The number of square meters in the helium flux tube, which in the plasma density diagrams appears to be approx. 5 times larger than the sun, is 758,733,957,500,000 sq/metres. This equals a average current density of any cross section of the flux tube at 1,896,834,893,750,000,000,000 amps assuming a homogenous cross section. If we use the voltage of Jupiter's moon, Io's flux tube as a guess(I would like to use the current values as well but I do not have an area for this value), 400,000v * 1,896,834,893,750,000,000,000A = 7,587,339,575,000,000,000,000,000,000 watts or
7.5 * 10^26 watts per interstellar helium flux tube, of which there are 2 of them crossing the sun. The potential across the sun is the difference of the 2 flux tubes assuming one is positive and one is negative, or one is in and one is out, which is 15 *10^52 watts. Each pole of the sun is responsible for 1/2 of the 4*10^26 watts of the total output of the sun, or 2 * 10^26 watts.
The sun power output is 4*10^26 watts, which is well within the potential of an individual flux tube.