A series ac circuit contains a 250-Ω resistor, a 15-mH inductor, a 3.5-μF capacitor, and an ac power source of voltage amplitude 45 V operating at an angular frequency of 360 rad/s.(a) What is the power factor of this circuit?

Video transcript

Hi, everyone in this practice problem, we're being asked to calculate or determine the cosine of PHI for the circuit, we will have a series RLC circuit driven by an AC generator with a peak voltage of 30 fold and an angular frequency of 314 radium per seconds. The resistance is 100 and 20 ohm. The capacitance is 3.60 micro Freud and the inductance is 650 milli. We're being asked to determine the power factor or the cosine of five for the circuit. And the options given are a 0.174 B 0.325 C 0.707 and D 0.866. So in order for us to determine the cosine of phi, we want to utilize the tangent of PHI equals to XL minus XC divided by R or the reactance of the inductor minus reactance of capacitor divided by resistance equals to tangent of phi. From here, we can then get the Phi by taking the inverse of tangent and then calculate the cosine of the phi. So we wanna also recall to calculate the reactants of the inductor and the reactants of the capacitor, we want to employ the formula where XL equals to L multiplied by omega or the uh angular frequency. While the XC is going to equals to one divided by C multiplied by Omega. In this case, we're given the LC and Omega value. So we can put that into our equation here. So let's start with our XL. So XL is L multiplied by Omega, L is given to be 650 milli Henry. So we want to take that and put here 650 multiplied by 10 to the power of negative three Henry because we definitely want to put them into the SI unit. And we want to multiply that with the Omega which is then going to be 314 radians per seconds given in the problem statement. So calculating this, we will then get the XL value to then equals to 204.1 ohm which is going to be the reactance of the inductor. So moving on to XCXC is going to equals to one divided by C multiplied by Omega. Where in this case, then C is given in the problem statement which is going to be 3.6 micro fat. So we have to convert that into SI. So we want to multiply the 3.6 with negative 10 to the power of negative six to get it into fat. And multiply that with the angular frequency value which is 314 radiance per seconds, that will give us an XC value of 884.64. 0. So now that we have our XL and RXC or the reactants of the capacitor and the inductor, we wanna substitute that value into our tangent of phi so that we can get our tension of five value. So XL is given to be 204.1 ohm. Let's start with that 304.1 ohm minus XC which is 884.64 ohm divided by R which is given in the problem statement to be 100 and 20 ohm. So all of this will give us the tangent of Phi to then equals to negative 5.67. So in order to get the tangent, uh in order to get the Phi, we will take the n first of the tangent or tangent to the power of negative one of negative 5.67 or equals to our tangent of negative 5.67. And that will give us a five value of negative 80 degrees. So now taking the cosine of Phi to get the power factor, then we will get cosine of negative 80 degrees which will equals to 0.174. So the cosine of Phi or the power factor is going to be 0.174 which will correspond to option A in our answer choices. So option A will be the answer to this particular practice problem and that will be it for this video. If you guys still have any sort of confusion, please make sure to check out our adolescent videos on particular topics that will be all for this one. Thank you.