In an L-R-C series circuit, the components have the following val...

Question

In an L-R-C series circuit, the components have the following values: L = 20.0 mH, C = 140 nF, and R = 350 Ω.The generator has an rms voltage of 120 V and a frequency of 1.25 kHz. Determine (a) the power supplied by the generator and (b) the power dissipated in the resistor.

Accepted Answer

Hi, everyone in this practice problem, we are being asked to calculate the power provided by the AC source and also the rate at which electrical energy is lost in a resistor. We'll have a series of RLC circuit driven by an AC power source of R MS 60 fold at a frequency of 100 and 20 Hertz. The resistor has a resistance of 100 and 50. The inductor has an inductance of 100 and 25 milli Henry and the capacitor has a capacitance of 100 and 12 mili fro we're being asked to calculate first, the power provided by the AC source and second, the rate at which is electrical energy is lost in the resistor. The options given are a 1st 10.8 watt and 2nd 10.8 watts B, 1st 10.8 watt and 2nd 17.2 watt. C, 1st 17.2 watt and 2nd 17.2 watt. And lastly D 1st 17.2 watt and 2nd 10.8 watt. So in order for us to calculate this problem, we wanna recall certain formulas that we're going to use. So for the power provided by the AC source, we're going to have to calculate or the first one, we have to calculate PAV, which will equals to VR MS multiplied by Ir MS multiplied by cosine of Phi or employing the sla we can then modify this equation so that BAV will equals to VR MS squared divided by Z multiplied by cosine of Phi. We are doing this because we do not have the Ir MS but we are given the resistance and the uh VR MS. So in this case, we will be able to calculate pav more easily. Second, we will have to calculate the rate at which electrical energy is lost in the resistor, which is going to be the Pr So Pr will equals to R multiplied by Ir MS squared or utilizing, utilizing the S law. Again, we can then uh modify this to get R multiplied by VR MS divided by Z and all of that is going to be squared. So if you notice the first step that we have to do is to calculate the Z and also the cosine of Phi or the power factor. So let's start by doing that. So in order for us to calculate the cosine of five, we have to determine the angle phi which is the angle between the full page and the current phasers. In order to do that, we want to utilize the equation of tangent of Phi equals to XL minus XC divided by R where XL is the um reactance of the inductor, XC is the reactance of the capacitance. R is the resistance and XL is going to equals to L multiplied by omega and XC is going to equals to one divided by C multiplied by omega. All of that is going to be divided by R to get tangent of Phi and L is the inductance C is the capacitance R is the resistance and omega is the angular frequency we have all of this value given in the uh problem statement. So let's actually cal uh input everything into this formula right here, Omega is going to equals to two pi F and the frequency is going to be 100 and 20 Hertz as it is mentioned in the problem statement also. So let's start tangent of five will then equals to L multiplied by Omega L is going to be 100 and 25 Mil Henry. So that will be 0.1 to five Henry in si multiplied by Omega which is going to be two pi F and the F is going to be 100 and 20 Hertz. So let's substitute that in minus one divided by C multiplied by Omega C is given to be 100 and 12 Mili fro which will then be uh 0.112 fro in si multiplied by two by FF is 100 and 20 Hertz. So let's substi substitute that in all of that is going to be divided by the resistance which is R which is given to be 100 and 50. 0 So what we wanna do next is to do the calculation for this and that will give us a tangent of five value to then equals to 0.6 to eight. And from there, we can then utilize the in first of tangent or tangent to the power of negative one of 0.6 to eight in order to get the five value or the arc tangent of 0.6 to eight, which will give us the five value of 32. degrees. So that will be our five value. Next, we want to determine our impotence of the circuit or the Z where we want to recall that Z will equals to the square root of R squared plus open brand C XL minus XC squared. This will then equals to let's substitute all of the value give uh that we know or the formula that we know. So R squared will remains to be IR squared plus open parenthesis. XL is L multiplied by omega minus XC is one divided by C multiplied by omega close parenthesis squared. So now let's substitute all of the information given in the problem statement in order for us to calculate the impotence of the circuit, so Z will then equals to the square root of first we have R squared which is 100 and 50 squared plus, we have L multiplied by Omega. L is given to be 0.125 Henry multiplied that with Omega which is two pi F, which is 100 and 20 Hertz minus one divided by C multiplied by Omega. So that will be 0.112 for rat multiplied by two pi multiplied by 100 and 20 Hertz for DF. And we have to put that in squared. So calculating this, we will then get the value for our impotence of the circuit to then equals to 100 and 77. 0 So next, we already calculated our impotence and our five. So we can substitute all of that uh values into the equation that we're going to use to calculate the first part and the second part. So let's start by doing that. Let's start with the first part. First. For the first part, we will have the average power provided by the AC source to then equals two or PAV to then equals to VR MS squared divided by Z multiplied by cosine of five. So VR MS uh is given to be 60 fold in the problem statement. So we will have 60 fold squared divided by impotence to be 100 and 77 multiplied by cosine of five, which is going to be cosine of 32.1 degrees. So calculating all of this, we will then get the PAV to then equals to 17.2. What? Awesome. So moving on to the second part which is calculating the Pr pr is going to equals to R multiplied by VR MS divided by Z squared. And let's uh substitute all of that information here. So Pr will equals to R multiplied by VR MS divided by A Z squared. So let's substitute everything R is given to be 100 and multiplied by VR MS which is given to be 60 fold in the problem statement divided by Z which is 100 and squared. So calculating this, we will then be able to get 17. watt for pr. So the answer to this press, this problem is PAV will equals to 17.2 watt and PR will also be equals to 17.2 watt or that will then correspond to option C in our answer choices. So option C will be the answer to this particular practice problem with the power provided by the ac source to be 17.2 watt. And the rate at which the electrical energy is lost in the resistor will also be 17.2 watt. So that'll be it for this video. If you guys still have any sort of confusion, please make sure to check out our other lesson videos on similar topic. That'll be it for this one. Thank you.

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In an L-R-C series circuit, the components have the following values: L = 20.0 mH, C = 140 nF, and R = 350 Ω.The generator has an rms voltage of 120 V and a frequency of 1.25 kHz. Determine (a) the power supplied by the generator and (b) the power dissipated in the resistor.

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