Recent content by brenfox

Hi, regarding question d. This is how I expressed my working out to get my answer (looking at the previous posts I now realize my answer is wrong but would like to know if my train of thought is correct) Total impedance = 45+j45 Voltage = 415v I =v/r* = 415(45-j45)/45+j45(45-j45) =...

Nah, its not my problem.

I am sure this has been discussed many times before but here goes. I was recently working on motors with a power factor cosine of 0.69. To counteract this the company installed banks of capacitors to get the power factor near 1. My question is this, the capacitors were installed downstream from...

Thanks Maged, Mark. 16.32 plugs in correctly.

If L-1 = 0.02 x 16 Then to get L on its own does the -1 turn into 1 on the other side of the equation? I can`t see the wood for the trees on this one.

I am starting to confuse myself here! The equation is -2.6 = 1/1.5 ln(L/Lo -1) multiply both sides by 1.5 -2.6x1.5 = ln(L/16 -1) -3.9= ln(L/16-1) L/16-1 e^-39 L/16 -1 = 0.02 L-1 =0.02 x 16 L = 0.02x16 +1 L= 1.32. I know i am going wrong somewhere because my answer does not...

L/16 -1 = e^-0.256 L -1 = e^-0.256 x 16 L = 1 e^-0.256 x 16 L = 1 + 0.774 x 16 = 28.3284. This is also wrong when i plug it in.

Yes. Just plugged in my answer and calculator shows error. So L must be bigger than 16.

Forgot to plug in my answer! I will have another look at it. Seems like I am way off so will have to do a bit more reading into it. Thanks for the reply.

Homework Statement w = 1/h ln(l/lo-1) w=-2.6, lo=16 and h =1.5 Find L. Homework EquationsThe Attempt at a Solution Plug in the values 2.6 = 1/1.5 ln(l/16 - 1) Make the log term the subject. ln(L/16 -1) = 1/1.5/-2.6 ln(L/16-1) = 0.666/-0.256 ln(L/16-1) = -0.256 Change the log statement to an...

Apologies jedishrfu, I have wrote it down wrong. Thanks haruspex, I should have used parentheses after the caret symbol. You live and learn!

Looked at the question again and I have wrote it correctly. 3.4 is in parentheses. Not sure why this is so. An equation icon makes my equations look neater if you know what I mean!

Homework Statement (3.4)^2x+3 = 8.5. Solve for x Homework EquationsThe Attempt at a Solution 3.4^2x = 8.5/3.4^3 ( divide the equation by 3.4^3. This bit i am not sure about) 2xlog3.4 = log (8.5/3.4^3 x = log (8.5)- 3log(3.4) / 2 log (3.4) =0.929-1.594/1.063 x = -0.626 This...

psparky, UK households universally use single phase. 3 phase is very rarely used in domestic situations unless the loadings on a big property require 3 phase. (extremely rare)..i think...

As an after thought (this is where i will get my technical *** whipped!). According to Kirchoff, current entering a node will leave a node in a closed circuit. So whatever current enters a load will leave this load via the neutral in a line to neutral system. The potential voltage will be...