
- Dead Short Definition: A dead short is when electrical current flows where it shouldn’t, with no resistance, often causing damage or hazard.
- Comparison with Short Circuit: Unlike a short circuit, which has some resistance and reduced voltage, a dead short shows zero voltage and resistance, indicating a more severe problem.
- Bolted Fault Similarity: A bolted fault, like a dead short, also exhibits zero resistance, but it specifically involves connections to ground.
- Ground Fault Differences: Ground faults involve some resistance and usually occur when a live wire touches a grounded surface, unlike the zero-resistance path in a dead short.
- Practical Example: Demonstrating a dead short with resistors, where shorting the terminals with zero resistance leads to a drastic increase in current, bypassing the resistors entirely.
What is a Dead Short?
A dead short is an unintended path that lets current bypass the load with (ideally) no resistance or impedance. The resulting current is limited only by the source and the shorting conductor, so it can destroy equipment and shock anyone nearby.
Finding a dead short is hard because the current rises so fast that an overcurrent device often trips at once, leaving little time to measure the live fault.
Typical causes are a solid metallic join between opposite polarities (or between phases) or a solid join from a live conductor to earth.
The hazard is the very large current the short forces through conductors, joints and anyone in the path.
Dead Short vs Short Circuit
A short circuit is any unintended low-impedance path. A dead short is the near-zero-impedance case. As a field check, take two points whose open-circuit voltage is 150 V.
Across those points in normal service the meter reads 150 V. A lower reading often means a short circuit path is loading the circuit, but that drop alone does not prove the fault impedance.
Many shorts still have some voltage across them because the fault path has leftover resistance or arcing.
If the meter across the same points reads 0 V, shops call that a dead short: the fault path is modelled as zero resistance. Real wire and joints are never exactly 0 Ω.
The figure below compares a normal reading, a short with leftover voltage and a dead-short reading of 0 V.

Dead Short vs Bolted Fault
A bolted fault is a study-case fault with zero fault impedance: the conductors are treated as solidly joined. It gives the largest calculated fault current for that source.
That solid join can be line-to-line, three-phase or line-to-ground. A bolted ground fault is only one of those cases, not the whole definition.
A bolted short and a dead short describe the same ideal: zero (or near-zero) fault resistance. Dead short is the shop name. Bolted fault is the calculation name.
Dead Short vs Ground Fault
A ground fault is an unintended path from a live conductor to earth, a grounded metal part or a grounding conductor.
The exposed metal can then sit at a hazardous voltage. Fault current depends on soil, bonding and any rust or arc in the path, so it is often far below the bolted value.
Many ground faults therefore have extra impedance. A bolted ground fault, with a solid metallic path to earth, is still a dead short to ground.
Example of a Dead Short
The numerical example uses three resistors in series, as shown in the figure below.

In normal conditions the circuit current is I amperes and the total resistance is REQ.
![]()
![]()
Ohm’s law gives:
![]()
![]()
![]()
With a 40 V source implied by the stored equation, the normal current is 1 A.
If the battery terminals are joined by a metal wire of ideally zero resistance, the circuit looks like the figure below.

Points A and B are then a dead short. Almost none of the current goes through the series resistors.
Almost all of the current takes the shorted path, because current shares in inverse proportion to resistance.
With the ideal zero-ohm short, the current through A and B is modelled as:
![]()
![]()
![]()
![]()
The stored latex writes I = 40/0 and I = infy (infinity). That is the ideal model. A real battery and wire have resistance, so the current is finite but far above the 1 A load current.
On a power network the same idea applies to a feeder or bus instead of three resistors. The circuit then looks like the figure below.






