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+### Statement
+
+$11.2.9.$ [Insert the problem statement]
+
+### Solution
+
+a) The area of the triangle:
+
+$$S = \frac{a^2 \sqrt{3}}{4}$$
+
+Denote by $v$ is the rate of change of flow through a unit area. Then the potential difference in the triangular frame is:
+
+$$U = \frac{a^2 \sqrt{3} v}{4}$$
+
+Frame resistance:
+
+$$R = 3\rho a$$
+
+Hence the current strength:
+
+$$I = \frac{U}{R} \approx 1.44 mA$$
+
+b) Considering our circle with a jumper as two circuits, the current in each circuit is the same and the current in the jumper goes towards each other, which means the current in the jumper is 0.
+
+Potential difference in the ring:
+
+$$U = \pi R^2 v$$
+
+Ring resistance:
+
+$$R = 2\pi R \rho$$
+
+The desired current in the arcs:
+
+$$I = \frac{U}{R} = 2.5 mA$$
+
+c) The potential difference to a greater extent and to a lesser extent, respectively:
+
+$$U_1 = (\frac{3}{4}\pi + \frac{1}{2})vR^2$$
+
+$$U_2 = (\frac{1}{4}\pi - \frac{1}{2})vR^2$$
+
+The resistances of the larger arc, the smaller arc, and the jumper, respectively:
+
+$$r_1 = \frac{3}{2}\piR \rho$$
+$$r_2 = \frac{1}{2}\piR \rho$$
+$$r_3 = \sqrt{2}R \rho$$
+
+So let's write down the Kirchhoff rule system.:
+
+$$I_1 - I_2 = I_3$$
+$$I_1 r_1 + I_3 r_3 = U_1$$
+$$I_2 r_2 - I_3 r_3 = U_2$$
+
+These equations are easier to solve by preparatory calculations of each voltage and resistance, I got the following currents:
+
+$$I_1 \approx 2.56 mA$$
+
+$$I_2 \approx 1.69 mA$$
+
+$$I_3 \approx 0.87 mA$$
+
+Savchenko's numbers are different, but very close to mine, if anyone has any ideas, write to me.
+
+#### Answer
+
+[Insert a concise answer or boxed result]