Answer 14.2

_psets/11.md

@@ -54,6 +54,38 @@ velocity v by $$IV = mgv$$. Using the inverse AC Josephson effect (equation 14.2
 voltage, and the quantum Hall effect (equation 13.41) along with the inverse AC Josephson effect to
 determine the current, relate the measurement to fundamental constant(s).
 
+In problem 6.5 in [problem set 4](04.html) we found that
+
+$$
+m = -\frac{I V}{g v}
+$$
+
+The AC Josephson effect gives us a relation between voltage and frequency that only depends on
+fundamental constants (and n, a positive integer).
+
+$$
+V = n \frac{h}{2 e} f
+$$
+
+The quantum Hall effect can give us a resistance that only depends on fundamental constants (and i,
+a positive integer).
+
+$$
+R_H = \frac{h}{i e^2}
+$$
+
+By Ohm's Law $$IV = V^2/R$$, so the Kibble balance equation can be written as
+
+$$
+mgv = -\frac{V^2}{R}
+$$
+
+Plugging in the values above we find
+
+$$
+2mgv = -h i n^2 f^2
+$$
+
 
 ## (14.3)
 
@@ -150,10 +182,6 @@ Let's zoom in on those spikes.
 ![reactance vs frequency](../assets/img/pset11_im_zoom.jpg)
 ![phase vs frequency](../assets/img/pset11_phase_zoom.jpg)
 
-$$
-\begin{align*}
-\end{align*}
-$$
 
 ## (14.5)