Question

Problem Two: You've got three capacitors, of size 100.0 1F, 220.0 F, and 470 uF. They are all rated at Vmax=200V. These are relatively large size of your thumb) electrolytic capacitors used in the power supplies of the most expensive computers and similar electronic equipment. Without being too specific about it yet, their job is to store and release enough energy to take out the ripple of our standard wall outlet power and provide steady voltage and current to the electronic computer components. a. How could you wire them up in combination (series, parallel, or a mix) to be able to take the largest voltage difference across that combination and how much voltage is that? How much energy are they storing for you at that point? b. If a was in series, how much energy in parallel ? Which is the better capacitor? c. If you drain that much energy in b at a steady rate during 1/60th of a second, what is that rate (or power, Joules per second, which is also called Watts).

Answer 1

100uF 2204F 470 MF Loov a = 59.9767 MF ♡ = Car = 59.9769810 x 200 = 11.9958103 C v (100) = 119.95 V. - U= 0.719 4 J. V(220) = 54.52 U V(450) = 25.52v U= 0.327 J = 0.1531 J = C = G+G+(3 =(100 + 220+470) MF = 790 MF .. v(100) = V(220) -V (470) = 200 v. U (100) = 2 CV ² = 4x100x10 #x200 U (220) = 4.4 J = 25 U(470) = 9.4J. first one is better. ① W & C V² = 4 x 790 x 15 x 200² = 15:8 I P = = 18,8 = 946 watte P = 948 W