Find the voltage UAB (t) for t≥ 0. t=on チー ix 10V (+ 252 m JAB (t) ¹3ix =1/5 F 2-22
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the questions here are for the follow up answer. the explanation is amazing but i have two questions. when the circuit is open if a capacitor is present can i always consider a voltage source of value 1v and solve accordingly or are there other conditions linked to it? the 2nd one is why is vc(0-)=vc(0+)=vab(0-)-3ix? and the last question is where did u get the relation in t= infinity? the oen for vc(t)=vc(infinity)..... and why is ic(t)=ix(t)=Cdvc/dt? and why is VAB(t0=3IX+vc(t) please answer all my concerns and i will give a thumbs up ofcourse thnk you!
by Bartleby Expertthis is incorrect. please re-solve. it should be ve(0-)=-5/2v vab(0+)=5V vab(+infinity)=10v tao=1sec vab(t)=(-5e^-2t + 10)V please resolve and get these values correctly and please solve it in a detailed and understandable manner
by Bartleby Expert
