In our example, the resistance of each circuit conductor is 0.20 ohms (100 ft of 12 AWG), the load has a resistance of 200 ohms, and our accident victim has a skin contact resistance of 1,000 ohms. To determine the voltage (potential) with our accident victim as part of the neutral current path, we need only perform four short calculations:
Step 1. Determine the total resistance of the series circuit using Ohm's Law.
RT=R1+R2+R3+R4
R1=Hot Wire=0.20 ohms
R2=Load=200 ohms
R3=Person=1,000 ohms
R4=Neutral Wire=0.20 ohms
RT=0.2 ohms+200 ohms+1,000 ohms+0.2 ohms
RT=1,200.40 ohms
Step 2. Determine the current of the series circuit.
I=E÷RT
I=120V÷1,200.40 ohms
I=0.0999667A
Step 3. Determine the voltage drop across each resistor.
E=I×R
E Line=0.0999667A×0.20 ohms=0.02V
E Load=0.0999667A×200.00 ohms=20.00V
E Person=0.0999667A×1,000 ohms=99.96V
E Neutral=0.0999667A×0.20 ohms=0.02V
Step 4. Using Kirchoff's Law, verify the sum of the voltage drops of the circuit is equal to the voltage source.
ET=E1+E2+E3+E4
ET=0.02V+20V+99.96V+0.02V=120V
By applying two fundamental laws in four math steps, we have proven:
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The voltage potential between the accident victim and the neutral wire is almost 100V.
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The electric shock hazard (100V) from the neutral is not greater than the shock hazard from a hot and neutral (120V), but it is just as dangerous.
You must realize that it's not necessary to get between two broken neutral conductors to receive a shock. You can receive the same shock if you contact the end of the neutral at the load and any grounded surface. And remember, even a ground fault circuit interrupter (GFCI) will not protect you against electric shock from a neutral-to-neutral (series) or a line-to-neutral (parallel) connection.