# Calculate ground fault current leak on human contact

I am sure this is very very simple and probably realy dumb but I've been banging my head for hours and can't figure this out. Or find an answer on google/books for that matter. Even with the diagram on the right i can't figure this out. This is supposed to be a demonstration of the function of how and when an rcd trips. To simulate it we used an appliance (in this case a washing machine) that is not properly insulated and the person touching the chassis.

D is the insulation fault resistance, T is the earth conductor resistance, and H is the body resistance.

I want to calculate Id (leakage current), the contact voltage and Ih (current in the person). Voltage and resistance values are given but I want to calculate the currents for multiple values.

• What have you figured out so far? – Spehro Pefhany Dec 19 '18 at 1:05
• I don't understand what you mean by "what have you figured out". I know that part of the current will go through It and the rest through Ih. How much depends on the resistance of T. That's about it. – Roaf Beast Dec 19 '18 at 1:12
• If the fault is the path for T, then the current is I= V(Ph)/R(D+H) but the value for H is not a resistor but rather R1+R2//C ok? – Sunnyskyguy EE75 Dec 19 '18 at 1:56
• This is an analogy to the "Human body model" that engineers use to calculate shock hazards. Use that as a search term to find out a lot more. – Sparky256 Dec 19 '18 at 1:56
• @Sparky256 almost, the HBM is 100pF used for ESD discharge which is different than the dielectric breakdown and impedance of a palm model at 50,60Hz , 100pF is only be 25 Mohm at line f but near 0 Ohms @ 300MHz. Leakage capacitance in a hand or clenched finger could easily exceed 10nF and drop to 50k – Sunnyskyguy EE75 Dec 19 '18 at 1:59