Skip to main content
Ch 31: Alternating Current

Chapter 31, Problem 31

An L-R-C series circuit is connected to a 120-Hz ac source that has V_rms = 80.0 V. The circuit has a resistance of 75.0 Ω and an impedance at this frequency of 105 Ω. What average power is delivered to the circuit by the source?

Verified Solution
Video duration:
4m
This video solution was recommended by our tutors as helpful for the problem above.
328
views
Was this helpful?

Video transcript

Hi, everyone in this practice problem, we're being asked to calculate the average power transferred to a circuit where we have a student connecting 100 resist an inductor L and a capacitor C in series across an ac generator of VR MS to be equals to 45 fold, the generator is operating at 100 and 50 Hertz and the circuit's impotence at 100 and 50 Hertz is 100 and 25. 0, we're being asked to calculate the average power transferred to the circuit. And the options given are a 0.36 watt B 10.4 watt C 16.2 watt and lastly D 20. watt. So for this practice problem, we want to recall that the expression for the average power into a general ac circuit is given by PAV to be equals to VR MS multiplied by Ir MS multiplied by cosine of Phi. In this case, VR MS is the R MS voltage Ir MS is the R MS uh current and Phi is the phase angle of the fault with respect to the current. Another um another formula that we will not recall as well. Is the, the power factor formula. Where in this case, the cosine of phi is equals to R divided by Z where in this case, R is the resistance of the resistor, which is given for from the problem statement to then be 100 and Z is the circuit impotence which is given in the problem statement at 100 and 50 Hertz to be 100 and 25. 0 So we will then just substitute those value into the R and the Z to calculate our cosine of five or to calculate the power factor where in this case, cosine of five will then equals to R divided by Z which is 100 divided by 100 and 25 which will then equals to 0.8. Next, you wanna utilize Os Law in order for us to actually rearrange and get a better formula that we can use for the PAV or the average power. So according to Os law, VR MS will actually equals to R multiplied by Ir MS. We wanna substitute our Rir MS into in terms of P and R in the PAV formula. So we wanna arrange so that Ir MS will then equals to VR MS divided by R. We wanna substitute this into our PAV formula so that our PAV will equals to VR MS multiplied by IRM si R MS is VR MS divided by R multiplied all of that by cosine of five and that will give us a PAV formula of VR MS squared divided by are multiplied by cosine of Phi. We are doing this because the only information given in the problem statement are the VR MS and the R and the cosine of phi. So we do not know what the Ir MS is. So we are substituting that with the S law in order for us to be able to calculate BAV. So now that we get the formula for the PAV that we're going to use. Let's substitute all of those values. So it is given in the problem statement that VR MS is 45 fold. So let's start with 45 fold squared divide that with R which is given to be 100 so 100 and then multiplied that with cosine of five, which is the power factor which is given to be 0.8. So calculating this, we will get PAV to then equals to 16. watt and 16.2 watt is then going to be the average power transferred to the circuit or PAV that will correspond to option C in our answer choices. So that will be the answer to distract this problem. And that'll be it for this video. If you guys still have any sort of confusion, please make sure to check out our adolescent videos on similar topics. That'll be it for this one. Thank you.