# DC Transients cap+resistor in series Theory

Discussion in 'Mathematics and Physics' started by Corky, Jan 8, 2015.

1. ### CorkyMember

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Hi Everyone,

I have a theory explination for a cap and resitor in series but the explination dissapears at the point on the photo.

i understand the theory of all the voltages must add to make the supply

could someone go through it starting with i=dq/dt im happy to answer any questions as i might know more than i remember

Cheers

2. ### RatchitWell-Known Member

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What is the question? As far as is shown, it looks like standard algebraic substitution. Where or what don't you understand?

Ratch

3. ### rumpfyActive Member

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the line you just cant read says;
"integrating both sides of the above equation"

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5. ### RatchitWell-Known Member

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Yes, that will solve the differential equation because the variables are separated. And the question is?

Ratch

6. ### CorkyMember

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Hi all,

im ok with algebra but im unsure what the letters represent as they dont state in the question.

does: dq=a given point of charge
dt= a given point in time
Q=CVc = charge is equal to capacitor value x voltage across the cap

if this is right i think ive got my head around it. its just a simple thing but im trying to get my head around what each letter represents

Cheers guys, P.S. sorry for the late reply

7. ### rumpfyActive Member

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Electric current is defined as; the rate of flow of charge. So, within an interval of time, if an amount of charge flows, then the current is (amount of charge /time interval). So mathematically, current i = q/t
For time varying circuit analysis, the time period is mathematically reduced to 'zero' and the incremental current is described as dq/dt and this means the (incremental change of charge/ a small increment of time). Summing the change of charge over time (integration) gives the total charge and hence the capacitor voltage from Vc= Qtot/C as you have said.

8. ### JimBSuper ModeratorMost Helpful Member

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Not really.

I am thinking that you have never studied calculus, defferentiation and integration.

dv/dt is the rate of change of voltage with time, and similarly
dq/dt is the rate of change of charge with time.

The letter d can be thought of as representing "a very small change in".

JimB

9. ### CorkyMember

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Hi,

I am terrible at explaining what im thinking, i knew that but i didnt realise d was "a small change in" i thaught it was a sample of a changing number and not a range

Cheers

Last edited: Jan 12, 2015
10. ### ColinActive Member

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yes. d is called or spoken as: "diddly"

11. ### MrAlWell-Known MemberMost Helpful Member

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Hi,

Often the lower case 'd' is taken to be "delta" which is used to indicate a small change in the variable that follows. So a few examples:
dt: a small change in t
dq: a small change in q
dv: a small change in v
di: a small change in i

and usually we'll see them in pairs being divided by another like:
dv/dt
which means "the change in voltage with respect to time", or more literally, "a small change in voltage as the time changes by a small amount".

When you integrate one of these by the respective variable you get the top variable as result, plus an as yet undefined constant:
Integral(dv/dt) with respect to time = v+K
Integral(di/dt) with respect to time = i+K