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| Figure 8-2 Schematic showing resistance to a fault current in a circuit. |

| that the resistance of the short length of conductors to the point of fault is on the order of 0.1 ohm. Then, by Ohm's Law (current in amperes equals voltage divided by resistance) the current flowing will be: |

| If this current were to flow through a breaker, it is obvious that enormous forces would be set up between the current-carrying parts of the breaker, perhaps large enough to destroy it. |

| Moreover, the magnetic fields set up by the currents in the conductors will induce or create currents in the metallic members of the circuit-breaker structure itself. These, in turn, will set up magnetic fields of their own, which will interact with the other magnetic fields, adding to the destructive forces created by the fault currents. |

| Since dead short circuits can literally never be realized, the magnitude of the short-circuit current flowing through any breaker will depend on the resistance (in a dc circuit) or the impedance (in an ac circuit) of the system back to the generating source (or sources) as well as on the actual contact resistance and nature of the short circuit. Generally speaking, the further away from the generating source, the lower will be the short-circuit current. |

| Rating |

| For the reasons stated, circuit breakers must be constructed mechanically strong enough to withstand the forces set up by enormous short-circuit currents. Conditions at the time of a short circuit are not very stable, that is, the voltage at the point of fault may be higher (from lightning or switching surges) or lower (because of greater loss of voltage from the larger current flow). For this reason, the ability of a breaker to withstand short-circuit forces is expressed in volt-amperes, that is, as the product of the nominal circuit voltage and the short-circuit current for which the breaker is designed. Since 1 VA |

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