Deriving capacitor voltage equation with initial condition

[JMFernandez]

Hi.

I don´t know if this question should be in the maths forum, but as it´s related with circuit analysis, I will post it here. I just would like to know how you get:

v(t) = 1/C ∫^t_t0 i(τ) dτ + v(t0)

From:

v(t)=1/C ∫^t_-∞ i(τ) dτ

I just know the basics of calculus and I don´t know how to operate the second equation to get the first one.

Thank you in advance.

 

[DaveE]

Hi.

I don´t know if this question should be in the maths forum, but as it´s related with circuit analysis, I will post it here. I just would like to know how you get:

v(t) = 1/C ∫^t_t0 i(τ) dτ + v(t0)

From:

v(t)=1/C ∫^t_-∞ i(τ) dτ

I just know the basics of calculus and I don´t know how to operate the second equation to get the first one.

Thank you in advance.

The equation describes how the cap voltage changes as current flows through it. Fortunately, all of the history of past current flow(s) is represented by the voltage at any time. That is what the initial voltage $v(t_o)$ is. Since that doesn't depend on the variable $t$, we can just call it a constant value, the "initial condition" of the capacitor. So,

$$v(t) = \frac{1}{C} \int_{-∞}^{t} i(\tau) \, d\tau = \frac{1}{C} \int_{-∞}^{t_o} i(\tau) \, d\tau + \frac{1}{C} \int_{t_o}^{t} i(\tau) \, d\tau \equiv v(t_o) + \frac{1}{C} \int_{t_o}^{t} i(\tau) \, d\tau$$

 

[JMFernandez]

Thank you. Very clear and concise explanation!!