A short circuit is an unintended connection of low resistance between two points of different potential, allowing a current far above the normal load current to flow. The current is limited only by the source impedance and the impedance of the fault path.
Learning objectives
- Define a short circuit and distinguish it from an overload
- Calculate a prospective fault current from voltage and loop impedance
- Explain why fault current heats conductors and sources
- Describe the ratings that protective devices must satisfy
Progressive explanation
Start with the plain-language idea, then move into the engineering detail. Each level stands on its own, so you can stop at the depth you need.
Beginner The core idea
The large current produces heat in the wiring, can start a fire and can damage the source. Fuses and circuit breakers exist to interrupt this current quickly, before the wiring is damaged.
Intermediate How it behaves in a circuit
Protective devices must have an interrupting rating above the prospective fault current and must clear within the time the conductors can withstand. The let-through energy I²t is the quantity that heats the cable.
Advanced Limits, assumptions, engineering detail
Formulas and variables
Units of measurement
| Quantity | Symbol | Unit |
|---|---|---|
| Fault current | Isc | ampere (A) |
| Loop impedance | Zloop | ohm (Ω) |
| Let-through energy | I²t | ampere squared second (A²·s) |
| Voltage | Voc | volt (V) |
Worked examples
Battery short circuit
Problem. A 12 V battery is shorted through a path whose total resistance is 0.02 Ω. What is the theoretical fault current?
Solution. Isc = Voc / Zloop = 12 / 0.02.
Mains fault loop
Problem. A 230 V circuit has a fault-loop impedance of 0.5 Ω. Find the prospective fault current.
Solution. Isc = 230 / 0.5.
Heat in a poor joint
Problem. A current of 100 A flows through a joint with 0.05 Ω of resistance. How much power is dissipated in the joint?
Solution. P = I² × R = 100² × 0.05 = 10 000 × 0.05.
Practical applications
- Protective device selection and coordination
- Automotive fusing and battery isolation
- Photovoltaic and battery energy storage systems
- Fault-level studies for industrial installations
- Arc-flash risk assessment
- Diagnosis of damaged wiring
Automotive applications
A metal tool dropped across battery terminals or a chafed harness can create a short circuit that melts insulation within seconds. Fuses protect the wiring, not the electronics, and the battery negative is normally disconnected first during service to reduce the risk.
Industrial applications
Installations require a fault-level study so that breakers and fuses can interrupt the maximum prospective current. Motor contribution, transformer impedance and cable length all affect the result, and arc-flash assessments use the same data to define safe work distances and protective equipment.
Measurement procedure
- Never measure prospective fault current directly with an ordinary multimeter.
- Use a dedicated loop-impedance tester, or calculate from the supply and conductor data.
- Measure insulation resistance before energising after a fault.
- After a fault, check the actual insulation and continuity rather than resetting the breaker repeatedly.
Common mistakes
- Thinking a fuse prevents all damage; it protects the wiring, and electronics may fail before it clears.
- Ignoring source and conductor impedance and assuming infinite fault current.
- Replacing a blown fuse with wire or a higher-rated fuse.
- Trying to measure fault current with a standard multimeter.
Safety considerations
- Remove metal jewellery and use insulated tools when working near battery or distribution terminals.
- A short-circuited battery can explode; wear eye protection and keep sparks away.
- Do not work on energised equipment where arc-flash energy may be present; use the required protective equipment.
- Isolate and lock off before inspecting damaged conductors.