Why R, L and C behave as they do

Holds for a bar of constant cross-section, current spread evenly over it (low frequency) and constant $\rho$. Derivation: $v$ across length $\ell$ gives $E=v/\ell$; with $J=i/A$, the law $J=E/\rho$ gives $v=(\rho \ell /A)i$. $R$ in $\Omega$, $\rho$ in $\Omega \text{m}$, $\ell$ in m, $A$ in m${}^2$.

Holds over the limited range around $T_0$ for which $\alpha$ was measured; outside it, use measured values of $\rho$. $\alpha >0$ for metals. ${\rho }_0$ is $\rho$ at $T_0$, $\alpha$ in per ${}^{\circ }$C, $T$ and $T_0$ in ${}^{\circ }$C. Neglecting thermal expansion, $R$ follows the same law: $R(T)\approx R_0[1+\alpha (T-T_0)]$.

Holds for plates of area $A$ and spacing $d$, one uniform dielectric filling the gap, and $d$ small against the plate size so the edge field can be ignored. Derivation: charge $\pm q$ gives $E=q/(\epsilon A)$ between the plates (Gauss's law), so $v=Ed=qd/(\epsilon A)$ and $C=q/v$ is as shown. $\epsilon ={\epsilon }_r{\epsilon }_0$ with ${\epsilon }_0\approx 8.85\times 10^{-12}$ F/m, so the dielectric multiplies $C$ by ${\epsilon }_r$. $C$ in F, $A$ in m${}^2$, $d$ in m.

Follows from $q=Cv$ and $i=dq/dt$, for constant $C$. Passive sign convention throughout: $i$ enters the $+$ terminal of $v$, so $p=vi$ is the power absorbed ($p<0$ means the element delivers power). $C$ in F, $v$ in V, $i$ in A.

Holds when every turn links the same flux $\Phi$, the core is linear and $L$ is constant. Faraday's law gives an induced emf $-d\lambda /dt$ in the direction of $i$; the minus sign (Lenz) says it opposes the change. The voltage drop in the direction of $i$ is its negative, $v=+d\lambda /dt$, as the passive convention needs. $N$ turns, $\Phi$ in Wb, $\lambda$ in Wb-turns, $L$ in H.

Holds for a long, tightly wound coil with a uniform field inside and a linear core. Ampère's law gives $B=\mu Ni/\ell$ inside, so $\Phi =BA$ and $\lambda =N\Phi =(\mu N^{2}A/\ell )i$. $\mu ={\mu }_r{\mu }_0$ is the core permeability in H/m, ${\mu }_0\approx 4\pi \times 10^{-7}$ H/m; $B$ in T, $A$ the core area in m${}^2$, $\ell$ the coil length in m.

Holds for constant $C$, with $v=0$ at $t_0$ and $v=V$ at $t_1$. Step: $i=Cdv/dt$ gives $vidt=Cvdv$, so the time integral becomes one in $v$ with limits $0$ and $V$. An ideal $C$ dissipates nothing, so this energy is recoverable. $V$ in V, $w_C$ in J.

Holds for constant $L$, with $i=0$ at $t_0$ and $i=I$ at $t_1$. Step: $v=Ldi/dt$ gives $vidt=Lidi$, with limits $0$ and $I$. An ideal $L$ dissipates nothing, so this energy is recoverable. $I$ in A, $w_L$ in J.

Integrate $i=Cdv/dt$ from $0^{-}$ to $0^{+}$. Holds when $i$ stays finite through switching, so the integral over the vanishing interval is zero. If an ideal voltage source at a different voltage is switched straight across C, with no resistance in the loop, the current is an impulse and $v$ does jump. The same steps on $v=Ldi/dt$ give $i(0^{+})=i(0^{-})$ while $v$ stays finite.