- #1
jeff1evesque
- 312
- 0
Statement:
Current by Ohm's law is written as
[tex]I = \frac{V}{R}[/tex] (#1)
Instantaneous current through a capacitor is written as
[tex]i = C \frac{dv}{dt}[/tex] (#2)Questions:
1. Why is equation #2 not written as [tex]i = C \frac{dv}{R dt}?[/tex]
*I am guessing it's because the resistance in a capacitor is negligible- is that a reasonable assumption?
2. Is equation #2 equivalent to taking the derivative of velocity to get acceleration?
3. I understand mathematically that current is 90 degrees out of phase with voltage for our equation #2 of instantaneous current. But can someone explain a more general physical explanation- in terms with the capacitor?
*Does this being out of phase mean that voltage lags current (90 degrees) or the other way around?
4. For some reason, I find that "current through a capacitor" is bad wording since current never goes through a capacitor. In DC, a plate becomes charged, and in AC, plates discharge- which is analogous with what most authors are trying to convey by saying "current through a capacitor".
Thanks,
JL
Current by Ohm's law is written as
[tex]I = \frac{V}{R}[/tex] (#1)
Instantaneous current through a capacitor is written as
[tex]i = C \frac{dv}{dt}[/tex] (#2)Questions:
1. Why is equation #2 not written as [tex]i = C \frac{dv}{R dt}?[/tex]
*I am guessing it's because the resistance in a capacitor is negligible- is that a reasonable assumption?
2. Is equation #2 equivalent to taking the derivative of velocity to get acceleration?
3. I understand mathematically that current is 90 degrees out of phase with voltage for our equation #2 of instantaneous current. But can someone explain a more general physical explanation- in terms with the capacitor?
*Does this being out of phase mean that voltage lags current (90 degrees) or the other way around?
4. For some reason, I find that "current through a capacitor" is bad wording since current never goes through a capacitor. In DC, a plate becomes charged, and in AC, plates discharge- which is analogous with what most authors are trying to convey by saying "current through a capacitor".
Thanks,
JL
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