Filter circuit. Figure 30–33 shows a simple filter circuit des...

Question

Filter circuit. Figure 30–33 shows a simple filter circuit designed to pass dc voltages with minimal attenuation and to remove, as much as possible, any ac components (such as 60-Hz line voltage that could cause hum in an audio system, for example). Assume V_in = V₁ + V₂ where V₁ is dc and V₂ = V₂₀ sin ωt, and that any resistance is very small. (a) Determine the current through the capacitor: give amplitude and phase (assume R = 0 and X_L > X_C). (b) Show that the ac component of the output voltage, V_2out, equals (Q/C) - V₁ where Q is the charge on the capacitor at any instant, and determine the amplitude and phase of V_2out (c) Show that the attenuation of the ac voltage is greatest when X_C << X_L, and calculate the ratio of the output to input ac voltage in this case. (d) Compare the dc output voltage to input voltage.

Accepted Answer

Welcome back everyone in this problem. A water filtration system treated as circuit is designed to allow clean water, the DC component to pass through with minimal loss while efficiently filtering out contaminants components that fluctuate over time. The input water consists of a mixture of clean water and contaminants where the clean water V one represents the steady desired component that needs to be passed through the filter. And the contaminants V two represent the fluctuating impurities such as dirt and chemicals that need to be removed. Where V two is the sound of Omega T with an amplitude and angular frequency omega. The filtration system consists of two key components. It has a coarse filter analogous to inductor. L sometimes I struggle with that word a little bit and it resists the flow of large contaminants effectively filtering out high frequency contaminants. It also has a fine filter analogous to capacitor C where the fine filter allows fine water to pass through while blocking smaller contaminants. The system is designed with the following assumptions that there is negligible resistance and the inductive reactance is greater than the capacity of reactant. The goal is to determine the behavior of the filtration system, particularly focusing on the following four aspects. First, the water flow through the filters, we want to determine the amplitude and phase of the water flowing through the system. And this will involve calculating the impedance of the system and the resulting current. Next, the output water quality. We want to show that the ac component of the output water quality is given by Q divided by C minus V. One, we accuse the charge accumulated on the fine filter and V one is the DC component or clean water. We want to determine the amplitude and phase of the AC component of the output contaminants. Third, we want to find the attenuation of contaminants. We do that by demonstrating that the attenuation of the contaminants or the AC components is greatest when the inductive reactance is much greater than the capacity reactants. We also want to calculate the ratio of the output to input AC voltage under this condition. And finally, the output, clean water compare the DC component, the clean water in the output with the DC component in the input. And we want to discuss how the system behaves with respect to steady water flow. Now, let's start here. You notice we also have a diagram showing our setup but let's start with the first part of our problem. We want to find the water flow through the filters. OK. Now, to do that, OK. Let's just list this as a here. We want to determine the amplitude and phase of the water flowing through the system. And the first thing we need to do is to find the impedance of the system. Now, since the system consists of an inductor and a capacitor with negligible resistance, then the impedance, yeah, the impedance Z is equal to the inductive reactance minus the capacitive reactance. Now, the current through the system, OK. That is representing which would represent the flow of the water. OK? By Ohm's Law, OK. That current would be equal to be not divided by XL minus one divided by Omega C. OK. Multiplied by the sine of Omega T minus a half of pi. And now, from my expression here, OK, we can tell that the current will have an amplitude, the current will have an amplitude that is our coefficient of the sine term. OK. So it's gonna have an amplitude here of V knot divided by XL minus one divided by Omega C and the phase of the current relative to the input voltage will be a half of pi since the inductor causes the current la voltage by a half of pi. So that means then that the amplitude of the water flow which represents our current will be Z not divided by XL minus one divided by Omega C and the phase shift. OK is equal to a half of pi meaning the current that is the water flow lacks the ac component of the input voltage by a half of pi. So let's just highlight both of our answers here to part A. Next, let's move on to part B of our problem. We want to find the output water quality that is the contaminant. We need to show that the output output contaminants, the AC component are given by the charge divided by the capacitance minus V one. Now, let's start to think about these