Recent content by The_daddy_2012

Seriously, how do you do that? how can you look at an equation and say, hey, I am going to do it this way. That method would never have occurred to me. Thank you very much Gneill, a legend as always!

sorry correction V^2/R = P

V^2R=P so V does equal 325.269sin(x) but it needs to be squared? i can't see how R is found in this situation though?

Just tried something different looking at other peoples solutions, but i have no idea why everyone is using sin^2? it makes no sense. the voltage wave is a sin wave. not a sin^2 wave.

I'm really stumped on this exact question. What i have so far is. I have found the peak voltage which makes sense as this gives the amplitude of the wave. Vpeak= 325.269V from this i find the equation for the voltage sine wave is: 325.269sin(x)V I now need to find the average...

R's x a = rs?

I have three answers for b. Can anyone point me in the right direction. Rs= 0.007 Rs= 4.92 Rs= 9.963x10^-6 Do any of these match anyone else's?

Ive got a feeling part (A) is wrong. VA/V1^2 X (R'p cos (theta) + X'p sin (theta)) x 100 is an approximation. Has anyone handed it in and got it correct this way? It just seems too easy. the exact equation is: (E2-V2)/V2 Where: E2= V2^2 + I2^2 (R's^2 + X's^2) +2 x V2 x I2 x (R's cos...

I don't know if you've solved it yet, but if you haven't here's some help. Write both of your equations down, then write down v=v underneath and see what you you spot about these three equations. Can you do some substitution? Then simply follow through using basic algebra.

I knew it would be something ridiculously simple. Thank you very much gneill. I'm laughing at myself at the moment.

Ok so far I have a few values, I have tried following some of the earlier instructions but I'm stuck for part (d) As everyone L=1/((2pif)^2)C L=25.33 microhenrys M=k sqrt L1L2 Or M=kL K=0.5 M=0.5L With rearranging Leq = (L^2 - m^2)/(2L -2m) Replace m with kL...

For (b) use the two equations you obtained in (a). Quoting the legend that is gneill. V=V right?

I am suffering on (c). I need to find I which is I1+I2 Therefore I need to find I1 and I2 From (b) working backwards by one step. i1 (L1-m) = i2(L2-m) Therefore: i1 = (i2 (L2-m))/ (L1-m) i2 = (i1 (L1-m))/ (L2-m) Adding them together I still have two values for I If I use substitution...

I don't understand the need to times by three again. Line voltage = 415v Therefore phase voltage = 239.6 239.6/22.5+j22.5 = 5.33+j5.33A P=VI* P= 240 x 5.33-j5.33 P=1279.2 -j1279.2 x3 because of the three phases. 3837.6-j3837.6 = total power 3837.6W is actual power. Why is there a...

Ok, what I have from the stage where the voltages are converted to polar/rectangular V1= j415 V2= 415 Using -v1 + (j4 x i) + (j6 x i) + v2 i = 41.5 + j41.5 Therefore: Vth = j6 x i + v2 Now I have: Vth = 166 + j249 Zth = j2.4 ZL = 35 + j 35.7 Current flowing through load is iL =...