"Inrush current," sometimes called locked-rotor current or starting current, refers to the electrical current flowing through the components of a motor in the fractions of a second after the motor's power turns on. In this brief time interval, current rapidly waxes and wanes before any of the parts of the motor actually begin to move and the system progresses toward a dynamic electrical equilibrium, at which time a steady-state current is achieved.

The repeated short spikes of current with amperage values many times that of the steady state represent a potential disruption to the system, as they can result in the unnecessary tripping of fault devices in a "false alarm" manner.

Characteristics of the motor, such as DC current, a change in power supply and lighting ballasts can all increase the magnitude of the inrush current. You may need to know the value of this current so you can equip your motor with the right inrush current limiter to protect it against the tripping noted above, rather like a surge protector in a household power strip.

## Sciencing Video Vault

You will usually need to know the motor's maximum output power and the input voltage. Other helpful pieces of information include reset time, whether the circuit is single-phase or three-phase, the magnitude of the capacitance, the resistance and the motor efficiency.

For these kinds of problems, you will usually make use of the relationships:

V_{peak} = I_{in}R, where V_{peak} = ā2(V)

and

E = CV^{2}/2

For example, assume you have a motor with an input voltage of 200 V and a minimum resistance of 15 Ī©.

## Step 1: Assemble the Variables

In this problem, you do not have energy or capacitance, but you do have voltage and resistance. Thus the equation of interest is the first one above, or:

ā2(200) = I_{in}(15)

## Step 2: Calculate the Inrush Current

This yields:

282.8/15 = I_{in} = 18.85 A

## Step 3: Interpret the Findings

This means that any elements of the motor that are triggered to stop at amperage values this high may cause problems at start-up and that you may need to alter parameters such as voltage and resistance.