Circuit Breaker Fault Power
P_fault = V × I_fault
Quick answer
This calculator solves P_fault = V × I_fault for Fault Power. With System Voltage 12 V, Fault Current 2 A, the result is 24 W. It solves in 3 directions, so you can enter the values you have and get the one you need.
Calculator
Formula
Description
The power dissipated during a short circuit or overload fault determines the thermal stress on wiring, busbars, and protection devices. Circuit breakers must interrupt the fault before the accumulated energy damages the installation. The breaking capacity (kA rating) of a circuit breaker must exceed the prospective fault current at the point of installation. Higher fault power means more energy must be safely interrupted, requiring beefier switching contacts and faster trip mechanisms.
Variables
- P — Instantaneous fault power (W)
- V — System voltage (V)
- I — Fault current (A)
Practical Notes
Residential prospective fault currents typically range from 3-10 kA depending on transformer size and distance. Industrial installations may see 25-65 kA. The arc energy during breaker opening is proportional to the arc voltage, current, and duration. Current-limiting breakers reduce fault energy by forcing the current to zero before it reaches its prospective peak, significantly reducing thermal and mechanical stress on the installation.
Symbols and units
| Symbol | Meaning | Default unit | Accepted units |
|---|---|---|---|
| P | Fault Power | W | µW, mW, W, kW |
| V | System Voltage | V | pV, nV, µV, mV, V, kV |
| I | Fault Current | A | pA, nA, µA, mA, A, kA |
Step by step
- 1. What you know: System Voltage 12 V, Fault Current 2 A
- 2. The relation: P_fault = V × I_fault
- 3. Substitute: V = 12 V, I = 2 A
- 4. Result: Fault Power = 24 W
Related quantities from the same inputs
- P = V × I = 24 W
Worked examples
Minimum range
Given: System Voltage 0.33 V, Fault Current 2 mA
Result: Fault Power = 0.66 mW
Low range
Given: System Voltage 3.3 V, Fault Current 20 mA
Result: Fault Power = 66 mW
Typical
Given: System Voltage 12 V, Fault Current 2 A
Result: Fault Power = 24 W
High range
Given: System Voltage 0.23 kV, Fault Current 100 A
Result: Fault Power = 23 kW
Maximum range
Given: System Voltage 2.3 kV, Fault Current 1 kA
Result: Fault Power = 2300 kW
Solving for each variable
Solve for Fault Power
System Voltage 0.33 V, Fault Current 2 mA → 0.66 mW
Solve for System Voltage
Fault Power 10 mW, Fault Current 2 mA → 5 V
Solve for Fault Current
Fault Power 10 mW, System Voltage 0.33 V → 30.3 mA
How Fault Power changes with System Voltage
| System Voltage (input) | Fault Power (result) |
|---|---|
| 1.2 V | 2.4 W |
| 3 V | 6 W |
| 6 V | 12 W |
| 9 V | 18 W |
| 12 V | 24 W |
| 18 V | 36 W |
| 24 V | 48 W |
| 48 V | 96 W |
| 84 V | 0.168 kW |
| 120 V | 0.24 kW |
The same result in other units
| Unit | Value |
|---|---|
| mW | 24000 |
| W | 24 |
| kW | 0.024 |
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<p><a href="https://vbatpower.com/formulas/circuit-breaker-power/">Circuit Breaker Fault Power</a> Calculator by vbatpower.com</p>Frequently asked questions
How do you calculate Circuit Breaker Fault Power?
Use P_fault = V × I_fault. With System Voltage 12 V, Fault Current 2 A, Fault Power works out to 24 W.
What is the formula for Circuit Breaker Fault Power?
The formula is P_fault = V × I_fault. The power dissipated during a short circuit or overload fault determines the thermal stress on wiring, busbars, and protection devices.
How do you find System Voltage when you know the other values?
Rearrange for System Voltage: with Fault Power 10 mW, Fault Current 2 mA, System Voltage is 5 V.
How do you find Fault Current when you know the other values?
Rearrange for Fault Current: with Fault Power 10 mW, System Voltage 0.33 V, Fault Current is 30.3 mA.
What units does Circuit Breaker Fault Power use?
Fault Power is shown in mW, W, kW. Values are held in SI base units internally, so you can mix prefixes without converting by hand.
Related Concepts
All Safety / Protection formulas →Need more features?
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