The small transformers that power many consumer products produce ...

Question

The small transformers that power many consumer products produce a 12.0 V rms, 60 Hz emf. Design a circuit using resistors and capacitors that uses the transformer voltage as an input and produces a 6.0 V rms output that leads the input voltage by 45°.

Accepted Answer

Everyone. Let's take a look at this practice problem dealing with ac circuits. So in this problem, we have a power supply that operates at 24 volts R MS and 80 Hertz. And what we need to do is design a circuit using a one, her micro ferric capacitor and some resistors to use this um power supply to generate a 12 volt R MS output that is leading the input voltage by 45 degrees. We're gonna start by drawing a circuit diagram. We're strong by having our input voltage or input ef which we label as E and we'll connect one of those leads to capacitor. We'll label that as C and we're gonna connect this capacitor in series with two resistors. We'll have R one as one resistor and are two as the second resistor. And we'll connect the second resistor back to the um other lead of the EMF now we're gonna want our output voltage to be coming off of R two. We'll connect that in parallel. And so we'll have those leads going across R two. We'll label that as B two. So that's gonna be our output voltage. We're gonna define a couple of quantities here. The first one is to define our total resistance of our system. This is gonna be capital R and that's gonna be equal to R one plus R two. And that's because those two resistors are in series. And so when you have resistors in series, you just add them together, we also need to recall our definition for reactants for a capacitor. So that's XD and that's equal to one divided by Omega C. Let's start off by looking at this um phase angle 45 degrees. So recall or a phase angle that the tangent of that phase angle which will cause phi that's gonna be equal to minus the XC divided by R. So that's minus the impedance or the um reactants of the capacitor divided by the total resistance R. And here um the phase angle is the angle at which the input voltage leads the current. And therefore the output voltage we actually want to lag the output voltage and the current that means our angle there is actually negative, we'll have the tangent of negative 45 degrees and this could be equal to minus XC divided by R. And so the tangent of negative 45 degrees is just negative one, the negative sides cancel, that leaves me with a relationship that R is equal to X. Now I have that relationship. Um Let's look at the total current that's gonna be running through this um circuit. And so for that, we're just gonna use Ohms Law. And so we're gonna have um the, er MS gonna be equal to the Ir MS multiplied by the impedance Z. So if you solve this for Ir MS, we'll have Ir MS is equal to the, er MS divided by. And for the impedance, you call our formula for the competence. That could be the square root of R squared plus XD squared. However, R is equal to X. So we can actually replace that in our formula. We'll have Ir MS will be equal to er MS divided by square root of XC squared plus XC squared. So we can actually simplify that a little bit. We'll have Ir MS equal to er MS divided by XC multiplied by the square root of Q. So now that we have the current, we can now actually calculate what the value for R two should be. And again, we're going to use Homes Law. You wanna have B two R MS is equal to Ir MS multiplied by R two. So now we can just plug in our, what we have calculated for the Ir MS will have B two R MS equal to, we have the er MS divided by XD multiplied by the square root of two. And that could be multiplied by R two. So now we can solve for R two, we'll have R two going to be equal to VV two R MS divided by er MS multiplied by the square root of two. And here I'm going to replace um XC with our formula that can be square two divided by Omega C. And at this point, we have values for all of those quantities, we can calculate what R two is. So we have R two gonna be equal to, we have for the V two R. MS is gonna be 12 volts divided by the er MS, that's gonna be the 24 volts and this could be multiplied by the square root of two divided by or mega. That's just gonna be pi multiplied by our frequency which is 80 Hertz and our capacitance, it is 100 micro ferret. So it'll be 100 100 multiplied by 10 to the negative six beads. And so we can plug all, all those values into our calculator, turns up with an R two of equaling 14.1 ohms. So at this point, we can go back and use our relationship between the total resistance and R one and R two to calculate what or one is. So for or one that's just gonna equal R minus R two. However, R is just one over Omega cr is equal to XC and that's equal to one over Omega C. So we have or one equal to one over Omega C minus are two. And so that leads me with, are one equal to and I can plug in values at this point. One divided by the two pi multiplied by the 80 Hertz multiplied by the 100 multiplied by 10 to the negative six bars. And then we'll have the minus 14.1 ohms. And so we plug those values into our calculator. This gives me 5.8 almost in order to get the output voltage that the question to ask for. We need to actually have two resistors in series. One is gonna be um 5.8 ohms and then the one we're gonna actually have the output voltage coming off of is gonna be 14.1 ohms. So just a quick little recap. Um We had to use the face angle. Do you find the relationship between the total resistance and the reactance of the capacitor? And then we had to use OHMS. So twice first to find the R MS current and then the second time to find the resistance for R two, which is the resistor or output voltage, we then use the total resistance and the relationship between total resistance and the reactance of the capacitor to find R one. So I hope that this has been useful and I'll see you in the next video.

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The small transformers that power many consumer products produce a 12.0 V rms, 60 Hz emf. Design a circuit using resistors and capacitors that uses the transformer voltage as an input and produces a 6.0 V rms output that leads the input voltage by 45°.

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