(a) Compute the reactance of a 0.450-H inductor at frequencies of 60.0 Hz and 600 Hz. (b) Compute the reactance of a 2.50-μF capacitor at the same frequencies. (c) At what frequency is the reactance of a 0.450-H inductor equal to that of a 2.50-μFcapacitor?

Video transcript

Hi, everyone in this practice problem, we're being asked to determine the inductive reactant, the capacitive reactant and the frequency where the inductive reactants will be equal to the capacitive reactants. We will have an inductive coil of L equals 300 mili Henry and a parallel plate capacitor of C equals 2.3 micro fro connected in series to a power source of variable frequency. The frequency of the power source is set at one kilohertz. And we're being asked to first calculate the inductive reactance XL of the inductor. Second, the capacitive reactance C of the capacitor at this frequency. And lastly third, the frequency of knot at which the inductive reactants will be equal to the capacitive reactants. The options given are a first 3002nd, 69.20 and lastly 3rd, 100 and 92 Hertz B 1st, 1000 880 0 2nd, 69.20 and 3rd, 100 and 92 Hertz C first A, 1000 882nd, 435 0 and 3rd 1200 Hertz. Lastly D first 3002nd 435 0 and lastly third, 1000 and 200 Hertz. So let's start with the first part of the problem where we are being asked to calculate the inductive reactants or XL. So for this 1st, 1st part, the inductive reactants or XL will equals to multiplied by omega where omega is given to be two by F, if we recall that, so that XL substituting Omega into our formula will equal to L multiplied by two by F. So we can actually substitute all of our information that is given in the problem statement to calculate XL. So L is given to be 300 Mili Henry. So um converting that into SI, that will be 0.3 Henry multiply that by two pi and multiply that by F which will be given to be 1 kg Hertz, converting that into SI, that will be 1000 Hertz calculating all of this. We will then get our XL value to then be equal to 1000 0.9 or rounding it down. We will then get 1000 880. 0 Awesome. Next, let's move on to the second part where we're being asked to calculate the capacitive reactants XC of the capacitor at this frequency XC will equals to one divided by C multiplied by Omega. And let's substitute the Omega into our equation where we get then XC to be equals to one divided by C multiplied by two pi F. So let's substitute all of our information that we know from the problem statement. So that XC will then equals 21 divided by C is given to be 2.3 micro fra. So mo uh converting that into si that will be 2. times 10 to the power of negative six fra multiplied by two pi multiplied by the frequency which is going to be 1000 Hertz N si calculating this, we will then get XC to be equals to 69.2. So the inductive reactant will be 880 the capacity reactants will be 69.2. 0 OK. Now, lastly, moving on to the third part of the problem where we are being asked to find the frequency F knot at which the inductive reactance is equal to the capacitive reactant. So at F NOTT XC will equals to XL. And we want to put in all of the formula that we know. So XC, we know that to be one divided by C multiplied by two by F and the F is going to be F knot and XL, we know that to be L multiplied by two by F and the F is going to be F not. So in this case, we can then get rear uh by rearranging all of this formula. If not can then be calculated to be equal to one divided by two pi multiplied by one divided by the square root of L multiplied by C. So let's substitute all of the known information that we have. So if not, well then equals to be one divided by two pi multiplied by one divided by the square root of L multiplied by ac where the L is given to be 300 Mili Henry in the problem statement and the C is 2.3 micro Fru and converting all of that into SI L will be 0.3 Henry multiplied by the C will be 2.3 times 10 to the power of negative six for. So that will be the F nought and calculating that F knott will then come out to a value of 100 and 91.6 Hertz. Awesome. So that will be all of the problem. So the inductive reactants or XL of the inductor will be 6 1000 880 0. And then the capacitive reactance of the capacitor will then equals to 69.20. And lastly, the frequency F not at which the inductive reactant will be equal to the capacitive reactants or F not will be 100 and 91.6 Hertz or rounding it up 100 and 92 Hertz. So all of that will actually correspond to option B in our answer choices. So option B 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 other lesson videos on similar topic on our website and that'll be it for this one. Thank you.