Electrical circuits are designed to allow a limited amount of current to flow through; this is done to prevent overheating of the wires due to I^{2}R losses. The limitation on the current is largely due to the resistance or impedance of the load to which it is connected.

**Short Circuit Definition**

If the resistance or impedance of the load is bypassed or shorted, then, according to **Ohm’s law**, an abnormally high current will flow through the circuit. This situation is called a **short circuit**.

Depending on the remaining resistance or impedance of the circuit, the short-circuit current could be up to 30 times as high as the normal current.

At this abnormally high level, most equipment and wiring will be ruined by the excessive amount of heat generated. Furthermore, there will most likely be the fire of combustible components within or in the vicinity.

**Short-Circuit Calculation**

To avoid damage from excessive heat and the magnetic force created by a short circuit, all electrical circuits, and equipment connected to the system must have an interrupting rating or interrupting capacity, equal to or greater than the calculated short-circuit capacity of the system.

The calculation of the short-circuit capacity of a power system is very involved and complex. Conservatively, it can be as high as 20–30 times the normal full-load current of the system.

**For example**, if a power system is designed to carry a full-load current of 2000 A, then the short-circuit capacity could be in the neighborhood of 20 * 2000 (40,000) A, or even 30 * 2000 (60,000) A. Computer programs are required to calculate a realistic level of short-circuit currents.

In its simplified form, the short-circuit current can be conservatively calculated by the following equations.

\[\begin{align} & \begin{matrix} {{I}_{S}}=\frac{E}{Z} & {} & \left( 1a \right) \\\end{matrix} \\ & or \\ & \begin{matrix} {{I}_{S}}=100\times \frac{I}{{{Z}_{p}}+{{Z}_{t}}} & {} & \left( 1b \right) \\\end{matrix} \\\end{align}\]

Where

I_{S} = short-circuit current, A

E = voltage, V

Z = total impedance, ohms

I = full-load current of the transformer, A

Z_{p} = impedance of the primary, %

Z_{t} = impedance of the transformer, %

Conservatively, the impedance of the primary on a large utility substation or a network may be considered negligible (near zero); thus, a short circuit near the secondary side of the transformer (or the system mains) is totally determined by the internal impedance of the transformer (or mains). **For example**, the short circuit current (I_{S}) of a large system connected to a transformer rated with 5 percent impedance could be 100 * I/5 or 20I.

**Interrupting Capacity**

To prevent the system and equipment from being destroyed by heat, arcing, and fire caused by a short circuit, all distribution wires and cables must be protected by devices such as fuses and circuit breakers.

All equipment or devices installed on the power system must also carry an interrupting capacity (I_{C}) greater than the calculated short-circuit current.

**Short Circuit Current Calculation Example**

The full-load current of a building power distribution system is 1200 A. The building is served by a single transformer having 5 percent impedance. The utility power service supplying the transformer is from a nearby substation with practically unlimited power. Determine the available short-circuit current at the main switchboard.

**Solution:**

From equation (1b)

\[{{I}_{S}}=100\times \frac{I}{{{Z}_{p}}+{{Z}_{t}}}=100\times \frac{1200}{0+5}=24,000A\]

(Impedance of the substation is considered to be near zero because the substation has an unlimited power supply)