Heating element rated 240v 4800W (20A) has been connected to 120/208v grounded "Y" system between two phases (208v). - Everybody egrees now 3600W / 17.3A. 1. Midpoint of the element is shorted to ground assuming 0 (zero) resistance of ground fault. What total power is rreleased during ground fault? 2. What current flow to ground?
What is your opinion on asked numbers? - Is it too hard for master electrician?
The Neutral of 208/120V system is intentionaly connected to Ground and you noted 0 restistance to ground, thus the circuit at midpoint is being operated as on 120V system.
1/2 of the 12 ohms is 6 ohms.
I on the 120v as now operating is;
I = E / R, I = 120V / 6R, I = 20A on each of the two sections of the ground faulted 240, 4800W, 20A heater operated on the 208V system.