Power Rating of a Resistor. The power rating of a resistor is the maximum power the resistor can safely dissipate without too great a rise in temperature and hence damage to the resistor. A 9.0 kΩ resistor is to be connected across a 120 V potential difference. What power rating is required?

Verified step by step guidance

1. First, understand the relationship between power, voltage, and resistance. The power dissipated by a resistor can be calculated using the formula: $P=\frac{V^2}{R}$, where $P$ is the power, $V$ is the voltage, and $R$ is the resistance.
2. Identify the given values in the problem: the resistance $R$ is 9.0 kΩ (or 9000 Ω) and the voltage $V$ is 120 V.
3. Substitute the given values into the power formula: $P=\frac{{120}^2}{9000}$.
4. Calculate the square of the voltage: ${120}^2$.
5. Divide the result from the previous step by the resistance value to find the power rating required for the resistor.

Key Concepts

Ohm's Law

Ohm's Law states that the current through a conductor between two points is directly proportional to the voltage across the two points, given by the formula V = IR, where V is voltage, I is current, and R is resistance. This principle helps determine the current flowing through the resistor when connected to a voltage source.

Power Dissipation in Resistors

Power dissipation in resistors is calculated using the formula P = IV, where P is power, I is current, and V is voltage. Alternatively, it can be expressed as P = V^2/R or P = I^2R, depending on known values. This concept is crucial for determining the power rating needed to ensure the resistor operates safely without overheating.

Resistor Power Rating

The power rating of a resistor is the maximum power it can handle without damage. It is essential to select a resistor with a power rating higher than the calculated power dissipation to prevent excessive heat buildup, which can lead to failure or damage. This ensures the resistor operates within safe limits under specified conditions.