An L-R-C series circuit with L = 0.120 H, R = 240 Ω, and C = 7.30 μF carries an rms current of 0.450 A with a frequency of 400 Hz. (a) What are the phase angle and power factor for this circuit?

Video transcript

Hi, everyone in this practice problem, we're being asked to determine the circuit phase angle. And the power factor of an electrical circuit consisting of a resistor of R equals to 300 0, an inductor of L equals to 0.2 Henry and a capacitor C equals to six 0.35 micro Ferrat connected in a series with an AC generator. When the frequency of the generator is set at 250 Hertz, the R MS current flowing in the circuit is 0.55 amp. We are being asked to first determine the circuit phase angle. And second, the power factor the options given are a first negative 54.5 degrees and 2nd 0.814 B first negative 35.5 degrees and 2nd 0.58 C 1st 35.5 degrees and 2nd 0.814. And lastly, D 1st 54.5 degrees and 2nd 0.58. So in order for us to calculate the circuit phase angle, we want to recall that the angle phi between the voltage and the uh current phasers is given by a tangent of five to be equals to XL minus XC divided by R. So in this case, we want to also then recall the formula for XL or the reactance XL of the inductor and the reactance XC of the capacitor where the XL is going to equals to L multiplied by Omega and XC is going to equals to one divided by C multiplied by omega. So Omega is going to equals to two pi F or frequency. And in this case, we can then substitute the Omega into our XL and XC equation and then substitute all of the information given in order for us to calculate the tangent of pi and get the five value. So let's start by doing that. Let's start with XL. So XL will equals to L multiplied by omega and that will equals to L multiplied by two by F. And let's substitute the given information in our problem statement where the L is 0.2 Henry or H multiplied by two by multiplied by the frequency which is given to be Hertz. In this case, this will then equal to 314. 0 for the reactants of the inductor. Now moving on to XCXC equals to one divided by C multiplied by Omega. And let's substitute the Omega site where XC will then equals to one divided by C multiplied by two by F substituting all of the given values we will then have one divided by the C or the capacitor is 6.35 micro fra. So we have to multiply that by 10 to the power of negative six to get to Ferrat multiply by two pi multiplied again by the frequency which is 250 Hertz. That will give us the XC or the reactants of the capacitor to then equals to 100.2026 0. So now let's substitute that values into our tangent of Phi so that we get tangent of phi equals XL divided uh minus XC divided by R. And in this case, XL is 314. minus XC, which is going to be 100.26 divided by RR is going to be 300 which is given in the problem statement here. And that will give us the tangent of five value to equals to 0.713. Now, to get the actual five value, we have to take the inverse of tangent which is tangent to the power of negative one of 0.713 or at the same time equals to arc tangent of 0.713 which will give us a five value of 35. degrees. So that will give us the certed phase angle which is going to be approximately 35.5 degrees. And next, what we wanna uh find is the power factor which is just cosine of phi. So cosine of phi equals to cosine of 35. degrees. And that will actually give us the value of 0.814. So that will be the power factor. So with the um current phase angle to be 35.5 degrees and the power factor to be 0.814 then the answer to this practice problem is going to be option C and option C will be the answer to this video. If you guys still have any sort of confusion, please make sure to check out our other videos on similar topics. But other than that, that will be it for this one. And thank you.