Ohms Law...

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xformer

Senior Member
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Dallas, Tx
Ohms Law teaches that voltage is equal to Current times Resistance. Since ohms law is an equation, is not the converse true as well?? For example, if I disconnect a ground wire while a ground fault is occurring, doesn't the voltage raise accordingly? If so, cant the voltage between the two ends of the grounding conductor become extremely high based upon the short circuit fault current and the resistance of the grounding conductor??????
 
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charlie b

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Seattle, WA
The driving force is the voltage. If you start with a 120 volt source, and create a fault condition by attaching the ground wire to an ungrounded conductor, that 120 volts will push current through a very low resistance, creating a very high current in the ground wire. But the voltage across the ground wire remains 120 volts. This condition does not cause an increase in the voltage of the source. If while this is happening you disconnect the ground wire, that will terminate the event, and the voltage of the source will still be 120 volts.
 

ggunn

PE (Electrical), NABCEP certified
Location
Austin, TX, USA
Occupation
Electrical Engineer - Photovoltaic Systems
Ohms Law teaches that voltage is equal to Current times Resistance. Since ohms law is an equation, is not the converse true as well?? For example, if I disconnect a ground wire while a ground fault is occurring, doesn't the voltage raise accordingly? If so, cant the voltage between the two ends of the grounding conductor become extremely high based upon the short circuit fault current and the resistance of the grounding conductor??????
Well, yes, but the resistance of a GEC is typically very low, so the voltage drop for even a high current flow is commensurately low. Your GEC needs to be sized so that it can handle the fault current. Looking at Table 8 in the NEC2008, the resistance of a #6 copper wire (a common size for a GEC) is about .5 ohms per 1000 feet. For a 100 amp fault at the end of a 100 foot run, V = IR = (100A)(.05 ohms) = 5V.
 

richxtlc

Senior Member
Location
Tampa Florida
Ohms Law teaches that voltage is equal to Current times Resistance. Since ohms law is an equation, is not the converse true as well?? For example, if I disconnect a ground wire while a ground fault is occurring, doesn't the voltage raise accordingly? If so, cant the voltage between the two ends of the grounding conductor become extremely high based upon the short circuit fault current and the resistance of the grounding conductor??????
If you remove the grounding conductor you most certainly would draw an arc, the arc resistance would decrease the fault current, as the distance increases, the arc would extinguish and no current would flow. No current flow would give you only line potential. This is to say if you could safely remove the ground wire during the ground fault. What you were using is the formula for a voltage drop, with no current flowing you only have a voltage rise (line voltage).
 

mull982

Senior Member
Basically the source 120V will be dropped across the series combination of the ungorounded conductor, and EGC when connecting the two to create a fault. There may be an impedance related to the fault itself (connection) that may have voltage dropped across it.

Ignoring any fault impedance and assuming that ungrounded and grounded conductor are same size then you will have aproximately 1/2 the supply voltage dropped across each.
 

charlie b

Moderator
Staff member
Location
Seattle, WA
The phrase "ground fault" is used in several different ways. If you are literally talking about a fault that involves planet Earth, then you need to keep in mind that the fault path will include not only the Grounding Electrode Conductor, but also some amount of dirt. I could, just for the sake of discussion, assign a value of resistance between the ground rod and planet Earth of 25 ohms. That will be in series with the GEC's resistance of 0.5 ohms, and in series with the voltage source, which we will continue to call 120 volts. So the ground fault current in this case would be 120/(25+.5), or about 4.7 amps. Here again, the voltage of the source is not changed.
 

