A resistor with R = 300 Ω and an inductor are connected in series across an ac source that has voltage amplitude 500 V. The rate at which electrical energy is dissipated in the resistor is 286 W. What is (a) the impedance Z of the circuit; (b) the amplitude of the voltage across the inductor; (c) the power factor?

Video transcript

Hi, everyone in this practice problem, we're being asked to calculate the impotence, the difference of potential. And also the circuit power factor, we will have a series of RL circuit containing a resistor with a resistance of 15 0 A coil of inducts L and an AC generator with an amputated voltage of 100 and 20 fold. The production rate of thermal energy in the resistor is 100 and 80 watt. And we're being asked to first calculate the impotence or C of the RL circuit. Second, the difference of potential across the coil or VL. And lastly third, the circuit power factor or cosine of phi the options given are a 1st 24.5 2nd, 46.7 fold and 3rd 0.43 B 1st 24.50 2nd, 94.9 fold and 3rd 0. C 1st 34.70 2nd 46.7 fold and 3rd 0.43 or lastly D 1st 34.7 2nd 94.9 fold and 3rd 0.612. So in order for us to be able to solve this problem. We wanna recall uh multiple different formulas that will be useful to solve this problem. So first, we are being asked to calculate the impotence or Z where Z will be equals to oops V divided by I, we know what the V is from the problem statement which is 100 and 20 fold, but we do not know what the I is so that we have to calculate beforehand. Second, we are being asked to calculate the difference of potential across the coil. And the formula that we're going to use is that V will equals to VR plus VL. And in this case, the angle between VR and VL is going to be 90 degrees. So V squared will be equals to VR squared plus VL squared and rearranging this then VL will then equals to B the square root of V squared minus VR squared. Awesome. VR is going to be equals to R multiplied by I. So VL will then equals to the square root of C squared minus open parenthesis R multiplied by I squared. Awesome. So lastly, we're being asked to calculate the cosine of Phi or the power factor for the third part. And for this one, we want to utilize the PAV to be equals to half V I multiplied by cosine of Phi. And we wanna rearrange this so that cosine of Phi will be equals to two multiplied by PAV divided by V I. So let's actually start by calculating the I. And in order for us to do that, we wanna recall a different formula for a pure resistor where the average power or PAV will be equals to half multiplied by I squared multiplied by R I is going to be our current amplitude and rearranging this equation, we will get an equation for I which will then be equals to the square root of two multiplied by PAV divided by R. Let's substitute all of the given information into this formula where R I will then equals to be the square root of two multiplied by PAV is given to be 100 and 80 watt in the problem statement divided by R which is given to be 15. 0 Also in the problem statement, calculating this, we will then get the current amplitude or I to then be equals to 4. amp. So now let's actually substitute that into our first part. So the first part is the impotence which is going to equal to V divided by I default age is given to be 100 and 20 fold in the problem statement and the is 4.9 amp. So calculating this, we will then get the impotence to then be equal to 24.5 also. So now let's move on to the second part of this problem. The second part is asking us to calculate the difference of potential across the coil or VL which we have developed the formula for. So that VL will then equals to the square root of V squared minus R I squared. Let's substitute all of our information that we know where V is given to be 100 and 20 fold. So V squared is going to be 100 and 20 fold squared minus R I. So R is given in the problem statement to be 15 and I, we just calculated to be 4.9 amp. So this will be 15 0 multiplied by a 4.9 amp squared. And calculating this, it will give us our VL value to then be equals to 94.9 fault. So that will be the second part of the problem. And now moving on to the last part where we are being asked to calculate the power factor of the circuit, we have developed the formula where cosine of five will equals to two multiplied by PAV divided by V I. So let's substitute the known values where Pav is and 80 watt. So two PAV will equals to two multiplied by 100 and 80 watt and V I will be 100 and 20 fold multiplied by 4.9 amp. And that will be giving us the cosine of five to be equals to 0.612. So the power factor of the circuit or cosine of five is going to be 0.612 and then the it across the inductor is going to be 94.9 fold. And lastly, the impotence is going to be 24.5. 0, and all of that will correspond to B in our enter choices. So option B will be the answer to this particular practice 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 and that will be it for this one. Thank you.