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Hi @anorlunda , So i know from the graph windage/friction losses constitute the power lost when the stator voltage=0V. I have actually reworked this question and got something I am more happy with. I've actually said the windage/friction loss is 220W from extending the graph to 0V. I know that...

So I've attached my graph and found that the windage/friction losses occur roughly around 230Watts. I just don't know how i further explain the magnetising current I0. I know I0 = input current on no load. So do i assumed from P = √3 x VL x IL x CosФ that CosФ=230/√3*220*6.8= 0.0849 or...

Attempt as above. image attached of Bytronic ICT3

Thank you, if No.1 is looking okay, i'll plod onto number 2 and get the rest answered for confirmation, thank you scott :)

Yeah you are correct, i meant IND 3 for rung 5 (N/O contactor 3/2). Also could you relink it, nothing there :) thanks!

Homework Statement Uploaded as image 1 I will show my workings for 2 if 1 are in the right direction I also don't understand what is meant as I/O Designation charts (nowhere in my study notes) Homework Equations None The Attempt at a Solution Uploaded as image 2 is my attempt.

@rude man so after searching through second order system responses I have a graph that looks like this. How does it look?

Apologies I think I've gotten myself confused... I think I am looking too deep into it. I'll be back when i have some form of graph! Thanks for your help @rude man

This is what I'm unsure of, if it mentioned one of the above i could work through it via my notes... I've only encountered step change input for second order systems. Thus i understand which ones are under-damped (a/b) and which are over-damped c if this is the case. But how do i draw this? only...

Homework Statement Sketch, on a set of common axes, waveforms to represent the transient response of circuits having transfer functions with the following parameters: a) ζ = 0.5, ω = 1×10^3 rad s^-1 b) ζ = 0.2, ω = 2×10^3 rad s^-1 c) ζ = 2, ω = 1×10^3 rad s^-1 Homework Equations wd=wo√(1-ζ^2)...

@jim hardy I'm still unsure on this question... sorry.

Apologies i didn't add the image. So are you saying to integrate the sine wave between T/4 & T? Sorry only thing i found in my notes was very little other than the statement above.

Homework Statement a) If the firing angle is set for alpha=(pi/3) estimate the power dissipated in the bulb if it is rated at 100 W and the voltage source is 230 V @ 50 Hz. b) An anomaly that can occur in controlled rectification is drift of the firing angle on one half cycle, so causing an...

Okay think i got it now :) thanks for your help.

Oh yeah! So by shifting it to the right by T/2 you would get a regular square wave? and thus making it have only sine terms. With regards to to whether they have symmetry I actually cannot tell now, it doesn't look symmetrial about the y-axis so it's not an even function but it doesn't have...

Homework Statement Q1. a) In relation to Fourier analysis state the meaning and significance of 4 i) odd and even functions ii) half-wave symmetry {i.e. f(t+π)= −f(t)}. Illustrate each answer with a suitable waveform sketch. b) State by inspection (i.e. without performing any formal analysis)...

Thanks @gneill! Only bit i was struggling on! But I hoped it'll be as simple as use the equation and plug in values. Helpful as always :)

https://www.wolframalpha.com/input/?i=%E2%88%9A((2%2B(16%CF%80)i)%2F(0.0005%2Bi(0.00046%CF%80))) Here's my workings using wolframalpha

l Following onto this question i got all the same workings however we needed to find Zo aswell. Using with the values above I got Zo=179.427+j26.5060 Ω or in polar Zo=181.375 /_+8.403° Ω (sorry don't currently have software to do the polar expression so used /_ to signify the angle).Does this...

Good! So in agreeance now that the new stress should be 8.68MPa and the percentage reduction is correct?

@CivilSigma Hey buddy, i realized i messed up on figures for e) i figured out the new maximum stress was 8.68MPa (somehow my calculated messed up) but (8.333*10^3*0.12)/(1.152*10^-4) wasn't 6.58*10^6 (6.58MPa) but 8.68MPa, fancy checking my math xD this gives me a new percentage reduction of...

phew! that question has killed me! it was a bit brutal. I have another one which my instructor has said will probably test myself (to do with column failure types and length of which buckling will happen etc. in a hollow cylindrical column)... i'll attempt it first and if not i guess i'll be...

So more like this picture? I’ve placed the 0 next to the neutral axis (just to show it’s at that point), and added a negative sign to the tensile stress. Does this now satisfy the question 2c)? If so is everything overall looking like a correct solution to the whole question? Ignore the...

Okay so below is my picture. After we’ve gone through c) can I just get a double confirm everything is good :) thanks again!

@CivilSigma that's brilliant this question baffled me for the best part. D I’m happy with but E just to confirm my thought of a 47.4 % reduction is correct? That seems quite a lot for only a 20mm increas in depth, but if you agree I’m happy! Lastly, sorry I’ve been busy but if I tag you and...

Homework Statement A beam of rectangular cross section 200 mm deep and 100 mm wide. If the beam is 3m long, simply supported at either end. And carries point loads R1 to far left, R2 to far right, 1m to left 5kN load, 2m to left 10kN load. (a) Calculate the maximum bending moment (b) Calculate...