Recent content by ifan davies

I will keep that in mind. Thank you

Or that yea lol 0.05 F

0.48 = 20/ 2 x 400 x C 0.48/20 = 2 x 400 x C 0.024 = 2x 400 x C 0.024/2= 400 x C 0.012= 400 x C C = 0.00003 F or 30 uF Not correct?

30 uF it is my bad! Thanks for the help, think il go and attempt some other questions now.

0.48= 200/2x400x C 0.00024= 2x 400 x C 0.0012= 400 x C C= 3 uF

When you put it like that its easy! Your awesome! Now to calculate a suitable value of smoothing capacitor C= i load / frequency x V 20/ (400x0.14) 0.357 F

this has to be it ... Kr x Vdc = Vrms 0.005 x 28 = 0.14

Vrms = 115/sqrt2 Vdc= 28 Kr= 81.3/28 Kr = 2.9 Please be right

its the average load voltage according to my notes? Thanks for sticking with me!

Kv= rms value of the a.c. voltage component/ average value of the load

Vpk= Vaverage x (pi/2) Vpk = 28 x (pi/2) Vpk= 43.98v Vrms= Vpk/sqrt(3) Vrms = 43.98/ sqrt(3) Vrms= 25.39

right let's start again... Vrms= Vripple/2sqrt3 Vripple= I(load)/2fc Vripple= 2/(2x400) = 0.0025 Vrms= 0.0025/(2sqrt3) vrms= 0.00072 this seems low..

Hey hey, its the first one you mentioned where it's considered to be equal to the ratio of the RMS ripple voltage to the DC component of the power supply output.

An aircraft electrical system operates at 115 volts at 400 hertz. It is to supply an average d.c. voltage of 28 volts and an average current of 20 A at a ripple factor of 0.005. Determine: The r.m.s. ripple voltage Here is my attempt. Vc= Vs-2Vd =28-2(0.7) = 26.6 Volts Vr= (1/2 x fs x C...