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Integrating Current-Voltage Relationship for a Capacitor

  • Thread starter vg19
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1. Homework Statement

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


3. 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
 

Answers and Replies

marcusl
Science Advisor
Gold Member
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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.
 

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