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Interesting Question ~ Charging up Capacitors

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Willt

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Hi friends,

I have an interesting question from a professor ~

We have two capacitors with the same capacitance (note: C1 is charged up by half and C2 is empty). Let's say, they are now charged up separately by the same source, i.e.

V1---R1---C1 & V2---R2---C2 where V1=V2, R1=R2, C1 is charged up by half initially before connecting to the source and C2 is empty before connecting to the source

Questions:
(1) Which capacitor has a higher initial dV/dt?
(2) Which capacitor will be charged up first?

Professor's answer:
(1) C2
(2) C1 <--- I doubt this answer. Maybe the professor was making joke :D

Any comments are very welcome !!

Will
 

Maybe he is right, c1=c2 and r1=r2 and v1=v2, so the time constant is equal in both circuits, if one already has an inicial charge it will be cansidered charged before the one without charge.
 

how can i find a book about semiconductor,cyrstallies,lattice types,lattice constants,distribution functions and their logic.
 

Dear the answer is logical and could be proven math.
i will answer physically

1- first he needs to know which one will have charging rate initilly faster : of course the non-charged C because it will not repulse the starting charge that will charge it as there was no starting voltage

2- of coruse the 2nd C will charged first as it is half charged while the first C needs time to be charged to the same voltage level of first the keep on charging which will takes longer time
 

Willt,
Your professor is correct. In the first instance, the initial voltage across the R1 = V/2. The initial current through R1, C1 = V/2R1
.
In the 2nd instance, the initial voltage across R2 = V. The initial current= V/R2.
.
Rate of change of voltage across a capacitor is proportional to the current. Since the instantaneous current thru C2 is twice the current thru C1, the rate of change of voltage across C2 is twice the value for C1.
.
.
For the 2nd answer, the time constants are equal. Since C1 already has an initial voltage, its voltage at any time will always be higher than the voltage on C2.
regards,
Kral
 

Dear friends,

Sorry that I've made the mistake. The professor said that C2, which is empty initially, will be charged up first. This means C1 with initial charge will be charged up slower than C2. How come??

It sounds like: You put 2 glasses of water into the refrigerator. One is just normal water (1) while another is boiling (2). Just imagine, you put them into the refrigerator at the same time. And what you get is that (2) becomes ice first. This is exactly what the professor means ~ but I have the same idea as you guys. :D

Will
 

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