What is the Relationship Between Charge and Current Over Time?

[matty204359]

I feel retarded because I'm not understanding this equation despite getting A's in all my math classes.

i(t) = \frac{dq(t)}{dt}

some how becomes

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

I'm guessing we apply the integral operator \int dt to both sides. so where is the q(t_{0}) 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.

 

[gopher_p]

I feel retarded because I'm not understanding this equation despite getting A's in all my math classes.

i(t) = \frac{dq(t)}{dt}

some how becomes

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

I'm guessing we apply the integral operator \int dt to both sides. so where is the q(t_{0}) 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.

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

is a special case of

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

which is a corollary to the Fundamental Theorem of Calculus, sometimes referred to as the Net Change Theorem.

 

[tiny-tim]

himatty204359! welcome to pf!

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

let's rewrite that as

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

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