Question about this Capacitor voltage integral equation

AI Thread Summary
The discussion focuses on deriving the capacitor voltage equation from the integral of current. The original equation, v(t) = 1/C ∫t-∞ i(τ) dτ, is transformed by recognizing that the initial voltage v(t0) represents the history of current flow. By splitting the integral into two parts, the relationship v(t) = v(t0) + 1/C ∫tt0 i(τ) dτ is established. This shows how the voltage at time t depends on both the initial condition and the current flowing from t0 to t. The explanation clarifies the connection between past current and present voltage in circuit analysis.
JMFernandez
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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 ∫tt0 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.
 
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JMFernandez said:
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 ∫tt0 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,

$$v(t) = \frac{1}{C} \int_{-∞}^{t} i(\tau) \, d\tau = \frac{1}{C} \int_{-∞}^{t_o} i(\tau) \, d\tau + \frac{1}{C} \int_{t_o}^{t} i(\tau) \, d\tau \equiv v(t_o) + \frac{1}{C} \int_{t_o}^{t} i(\tau) \, d\tau$$
 
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Thank you. Very clear and concise explanation!!
:smile::smile:
 
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