Fault Loop Impedance Calculation ((new)) -

When a fault occurs—specifically a short circuit between a live conductor and earth—a current flows back to the source through the "fault loop." If the impedance of this loop is too high, the fault current will be low. If the fault current is too low, the protective device (circuit breaker or fuse) may not trip quickly enough. This creates a deadly risk of electric shock, fire, and equipment damage.

$$I_f = \frac{U_0}{Z_s}$$

Or more practically, the measured/calculated ( Z_s ) must be less than the (( Z_{s(max)} )) given in tables for each protective device. For a circuit breaker, this is derived from its instantaneous trip threshold. For example, a Type B breaker trips at 3–5 times its rated current. If a 20A breaker requires 100A to trip instantly, then ( Z_{s(max)} = U_0 / 100 ). For a 230V supply, this yields 2.3 ohms. Any calculated ( Z_s ) above this value fails the safety requirement. fault loop impedance calculation

Consider a simple Single-Phase (230V) installation:

Fault Loop Impedance Calculation: A Complete Guide to Electrical Safety When a fault occurs—specifically a short circuit between

You need to calculate the resistance of the line conductor and the circuit protective conductor (CPC).

The calculated $Z_s$ must be lower than the maximum allowed value ($Z_{max}$) found in the regulations. If a 20A breaker requires 100A to trip

Consider a final lighting circuit protected by a 6A Type B circuit breaker, with a supply voltage of 230V. The external impedance ( Z_{source} ) from the utility is 0.35 ohms. The phase conductor (2.5 mm² copper) and earth conductor (1.5 mm² copper) run 40 meters. The resistance per meter for 2.5 mm² is 0.00741 ohms/m, and for 1.5 mm² is 0.0121 ohms/m.