quantities here. Now, when it comes to the current and charge and the capacitor, we know the current through the system represents the rate of change of charge. And the capacitor, we already have an equation for our current. So using our equation, the charge QT OK accumulated on the capacitor is going to be the integral of the current with respect to time substituting our expression. That means it will be the integral of V knot divided by XL minus one divided by Omega C multiplied by the sine of Omega T minus a half of pi with respect to T. Now, the integral of the sine of omega T minus a half of pi is a negative value of one divided by omega multiplied by the cosine of omega T minus a half of pi. So no, this tells us then that our charge with respect to time the child accumulated with respect to time equals the negative value of V knot divided by Omega multiplied by XL minus one divided by Omega C multiplied by the cosine of Omega T minus a half of pi. Now, the total output voltage is the sum of the clean water, the DC component V one and the contaminants, the AC component. However, the ac contaminants are what we're interested in. So the output contaminants are the ac voltage across the capacitor. And if we substitute or charge into the expression that we had earlier, OK, then you get V two to be equal to the negative value of V not divided by Omega C. OK. Multiplied by XL minus one, divided by Omega C OK. Multiplied by the cosine of Omega T minus a half of pi OK minus V one. No. Since the cosine of Omega T minus a half of pi equals the sine of Omega T, then that means this will be equal to the negative value of the knot divided by Omega squared LC minus one multiplied by the sine of Omega T. OK. And this is the expression for the output output contaminants. Now, the negative sign in the equation is eliminated. OK? Because the magnitude of the voltage is considered and we focus on the absolute value of the output AC voltage. OK. So that means no, then OK, that basically we're gonna take this expression, I'm just going to copy it in inches of time. And now we know that the AC two out then is going to be equal to the positive value of this expression. Uh doesn't seem like I have much space here. So I'll need to write it out. OK. And now, from this expression, we can tell that the amplitude of the output AC voltage, the amplitude of the output AC voltage is V not divided by omega squared LC minus one. OK. And the phase, the fears of the output AC voltage is the same as the source. So what does that tell us? Then it is the sine of omega T And that means the phase of the output voltage is identical to the input AC component with no phase shift. Thus, both the amplitude and phase of the output AC voltage are the same as the source. But the amplitude is scaled by a factor of one divided by Omega squared LC minus one. Let's move on to the third part of our problem. We want to find attenuation of contaminants. OK. Now let's put that in blue here. Now, the attenuation to find the attenuation, we need to show that the, that it is maximized when the inductive reactance is much greater than the capacity reactant and then calculate the ratio of the output to input a voltage. So we know that at our attenuation condition is that VA two out divided by V AC two N is our attenuation. And from the expressions we derived earlier, that's going to be VNO divided by Omega squared LC minus one all divided by VNO. And now when we work that out, we get it to be one divided by omega squared LC minus one and no, the attenuation is greatest when the denominator is large, that we can tell from our expression. So now that occurs when Omega squared LC is much larger than one, which tells us if we do some transposition here, Omega L or inductive reactants, this has to be much greater than one divided by Omega C which is our capacitive reactance. So that means then that XL has to be much greater than XC. Our condition is satisfied. This shows that the attenuation is greatest when XL is much greater than XC. Now, for the final part of our problem D, we want to compare the output clean water component to the input clean water. Since the inductor does not drop any DC voltage, the clean water will pass through the system unaffected. Therefore, the output DC voltage is going to be the same as the input DC voltage. The clean water DC component is not attenuated by the system. So to summarize our solution, the output clean water DC component is not attenuated by the system. OK. Uh The attenuate to show the attenuation of contaminants here, we know that that is going to be one divided by omega squared LC minus one. Next, we also know that uh or let me just get that point here. We also know that the output water quality is going to be equal to V not divided by omega squared LC minus one multiplied by the sine of Omega T. And finally, we also had an expression for amplitude and phase shift for the water flow through the filters. And we had our expression here as well. Thanks a lot for watching everyone. I hope this video helped.

Key Concepts

Capacitive Reactance (XC)

Charge on a Capacitor (Q)

Voltage Attenuation

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