A100Q resistor must have what minimum power rating if it carries a maximum current of 400 mA?

To determine the minimum power rating for a resistor that carries a maximum current of 400 mA, you can use the formula for electrical power, which is calculated as the product of the current squared and the resistance (P = I²R). However, if the resistance value is not provided, we can simplify the approach by determining the required power rating based solely on the current.

Power can also be calculated using the formula P = I × V, where voltage (V) can be replaced by Ohm’s Law (V = I × R). If we rearrange this in terms of current, we can derive that for purely resistive loads:

P = I²R.

To ensure safety and reliability, the resistor's power rating must be greater than what is calculated, so a factor of safety is often considered.

Given the maximum current of 400 mA (which is equivalent to 0.4 A):

1. If we assume a voltage drop across the resistor, P = I²R provides us with the implications of this current through resistance.

2. The minimum power rating to avoid overheating or damage from the current can be calculated as:

P = (0.4 A)² × R

Since we typically don't have the resistance