Recent content by oph

On an other site I have found some information but I don´t know if they are useful to solve the problem. For the continuous and steady operation the law of induction applies ΔIL = \frac{1}{L}Ve ⋅ t1 = \frac{1}{L} (Va - Ve) ⋅ (T-t1) Va = Ve \frac{T}{T-t1} The output voltage depends...

Yes, you are right. I should determine the maximum of Vc and i`m looking for a general expression. I have really no idea where I should start, can you please give me a tip?

but where should I start?

task 2: Derive an expression for the maximum capacitor voltage in dependence on the occurring sizes. the task has to do with an up-converter and I should determine an equation for the maximum capacitor voltage this are the new details: Since the voltage Vo of the battery is lower than...

i really don`t know how should solve this task but, maybe with this equation: law of induction: ΔIL = (1/L) * Vi * t1 = (1/L) * (Vi-Vo) * (T-t1) Vi= input voltage Vo= output voltage Vi=Vo ·T/(T-t1) can somebody say me if this is the right way?

Thank you for all your Great help! Have anyone an Idea how i can solve task 2?

Vc(t) = Ve^-t/(RC) Vc(150ms)/V = e^-t/(RC) ln (0.05) = -t/(RC) ln (0.05) * (RC) = -t C = -t/(ln(0.05) * R) = -0.150/ln(0.05) * 100 Ω = 5.007 * 10^-4 F is this right?

V = 895.79 Volt Is this right? Task 1 was: Estimate the capacitance of the capacitor which must have and what voltage it needs to be charged for the operation at least. Now I have got the voltage but what is with the capacitance?

(5) V = √ln(0.05) * 2Uc * 100Ω/(-150ms) -150ms = -0.150s V = √ln(0.05) * 2 * 200J *(100Ω/(-0.150s)) = √-2,99 * 400J * (-666.67) = √ 798861.94 = 893.79

(5) V = √ln(0.05) * 2Uc * 100Ω/(-150ms) V = √ln(0.05) * 2 * 200J *(100Ω/(-150ms)) = √-2,99 * 400J * (-2/3) = √ 798,86 = 28,26 what is wrong?

V = √798.86 = 28.26 is this right? and then?

(3) ln(0.05) * 2Uc = -150ms/100 Ω * V^2 (4) V^2 = ln(0.05) * 2Uc * 100Ω/(-150ms) (5) V = √ln(0.05) * 2Uc * 100Ω/(-150ms) is this right?

(1) ln(0.05)=-150ms/(100 Ω*(2Uc/V^2)) (2) ln(0.05) * (2Uc/V^2) = -150ms/100 Ω are this two equations right?

ln (0.05) = ln(-150ms/(100 Ω*(2Uc/V^2))) (2Uc/V^2)*ln (0.05) = ln(-150ms/100 Ω) (2Uc/V^2) = ln(-150ms/100 Ω) / ln (0.05) (2Uc/V^2) = (-150ms/100 Ω) / 0.05 (2Uc/V^2) = -3000ms/2000Ω is this right? and then?

yes, i would agree but how should i go on? should i insert it in the equation? 0.05=e^-150ms/(100 Ω*(2Uc/V^2)) and then??thank you for your great help!