# Physical explanation for capacitive circuit

In capacitive circuit, the voltage and current are out of phase. the current leads the voltage by 90 degrees. this can be explained by these two equations:

$$V = V_{\max} \sin \omega t$$

$$I = I_{\max} \cos \omega t.$$

Now, my question is that what does this statement mean physically that "current leads the voltage by 90 degrees? It is clear mathematically that if $V = 0$ then $I$ is maximum but how do we explain this physically? hope you understood my question

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Capacitor charge $q$ is just the accumulation of the current $i$ into the capacitor:

$$q = \int i \, dt \quad \text{or} \quad i = \frac{dq}{dt}$$

So, starting with an uncharged capacitor, a current $i$ does not instantaneously lead to a charge $q$; instead charge builds up with time, lagging the current. The formulas you quoted are direct consequences of this behavior.

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Not only of. A capacitor hasn’t a specific provision to operate with sine wave signals, but the original poster presumed such waveform. So, each one of the formulas in the original post is a consequence of another one and the equation with “dt”. –  Incnis Mrsi Oct 24 '14 at 20:37
@IncnisMrsi, I take your point that my answer could be elaborated. Would you care to add an answer that does so? –  Art Brown Oct 24 '14 at 20:47

The “if $V = 0$ then $I$ is maximum” statement does not show a distinctive character of a capacitor and depends on external conditions, namely an exactly sinusoidal waveform without any DC bias.

Indeed, variables should be swapped. When a smooth voltage function $V$ is at its extremum and, hence, has zero time derivative (neither grows nor diminishes), then so does the charge $q$ and, consequently, the current at such moment doesn’t flow. This feature is not restricted to sine waves; continuous differentiability ($C^1$) of $V$ is the requirement, and also the capacitor is assumed to be ideal (it must have no leaks, i.e. the space between and around conductors must have zero conductivity, whereas conductors must have zero voltage drop).

Qualitatively this “current leading by a quarter of period”, not necessarily for a sine waveform, can be explicated with the following table:

$V$         $I$        power

   increasing           same sign        consumed
(by absolute value)     as voltage      (like a resistor)

extremum              0                0

decreasing          opposite sign     released
(by absolute value)    than voltage     (like a battery)

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My bad: non-zero series resistance, of course, adds a current-dependent voltage drop to $V$ and the nice picture becomes spoiled. Thanks! –  Incnis Mrsi Oct 25 '14 at 6:58