Hameedulla-Ekhlas

Senior Member
Location
AFG
Ohms Law teaches that voltage is equal to Current times Resistance. Since ohms law is an equation, is not the converse true as well?? For example, if I disconnect a ground wire while a ground fault is occurring, doesn't the voltage raise accordingly? If so, cant the voltage between the two ends of the grounding conductor become extremely high based upon the short circuit fault current and the resistance of the grounding conductor??????
Suppose you have a three phase system and when single line to ground fault occurs in one phase then the current the other two phases goes to zero. If you draw its circuit diagram it becomes very simple closed circuit in which current fault current flows. If you disconnect the ground wire, it becomes open circuit and there will not be current flow. Now per law formula see

after Disconnecting

V = I * R
V = 0*R
V = 0

Current I = 0
Resistance = infinite
Voltage = 0
 

mbeatty

Senior Member
Location
Illinois
If you remove the grounding conductor you most certainly would draw an arc, the arc resistance would decrease the fault current, as the distance increases, the arc would extinguish and no current would flow. No current flow would give you only line potential. This is to say if you could safely remove the ground wire during the ground fault. What you were using is the formula for a voltage drop, with no current flowing you only have a voltage rise (line voltage).

I agree that if you open the only active fault, the potential would go back to what is provided by the source.
 

K2500

Senior Member
Location
Texas
If you opened the fault while the current was near peak, wouldn't the field collapse cause relatively high transient voltages?
 

glene77is

Senior Member
Location
Memphis, TN
Hameed,
Why would the un-grounded phases show zero current?

Each poster seems to have a different scenario in mind, without clarification.
Without getting into equations involving impedance,
I would summarize with this:

(1) Line input voltage would be he max voltage to drive the circuit.

(2) If the grounded circuit involves a transformer, with its inductive magnetic field,
then breaking the short circuit would cause the resulting Inductive spike.
Slowly breaking he short circuit, allowing for arcing and the impedance of the arc,
would diminish the resulting Inductive spike.
This Inductive spike could be much higher than the applied Line Voltage.

Comments are always welcome.
I am out of town, and may not respond.
 

SAC

Senior Member
Location
Massachusetts
If you opened the fault while the current was near peak, wouldn't the field collapse cause relatively high transient voltages?
Yes - if there is a large inductance the voltage can become high if the resistance is changed "quickly" by introducing an air gap in the conductor. The stored energy in the magnetic field of the inductor resists a change in the current flowing through it. Since the inductance keeps the current flowing even as the resistance in the circuit is changed, the voltage across the gap can become high relative to the source voltage. "R" is increased while "I" stays the same, so in V = IR, "V" increases proportionally with the increase in "R" as long as the inductor in able to keep "I" constant (at least for a short period of time). However, in typical wiring systems the inductance isn't large enough relative to the current and source voltage for this to be a noticeable effect.

I was avoiding calculus, but the effects can be calculated using v = di/dt * L, where "v" the voltage across the inductor is equal to the rate of change of the current "di/dt" times the inductance, "L".
 

jghrist

Senior Member
Why would the un-grounded phases show zero current?
Why would the ungrounded phases show anything other than zero current (assuming no load)? If only one phase is connected to anything, there is nowhere for current to flow in other phases.
 

jghrist

Senior Member
after Disconnecting

V = I * R
V = 0*R
V = 0

Current I = 0
Resistance = infinite
Voltage = 0
Zero times infinity is not zero. If you disconnected the ground wire, after the arcing stopped and the current went to zero, then the voltage to ground would be the same as it was before the fault. That is, assuming that you didn't burn up the transformer or yourself.:mad:
 

kingpb

Senior Member
The situation in which you are referring is solved by using symmetrical components.

Faults, in general, are unsymmetrical in nature, and occur as single line-to-ground faults, line-line faults, or double line-to-ground faults. Unsymmetrical faults will cause unbalanced current to flow, therefore the method of symmetrical components are necessary to determine currents and voltages in the system after the fault occurs. Assuming that the impedance at the fault is zero, will result in the highest fault current level, and is the conservative approach, however in practicality the fault impedance is seldom zero since even an arc has impedance.

The math gets complex, but the results can only be determined using thoese formulas.
 
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