Recent content by magician

And PF is lagging... 50< - cos^-1 . 0.7

Try: Vth = i * 2.4j + i * Z_load

Ok. I haven't checked your maths, but using that, you will now have a Voltage, Vth connected to j2.4, connected to Z_load, with current, i through the load. Calculate Z_load (pretty simple) Hence current in load is: Vth = ...

Also, Vth = j6 * I + V2

Then if you have the current, continue to find the Thevevin equivalent voltage. Once you have that, then you should be able to sketch a very simplified drawing. From that calculate Z(load) and find the current in the load. This can be expressed in the same format as the beginning voltages...

Remember. The PF is lagging so the circuit equation (for calculating the current) will begin like this: CIRCUIT EQUATION; -V1 + j4 * I + j6 * I + V2 =0 Thus; -415<90° + j4 * I + j6 * I + 415<0° = 0 Continue with the above until you get your result for current, I. Then you can quite easily...

Completly understand Ebies. Put up what you've got. Ive had mine marked and it is correct

Hey Ebies, As gneill has already said, "The 'j' represents the square root of negative one." To help you get started, below is the calculations I used; 1/Z_th = 1/j4 + 1/j6 1/Z_th = (4+6)/j24 1/Z_th = 10/j24 Therefore; Z_th = j24/10 Z_th = j2.4 Now following the procedure I did above, [post...

Ah, damn it! That makes sense when you put it like that! OK. I will re calculate using; Z_L=50∠45.572°Ω Thank you

OK. This has been an absolute nightmare for the last month or so! This is what I have so far... Thevenin equivalent reactance; 1/Z_th = 24j/10 = 2.4j v1 = sqrt2 * 415sin(100pi.t +90) v2 = sqrt2 * 415sin(100pi.t) converting to rms value: v1 = 415∠90 v2 = 415∠0 circuit eq now; -v1 + j4 * i...

OK. So cosine is phase shifted by 90°. V2 = √2 * 415 Cos(100π t - 90)

Thanks again to you both for the above post's. It's been a great help so far. So, I believe I am at the point of calculating the equivalent impedance. Which I have calculated as; Z_t = (j4 * j6) / (j4 + j6) Z_t = -24 / j10 Z_t = j2.4 Then I need to find the equivalent voltage. So the...

But, then the circuit would be in 2 parts, split down the middle. ##Edited## Oh dear. I believe I was taking it far too literally. Ignore the above statement! Would the next step then be; To work out the current by adding the resistance in series. Then reconnecting the load, and work out...

Hi guys - Sorry for the late reply, but I went back to the drawing board as I just wasn't understanding it. And to be honest - I'm still lost. From the notes I have: "First find the equivalent resistance. To do this we remove the load..." OK That bit makes sense. "Replace all sources with...

In rectangular form, would it be written as; ##Z_L = 35 \Omega + j35.71## ? Then would I; Disregard the right portion of the circuit and and find the total impedance of the coil and load. Find the equivalent Voltage across the open circuit terminals (disregarding the j6 coil?) Really...