# How would I explain a farad being a second per ohm?

The farad has many alternative representations in SI. I can comprehend some of them. For example, a capacitance of 1 farad means that it will take a single coulomb to create a single volt between the capacitor plates. It could also mean that it will take one second of continuous 1A current to create that volt between the plates. In this manner, how would I imagine the second/ohm representation?

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If you subject a capacitor of capacitance $C$ to a voltage $v(t)=V_0\cos(\omega t)$, then the current $i(t)$ leading to its plates, which have charge $q(t)$, will obey $$i(t)=\frac{dq}{dt}=C\frac{dv}{dt}.$$ This is best analyzed by considering the (slightly fictitious) complex voltage $v(t)=V_0e^{j\omega t}$, where $j^2=-1$ by the electrical engineers' convention, and all physical results can be obtained by taking the real part of the relevant, physically measurable quantities.
Since now you have $\frac{dv}{dt}=j\omega v(t)$, the differential equation above comes down to the algebraic equation $i=j\omega C v,$ or $$v(t)=\frac{1}{j\omega C}i(t).$$ This almost has the form of Ohm's law, except that the "resistance" is now complex. This is known as the reactance of the capacitor as a circuit element, and denoted $$X=\frac{1}{j\omega C}.$$ Reactances are nice! They let you talk in Ohm's-law terms even when the real voltages and currents are not, strictly speaking, proportional (since they're out of phase). They treat resistors, capacitors and inductors on an equal footing. They combine like resistances when the circuit elements are connected in series, in parallel, or even in the weirder star and delta configurations. They embody the most general linear response to an alternating current. They have an apparent downside: if your signal is not sinusoidal then you need to decompose it into Fourier components and treat each contribution separately. This is actually an upside: they let you treat complicated time-varying signals in the spectral terms you really should be talking about.
How, precisely, does this relate to your question? You really should see the identity as $$\frac{1}{ \text{ Hz F}}=1\,\Omega,$$ which lets you calculate the "resistance" (reactance) of your capacitor to voltage of a given frequency. Thus a 1 F capacitor will let through one ampere per applied volt at 1 Hz AC. (Up to factors of $2\pi$! Be careful of where it should be frequency and where it should be angular frequency. What's the correct version of my statement?)
A good exercise is to repeat all this exercise for the corresponding identity for inductance: $$1\text{ H}=1\text{ s }\Omega.$$ The important concept there is impedance, which you've probably heard about.