What is the Relationship Between Charge and Current Over Time?

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In summary, the equation i(t) = \frac{dq(t)}{dt} can be rewritten as q(t) = \int^{t}_{t_{0}} i(t)dt + q(t_{0}), where q(t) represents charge and i(t) represents current as a function of time. This is a special case of the Net Change Theorem, which is a corollary to the Fundamental Theorem of Calculus. The equation q(t) - q(t_{0}) = [q(t)]^{t}_{t_{0}} = \int^{t}_{t_{0}} q'(t)dt can also be used to represent this relationship.
  • #1
matty204359
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I feel retarded because I'm not understanding this equation despite getting A's in all my math classes.

[tex]i(t) = \frac{dq(t)}{dt}[/tex]

some how becomes

[tex]q(t) = \int^{t}_{t_{0}} i(t)dt + q(t_{0})[/tex]

I'm guessing we apply the integral operator [tex]\int dt[/tex] to both sides. so where is the [tex]q(t_{0})[/tex] term on the RHS coming from?

Its a charge(q) and current(i) formula as a function of time(t) if that helps make more sense.
 
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  • #2
matty204359 said:
I feel retarded because I'm not understanding this equation despite getting A's in all my math classes.

[tex]i(t) = \frac{dq(t)}{dt}[/tex]

some how becomes

[tex]q(t) = \int^{t}_{t_{0}} i(t)dt + q(t_{0})[/tex]

I'm guessing we apply the integral operator [tex]\int dt[/tex] to both sides. so where is the [tex]q(t_{0})[/tex] term on the RHS coming from?

Its a charge(q) and current(i) formula as a function of time(t) if that helps make more sense.

[tex]q(t) = \int^{t}_{t_{0}} i(t)dt + q(t_{0})[/tex]

is a special case of

[tex]f(t) = \int^{t}_{t_{0}}f'(t)dt + f(t_{0})[/tex]

which is a corollary to the Fundamental Theorem of Calculus, sometimes referred to as the Net Change Theorem.
 
  • #3
himatty204359! welcome to pf! :smile:
matty204359 said:
[tex]q(t) = \int^{t}_{t_{0}} i(t)dt + q(t_{0})[/tex]

let's rewrite that as

[tex]q(t) - q(t_{0}) = [q(t)]^{t}_{t_{0}} = \int^{t}_{t_{0}} q'(t)dt[/tex]

(btw, you really ought to have a different variable, eg τ, inside the ∫ :wink:)
 

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