# HW#11 - Close Window Viewing PHY 184B Message Thread...

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Sheet1 Page 1 Close Window Viewing PHY 184B Message Thread Previous Message Next Message HW #11 1) Posted by HaloGod on 07-29-2005 07:37 PM Ok heres set 11. I actually wasn't planning on doing it today but i was on capa and noticed it was all AC circuit shit which im d o 1)AC Generator and a Resistor Part A: RMS Voltage = (.7)*(Peak Voltage) Our voltage is a sinusoidal waveform that reaches it peak when sin(@t) = 1. So our peak is simply the value in front of the sine (or cosine) function. Part B: Frequency = w/(2pi) ---> w = omega Your omega is the value inside the sine function Part C: Just plug in the time into your voltage equation. Make sure your calculator is in radian mode as omega is in rad/sec. Part D: I = V/R V is going to be your peak value. 2)Reactance of a Capacitor Step 1: Calculate our capacitor value 1/(C*w) = reactance So first find your omega(w) with the frequency they give you (w = 2pi*freq). Then solve for C using the reactance they give you.

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Sheet1 Page 2 Step 2: Now that you know C, use the same equation as above but now we're solving for the reactance. You need to calculate the omega once again for this frequency before you plug in your numbers. 3)Current in a Purely Capacitive AC Circuit i = wCVcos(wt + 90) The current through a capacitor leads the capacitor voltage by 90deg. This is insignificant in this problem though all we need is the magnitude of the equation above(wCV). This is derived from the fact that the current through a capacitor is the derivative of the voltage. So in these sinusoidal cases, the voltage is given by a sine function. Anyways, Answer = i(RMS) = (.7)*(w)*(C)*(Vmax) Omega is also known the angular frequency because it describes the voltage(or current) in terms of rotation. They give you its value in the problem. 4)Inductor to Limit Maximum Current Because we are working in frequency domain (phasors), you can consider a capacitor or inductor to just cause an impedance(think resistance) or admittance(think 1/R). So the point is that Ohms law works here not just for a resistor, but for our inductor in this problem. I = V/Z ------>Z is inductance Z = L*w (only for an inductor) Step 1: Find your peak voltage with RMS/.7 = Vmax Find your omega value with 2pi*freq = w Step 2: (I = V/Z)--->plug in your values--->(I = Vmax/(L*w)) Simplify: L*w = Vmax/I
Sheet1 Page 3 L = (Vmax/I)/w = answer (they give you I) 5)AC Circuit with R and C Part A: Step 1: calculate your omega: w = 2pi*freq calculate Vmax: Vmax = Vrms/.7 Step 2: Zeq = R - j(1/Cw) So we have our equivalent impedence in rectangular form. You need to convert this into polar form using a calculator. Its not hard you just have to know how to do it. That will spit out two different values to you, the magnitude and angle. We need the magnitude: I = Vmax/magnitude Step 3: Answer = Irms = (.7)*I Part B: Vrms = (Part A)(R) Part C: (Part A)(1/Cw) = V(c) Part D:

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HW#11 - Close Window Viewing PHY 184B Message Thread...

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