Inductor vs. capacitor discharging - how a low resistance affects the two?

Question

I'm confused about why the two discharging patterns occur in regards to resistance. If I am understanding correctly:

An inductor: A lower resistance means that it will take longer for the inductor to discharge. Theoretically, infinitely small resistance means that the inductor will discharge forever (source) Higher resistance means it will take less time for the inductor to discharge. But then I was thinking that the lower the resistance, the higher the power dissipated ($I^2R$) and the faster it will discharge. But this is not the case.

A capacitor, on the other hand, with a lower resistance, will discharge faster. Infinitely small resistance theoretically would mean the capacitor discharges instantly. This aligns with the power dissipated idea, but I completely don't get why the two work opposite, and why the change in resistance causes these shifts in discharge times.

Answer 1

The energy stored in a capacitor is $\dfrac 12 \dfrac{Q^2}{C}$ where $Q$ is the charge stored and $C$ is the capacitance of the capacitor.

To remove the stored energy as quickly as possible you need to reduce the charge $Q$ as rapidly as possible.
So what you need is to make $\dfrac {dQ}{dt} = I$, the current as large as possible.
You do that by making the resistance in the circuit as small as possible.

The energy stored in an inductor is $\dfrac 1 2 L I^2$ where $L$ is the inductance of the inductor and $I$ is the current passing through the inductor.
So what you need is to make $\dfrac {dI}{dt}$, the rate of change of current as large as possible.
In this case it is the rate of change of current which is important not the current itself.

That change of current will be opposed (Lenz) by the induced emf (really induced current - Faraday) and for a given induced emf the larger the resistance in the circuit the smaller will be the induced current opposing the change in current.
If there is less opposition to the change in the current then $\dfrac{dI}{dt}$ will be larger.

Another way of looking at the inductor and resistor is consider what happens soon after a resistor $R$ is connected to the terminals of the inductor which has a current $I$ flowing thought it.

The voltage across the resistor is $IR = L \dfrac {dI}{dt} \Rightarrow \dfrac {dI}{dt}=\dfrac{R}{L}$.

So if you want the rate of change of current to be large you make the resistance of the resistance large.