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00:48

Averell H.

(I) What force is needed to accelerate a sled (mass = 55 kg) at 1.4 m/s$^2$ on horizontal frictionless ice?

04:27

Kai C.

(I) A 110-kg tackler moving at 2.5 ms meets head-on (and holds on to) an 82-kg halfback moving at 5.0 m/s. What will be their mutual speed immediately after the collision?

03:09

Aditya P.

(I) A 7150-kg railroad car travels alone on a level frictionless track with a constant speed of 15.0 m/s. A 3350-kg load, initially at rest, is dropped onto the car. What will be the car's new speed?

0:00

Muhammed S.

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welcome to our first example Video looking at power dissipated in alternating current circuits. This video, we're going to consider the circuit we have here with a source that has nine volts applied E m f the maximum nine multiplied e m f And that oscillates is 60 hertz a 50 ohm resistor attend Micro Farid capacitor and a 15 million Henry in Dr. Given this trick it, I'd like to find out what is going to be the power dissipated by the resistor here. Well, we know that the average term is going to be equal to P equals I r m s squared times are, uh, now in order to find I rms here, remember that we need to find the maximum current through the resistor divided by route to okay. Eso to find this maximum current. Remember that because they're in Siris here. The current through each individual piece will be the same. Which means we can use our equation from before which found that we had Enoch divided by the square root of r squared plus x l minus x c quantity squared. Now, in order to calculate, Exelon next to remember excels equal to Omega times. L xsi is equal to one over Omega Times, see, and that omega is equal to two high times F So there's a lot of different things that we need to add in here In order to write this out correctly. Eso First of all, we can expand this to be I r squared times are divided by two and then to get our i r squared that will be he is equal to e not squared, divided by r squared plus x l minus x c squared and that is multiplied by our over to. And then we can expand it further by plugging in omega L and one over Omega Psi for our inductive reactions and are capacitive reactions. But you get the idea. We calculate the current. And then we used the equations that we found in order to find the average power delivered to the resistor.

Electromagnetic Waves

02:33

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02:08