Problem recently posted for my class:

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?

Andre

Find ohms of the original.

R = E / I, R = 240V / 20A, R = 12 ohms.

Now this is used on a 208/120V system.

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.

Yes, thanks. Now a little harder second question. Find ground current.

20 Amps.

Ichabod

Should I draw this one up and post it in the Tech. Reference section?

Scott

Yes Scott, please do.

Ichabod