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if you can dumb it down for me In an L-R-C series AC circuit L = 0.660 H, R =380.0 ohms, and C = 1.60 μF. The AC source is operating at the resonance and has a voltage amplitude of 250.0 V . Find (a) the instantaneous voltage across R, L, and C respectively; (b) find the average power of the AC circuit.

          if you can dumb it down for me
In an L-R-C series AC circuit L = 0.660 H, R =380.0 ohms, and C = 1.60 μF. The AC source is operating at the resonance and has a voltage amplitude of 250.0 V . Find (a) the instantaneous voltage across R, L, and C respectively; (b) find the average power of the AC circuit.
        
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University Physics with Modern Physics
University Physics with Modern Physics
Hugh D. Young 14th Edition
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if you can dumb it down for me In an L-R-C series AC circuit L = 0.660 H, R =380.0 ohms, and C = 1.60 μF. The AC source is operating at the resonance and has a voltage amplitude of 250.0 V . Find (a) the instantaneous voltage across R, L, and C respectively; (b) find the average power of the AC circuit.
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Transcript

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00:01 I'm going to write down the pertinent information.
00:05 V -sab -0 is 100 volts.
00:11 F equals 500 hertz, which would mean that omega equals 2 times 500, which is 1 ,000 pi radiance per second.
00:28 R is 500.
00:32 Holmes, l is 0 .20 henry.
00:40 In part a, we have a capacitor also, which is 2 times 10 to the negative 6th power, ferrets.
00:55 So i would start by calculating the impedance.
00:59 It's going to be the square root of r squared plus xl, which is just going to be omega l, minus xc, which is one over omega c, squared.
01:14 So let me put that into a calculator, sqrt of r squared, r is 500, minus, no, plus in parentheses so it's going to be 1 ,000 pi times .2 minus 1 over 1 ,000 pi times 0 .2, minus 1 over 1 ,000 high times c, which is 2 times 10 to the negative 6th power.
01:59 That's going to give me to two significant digits, 500 oms.
02:10 I forgot to square it.
02:13 Sorry about that.
02:14 It's not 500 oms.
02:16 It is 690 oms.
02:30 Okay, so we're trying to find the average power dissipated by the resistor.
02:37 All right.
02:39 I believe we're going to have to figure out the current.
02:41 Which is going to be v -0 over z, 100 over 690.
02:54 Let me put that into a calculator.
03:03 100 divided by 190 is 0 .15 amps.
03:21 Now, the power dissipated over the resistor power average is going to be one -half i -0 v -0.
03:49 So i think i'm going to deviate a little bit from the book because this voltage is just over the resistor, not the entire voltage...
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