Integrating Current-Voltage Relationship for a Capacitor

[vg19]

Homework Statement

To find the voltage-current relationship of a capcitor, integrate both sides of
i = C(dv/dt)

The Attempt at a Solution

In the book they say, v = (1/C)(The Integral from -tve infitity to t) of i dt.

or

v = (1/C)(The Integral from tnot to t) of i dt + v(tnot)


I am trying to understand why they integrated from negative infinity. Why not start the integration at 0. I also do not undersstand in the second equation where the + v(tnot) came from.

Thanks in advance

 

[marcusl]

An ideal capacitor will store all the charge that's ever applied to it. What you call 0 time is arbitrary (it could be a certain date and time, or when you start taking data, or anything else)--but the capacitor voltage also reflects currents that flowed before that time, all the way back.

BTW, the subscript is "nought" which means zero, not "not".

V_t0 is the integration constant (look back at your calculus book for an indefinite integral). The capacitor might have started with a voltage on it before it was hooked into your circuit.

 

[piercas]

I would approach this problem by first understanding the physical meaning of the equations and then analyzing the mathematics. The equation i = C(dv/dt) represents the relationship between the current and voltage in a capacitor, where C is the capacitance. This equation is known as the "charging equation" and it describes how the current changes with respect to time as the capacitor charges.

To find the voltage-current relationship, we need to integrate both sides of the equation. Integration is a mathematical operation that helps us find the total value of a quantity over a certain period of time. In this case, we are integrating the current with respect to time, which will give us the voltage.

Now, let's look at the first equation v = (1/C)(The Integral from -tve infitity to t) of i dt. The integral from -tve infitity to t represents the total value of the current from negative infinity to a certain time t. This means that we are considering the entire history of the current, starting from the moment the capacitor was connected to a voltage source. This is important because the voltage across a capacitor depends not only on the current at a given time, but also on the previous values of the current.

On the other hand, the second equation v = (1/C)(The Integral from tnot to t) of i dt + v(tnot) considers the current only from a certain initial time tnot to a later time t. This means that we are only looking at a specific time interval and not the entire history of the current. The term v(tnot) represents the initial voltage across the capacitor at time tnot.

So, to answer your question, we integrate from negative infinity to take into account the entire history of the current and its effect on the voltage across the capacitor. And the term v(tnot) is added to consider the initial voltage at a certain time tnot.

In summary, integrating the current-voltage relationship for a capacitor helps us understand how the voltage across the capacitor changes over time. By integrating from negative infinity, we take into account the entire history of the current, and the term v(tnot) accounts for the initial voltage at a certain time.