2 Fixed typo.
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Talk about "current leading voltage" or "phase difference" only applies to AC analysis. In the more general case, one could say what a capacitor really does is differentiate voltage, according to:

$$ i = C\frac{dv}{dt} $$

From this, you can derive all sorts of well-known things about capacitors. Such as, if you want a linearly changing voltage across a capacitor, you must apply a constant-current source to it. As an example, consider a 1 ampere current source connected to a 1 farad capacitor:

$$ \require{cancel} \begin{align} 1A &= 1F \frac{dv}{dt} \\ 1A &= \frac{1 A \cdot s}{V} \frac{dv}{dt} \\ \frac{1\cancel{A}\cdot V}{1\cancel{A}\cdot s} &= \frac{dv}{dt} \\ \frac{1V}{s} &= \frac{dv}{dt} \end{align} $$

If you consider the case where the applied voltage is sinusoidal, then so too is the current:

$$ \begin{align} i &= C\frac{dv}{dt} \\ i &= C\frac{d\sin(t)}{dt} \\ i &= C\cos(t) \end{align} $$

because \$\cos\$ is the derivitivederivative of \$\sin\$.

You will also see if you graph these functions, that \$\cos\$ (current) leads \$\sin\$ (voltage) by 90 degrees, as an electrical engineer would put it:

plot of sin and cos

Talk about "current leading voltage" or "phase difference" only applies to AC analysis. In the more general case, one could say what a capacitor really does is differentiate voltage, according to:

$$ i = C\frac{dv}{dt} $$

From this, you can derive all sorts of well-known things about capacitors. Such as, if you want a linearly changing voltage across a capacitor, you must apply a constant-current source to it. As an example, consider a 1 ampere current source connected to a 1 farad capacitor:

$$ \require{cancel} \begin{align} 1A &= 1F \frac{dv}{dt} \\ 1A &= \frac{1 A \cdot s}{V} \frac{dv}{dt} \\ \frac{1\cancel{A}\cdot V}{1\cancel{A}\cdot s} &= \frac{dv}{dt} \\ \frac{1V}{s} &= \frac{dv}{dt} \end{align} $$

If you consider the case where the applied voltage is sinusoidal, then so too is the current:

$$ \begin{align} i &= C\frac{dv}{dt} \\ i &= C\frac{d\sin(t)}{dt} \\ i &= C\cos(t) \end{align} $$

because \$\cos\$ is the derivitive of \$\sin\$.

You will also see if you graph these functions, that \$\cos\$ (current) leads \$\sin\$ (voltage) by 90 degrees, as an electrical engineer would put it:

plot of sin and cos

Talk about "current leading voltage" or "phase difference" only applies to AC analysis. In the more general case, one could say what a capacitor really does is differentiate voltage, according to:

$$ i = C\frac{dv}{dt} $$

From this, you can derive all sorts of well-known things about capacitors. Such as, if you want a linearly changing voltage across a capacitor, you must apply a constant-current source to it. As an example, consider a 1 ampere current source connected to a 1 farad capacitor:

$$ \require{cancel} \begin{align} 1A &= 1F \frac{dv}{dt} \\ 1A &= \frac{1 A \cdot s}{V} \frac{dv}{dt} \\ \frac{1\cancel{A}\cdot V}{1\cancel{A}\cdot s} &= \frac{dv}{dt} \\ \frac{1V}{s} &= \frac{dv}{dt} \end{align} $$

If you consider the case where the applied voltage is sinusoidal, then so too is the current:

$$ \begin{align} i &= C\frac{dv}{dt} \\ i &= C\frac{d\sin(t)}{dt} \\ i &= C\cos(t) \end{align} $$

because \$\cos\$ is the derivative of \$\sin\$.

You will also see if you graph these functions, that \$\cos\$ (current) leads \$\sin\$ (voltage) by 90 degrees, as an electrical engineer would put it:

plot of sin and cos

1
source | link

Talk about "current leading voltage" or "phase difference" only applies to AC analysis. In the more general case, one could say what a capacitor really does is differentiate voltage, according to:

$$ i = C\frac{dv}{dt} $$

From this, you can derive all sorts of well-known things about capacitors. Such as, if you want a linearly changing voltage across a capacitor, you must apply a constant-current source to it. As an example, consider a 1 ampere current source connected to a 1 farad capacitor:

$$ \require{cancel} \begin{align} 1A &= 1F \frac{dv}{dt} \\ 1A &= \frac{1 A \cdot s}{V} \frac{dv}{dt} \\ \frac{1\cancel{A}\cdot V}{1\cancel{A}\cdot s} &= \frac{dv}{dt} \\ \frac{1V}{s} &= \frac{dv}{dt} \end{align} $$

If you consider the case where the applied voltage is sinusoidal, then so too is the current:

$$ \begin{align} i &= C\frac{dv}{dt} \\ i &= C\frac{d\sin(t)}{dt} \\ i &= C\cos(t) \end{align} $$

because \$\cos\$ is the derivitive of \$\sin\$.

You will also see if you graph these functions, that \$\cos\$ (current) leads \$\sin\$ (voltage) by 90 degrees, as an electrical engineer would put it:

plot of sin and cos