• JMFernandez
In summary, the conversation discusses how to derive the equation v(t) = 1/C ∫tt0 i(τ) dτ + v(t0) from v(t)=1/C ∫t-∞ i(τ) dτ, which describes the change in capacitor voltage as current flows through it. The initial voltage is represented by v(t0) and the two equations are equivalent due to the property of integration.
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 ∫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.

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.

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,

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

Last edited:
berkeman
Thank you. Very clear and concise explanation!!

berkeman and DaveE

## What is the integral equation for the voltage across a capacitor?

The integral equation for the voltage across a capacitor is $$V(t) = \frac{1}{C} \int_{0}^{t} I(\tau) \, d\tau + V(0)$$, where $$V(t)$$ is the voltage across the capacitor at time $$t$$, $$C$$ is the capacitance, $$I(\tau)$$ is the current through the capacitor as a function of time, and $$V(0)$$ is the initial voltage across the capacitor.

## How do you derive the capacitor voltage integral equation?

The capacitor voltage integral equation is derived from the relationship between current and voltage in a capacitor. The current $$I(t)$$ through a capacitor is related to the rate of change of voltage $$V(t)$$ by $$I(t) = C \frac{dV(t)}{dt}$$. By integrating both sides with respect to time, we obtain $$V(t) = \frac{1}{C} \int_{0}^{t} I(\tau) \, d\tau + V(0)$$.

## What does each term in the capacitor voltage integral equation represent?

In the equation $$V(t) = \frac{1}{C} \int_{0}^{t} I(\tau) \, d\tau + V(0)$$, $$V(t)$$ represents the voltage across the capacitor at time $$t$$, $$C$$ is the capacitance of the capacitor, $$I(\tau)$$ is the current flowing through the capacitor as a function of time, and $$V(0)$$ is the initial voltage across the capacitor at time $$t = 0$$.

## How does the initial voltage $$V(0)$$ affect the capacitor voltage over time?

The initial voltage $$V(0)$$ sets the starting point for the voltage across the capacitor. As the current flows through the capacitor, the voltage changes according to the integral of the current over time. The term $$V(0)$$ ensures that the voltage at time $$t = 0$$ is accounted for in the overall voltage $$V(t)$$ at any later time.

## Can the capacitor voltage integral equation be used for both AC and DC circuits?

Yes, the capacitor voltage integral equation can be used for both AC and DC circuits. For DC circuits, the current $$I(t)$$ is typically constant or a step function, making the integration straightforward. For AC circuits, $$I(t)$$ is usually a sinusoidal function, and the integral will reflect the time-varying nature of the current. In both

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