Recent content by JoeMarsh2017

so, since we know that sqr rt2(532 Vrms)/20*pi = 12A which is our average current I would compare 532vrms/240 vrms to get my turns ratio..its basically a 1:2.2 step up transformer

240rms X sqr rt2 =339 volts

OK the rectifier is ideal, the loss is no loss (voltage is high enough we can we neglect it... The voltage after the rectifier is DC Voltage

Re-doing this post since I am starting to figure it out... 754 Volts is the Vmax I can take 75/Sqr rt 2 which will give me my rms Volts =532.88

OK CWatters ... I beielve I misundestood the relationship between the Iavg, Irms and Ipeak current

Oh man...I just found what I think I was missing..since I know the power in the secondary, 12A times 754 Volts...I get 9,048 Watts 9,048 divided by 240 Volts on the primary would give me the Current in the Primary...right? I am still getting thrown for a loop here because the Resistance is...

Homework Statement A half wave rectifier circuit has a transformer inserted between the source and the remainder of the circuit. The source is 240Vrms and 60 Hz, and the load resistance is 20ohms. 1) Determine the required Turns Ratio of the transformer such that the average load current is...

210.13 tan 46 gets you 217.59 which is what I was looking for.. Answers from the book are 210.1 kW Active Power 217.6 kVAR Reactive Power This is per the power triangle

GUYS, I figured it out...My problem was that once I solved for "S"..and I had the angle @ 46°...then I can solve for active and reactive power Power triangle means s=302.5 theta=46° Active=210.13 =302.5cos 46° Reactive=217.34 =210.13tan 46 Answer

210.1 kW 217.6 kVAR These are the answers from the book

Homework Statement A 7200-240 V, 60 Hz transformer is connected for step up operation, and a 144∠46° Ω Load is connected to the secondary. Assume the transformer is ideal and the input voltage is 220 V at 60 Hz. Determine a)secondary voltage b)secondary current c)primary current d)input...

Yes, we are working on Unit Step functions, ramps, and combining signals right now. I am stuck on another part because I am still learning how to use MATLAB. i(t)=u(5t)+u(3t^2) I am still confused but is this where it becomes a "power function" like A^2/n! ? Joe

My instructor is actually making his own textbook for the course, so we don't have a textbook to reference to this semester. He is cutting problems out of another book and posting them in a PDF for us to work on...no clue what the book is...

Homework Statement Determine the "Damping Constant and Time Function" v(t)= 8ε^-106t Homework Equations ∝=Damping Constant = 1/∝=Time Contstant = 1/DC The Attempt at a Solution -848ε^-106t = DC then 1/DC = *1.087x10^-49 It does not look right...Is the e in my calculator the same...

Homework Statement [/B]Homework Equations i(t)=5u(t)+ 3u(t^2)The Attempt at a Solution