= Solution
The <AC Josephson effect> is $V=(\Phi_0/2\pi)\dot\delta$. Since the imposed flux is held fixed when defining a small-signal <inductance>, differentiate <8/c> to obtain
$$
\dot I=2I_C\cos\phi\cos\delta\,\dot\delta,\qquad
\boxed{L(\delta,\Phi)=\frac{V}{\dot I}
=\frac{\Phi_0}{4\pi I_C\cos\phi\cos\delta}.}
$$
On the stable small-current branch with $\cos\phi>0$, $\delta\simeq0$ and the result reduces to the printed expression,
$$
\boxed{L_{\rm eff}=\frac{\Phi_0}{4\pi I_C\cos(\pi\Phi/\Phi_0)}.}
$$
The <stable small-signal inductance of a symmetric DC SQUID> requires $|I|\ll2I_C|\cos\phi|$, stronger than $I\ll I_C$ near a half-integer flux quantum. If $\cos\phi<0$, the stable zero-current phase is $\delta\simeq\pi$ and $\cos\delta\simeq-1$. Its <inductance> is positive:
$$
\boxed{L_{\rm stable}=\frac{\Phi_0}{4\pi I_C|\cos(\pi\Phi/\Phi_0)|}.}
$$
For example, $\Phi=\Phi_0$ makes the printed signed formula negative when expanded about $\delta=0$, but that phase is an energy maximum. Expanding about the stable $\delta=\pi$ corrects the sign. At ideal half-integer flux the critical current vanishes and the linear <inductance> formally diverges; a finite transport current cannot satisfy the small-current approximation there. Geometric <inductance> and other nonidealities then have to be retained instead of extending the simple formula through its pole.
Back to article page
