I just dont get why my answer is not correct
>>27229141
Rotate it and I will help you.
>>27229141
because it's sideways
>>27229172
done
hows this?
>>27229207
Oops sorry. I thought it was a geography homework. I have no idea about wire technology.
What course is this?
Inb4 mute.
i mean, the units would already be wrong
answer is 1/5 cv
Calculate the Ceq of the whole thing on the right side. Then you will have. C1 and Ceq. You will know how much V falls in C1 that way really easily because you know that V has to be the same in the node. Afther that you know how much it's in that node and then you need to calculate how much it falls in C2 which you can do simply knowing how much it falls in both side of the parallel thing. Then you make the ratio with C3.
t. EE
>>27229384
c23 = 1/[(1/c)+(1/c)] = 0.5 c
c234 = c23 + c4 = 1.5c
c2345 = 1/[(1/c234) + (1/c5)] = 3/5 c
q2345 = 3/5 cv = q234
q23 = 1/5 cv = q2
You have to account for the transients.
>>27229276
Probably the first week of a first-year general physics course.
Certified EE here. What seems to be the problem?
>>27229141
i studied that when i was 15
why the hell are the betas helping this kid?
>>27230267
You're a smart guy...
Original
>>27229141
Just find the equivalent Capacitance.
then q = CV
To find the equivalent capacitance use the parallel formula for the top right ones. Then add the result to the bottom right one. And then use the result as parallel to the far left capacitor. That would be the equivalent capacitance.
