? Uh, because, uh, the this resistor would get very hot because it&#x27;s basically turning energy into heat, as opposed to storing it in a magnetic field. Liken, induct. Er does. So you&#x27;re not only sort of wasting power, you&#x27;re also, um, making life more difficult for yourself because now you have toe figure out some way of keeping everything from overheating, which is, uh, sort of a serious problem in lots of electrical engineering problems involved. Large amounts of power is how do you get the heat that&#x27;s inevitably generated, You know, out of there. And so if you could just avoid making those problems for yourself in the first place, you really want to

Problem

Consider a circuit consisting of a lightbulb and …

Problem 73 Medium Difficulty

Answer

a) see explanation
b) see explanation

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