The RC circuit shown in Fig. 30–39 is a low-pass filter becaus...

Question

The RC circuit shown in Fig. 30–39 is a low-pass filter because it passes low-frequency ac signals with less attenuation than high-frequency ac signals. (a) Show that the voltage gain is A = Vₒᵤₜ/Vᵢₙ = 1/ (4π²f²R²C² + 1)¹/² (b) Discuss the behavior of the gain A for f → 0 and f → ∞.

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Accepted Answer

Welcome back everyone in a lab experiment. A physics student is tasked with constructing an RC low pass filter using a resistor and a capacitor. The resistor has a value of one ohm and the capacitor has a value of two micro fats. The student needs to determine the voltage gain given by a equals V not divided by V as a function of the frequency F what is the correct expression for the voltage gain? And we want to also determine the voltage gain for a frequency of 200 kilohertz. Here we have a diagram of our circuit. And for our answer choices. A says that the voltage gain equals one divided by the square root of two pi F squared. And when F equals 200 klotz, the voltage gain is 0.37. B says it's one divided by the square root of two pi F squared and 0.15 C says it's one divided by the square root of 1.6 multiplied by 10 to the negative 10th second squared multiplied by F squared plus one. And it's 0.37 and D says it's one divided by the square root of 1.6 multiplied by 10 to the negative 10th second squared multiplied by F squared plus one. And that is 0.15. No, we're given quite a bit of information here. OK. We, we, we're given the value of our resistor as one home, the value of our capacitor to micro Frates. And now we need to figure out what our voltage gain is. The expression for our voltage gain and the value of our voltage gain if we're given our frequency of 200 k. So let's start with the first part of our problem. What is the correct expression for the voltage gain? Now, so far notice that for our RC low pass filter, the voltage gain A is equal to our output voltage divided by our input voltage. What do we know about our output voltage and input voltage? Well, we know that the output voltage is taken across the capacitor terminals. So vo equals the current multiplied by the capacity of reactants. Now, we also know that for our input voltage, the current is going to be equal to the input voltage divided by the square root of R squared plus XC squared. Now here if we substitute the value of our current I into our formula for our output voltage, eventually we will be able to solve for the ratio of V to V I which is the voltage gain. So let's go ahead and try to do that. So Now, by substituting I into V, that means V equals V divided by the square root of R squared plus XC squared, multiplied by XC. Now here, if we think about it, if we divide through by XC, OK, then our output voltage equals V. OK. Divided by the square root of R divided by XC or squared. OK, plus one. Now what do we know about XC or capacitive reactants? Well, we know that the capacity reactance is equal to one divided by Omega C. OK. Where Omega is equal to two pi F. OK. And remember we want to get an expression for our voltage gain in terms of the frequency because it's written as a function of the frequency F So now if we substitute this, OK? Because if this is our capacity reactant that tells us OK, that our capacity reactant equals one divided by two pi FC. If we substitute that, let me just clean this up a bit. OK. If we substitute that into our formula for the output voltage, that means our output voltage then is gonna be equal to our input voltage. I'm sorry, our ratio. And then let me simplify it from no the ratio of our output voltage to our input voltage. If we divide through by our input voltage is going to be equal to one divided by R multiplied by the reciprocal of XC. OK. So that's the square root of R multiplied by two pi FC or you can write that as two pi FRC all squared, OK plus one. Now here, if we expand our square term, that means VO to V I is going to be equal to one divided by the square root of four pi squared, F squared R squared C squared plus one. Now remember what is V or divided by V I, that's our voltage game. So we could replace this here with A OK. And we can substitute the values that we know because no, it's gonna be a big fraction here. But no, that means A is gonna be equal to one divided by the square root of four pi squared F squared multiplied by R which is one ohm squared multiplied by C which is two micros, two multiplied by 10 to the negative sixth fats squared plus one. And now when we go ahead and simplify, then we get a to be equal to one divided by the square root of 1.579 multiplied by 10 to the negative 10th second squared or to two significant figures, 1.6 multiplied by 10 to the negative 10th seconds squared multiplied by F squared plus one. Thus, this is an expression for our voltage gain as a function of our frequency. No, remember we said that we wanted to figure this out when the second in sorry. In the second part of our problem, we wanted to figure this out when our frequency is 200 kilohertz. So let me just quickly make some space. Well, I don't have much more space, but here, let me put it in blue. OK? This means then that when F equals 200 kilohertz or 200 multiplied by 10 to the third Hertz, OK. Then the voltage again, A is gonna be equal to one divided by the square root. OK? Of 1.6 multiplied by 10 to the negative 10th second squared multiplied by 200 kilohertz squared plus one. And now when we go ahead and substitute and solve, we get a to be 0.3697 or to two significant figures, approximately 0.37. Thus, our voltage gain equals one divided by the square root of 1.6 multiplied by 10 to the negative 10th second squared multiplied by F squared plus one. And when the frequency is 200 klotz, the voltage gain is 0.37. If we look back at our answer choices that tells us, then that C C is the correct answer. Thanks a lot for watching everyone. I hope this video helped.

Key Concepts

RC Circuit

Voltage Gain

Frequency Response

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