Problem 34 - College Physics

Question

Power rating of a resistor. The power rating of a resistor is the maximum power it can safely dissipate without being damaged by overheating. (a) If the power rating of a certain 15 \(\mathrm{k} \Omega\) resistor is \(5.0 \mathrm{W},\) what is the maximum current it can carry without damage? What is the greatest allowable potential difference across the terminals of this resistor? (b) If a 9.0 \(\mathrm{k} \Omega\) resistor is to be connected across a 120 \(\mathrm{V}\) potential difference, what power rating is required for that resistor?

Step-by-Step Solution

Answer

(a) Maximum current: 18.3 mA; Maximum voltage: 273.9 V. (b) Required power rating: 1.6 W.

Step 1: Calculate Maximum Current for Part (a)

Use the power formula, \(P = I^2 R\), to find the maximum current, \(I\), for the 15 k\(\Omega\) resistor with a power rating of 5 W. Rearrange the formula to solve for \(I\): \(I = \sqrt{\frac{P}{R}}\). Substitute \(P = 5\) W and \(R = 15000\ \Omega\). Calculate: \[I = \sqrt{\frac{5}{15000}} = \sqrt{0.000333} \approx 0.0183\,\text{A} = 18.3 \,\text{mA}.\]

Step 2: Calculate Maximum Voltage for Part (a)

To find the maximum voltage \( V \), use the power formula \( P = \frac{V^2}{R} \). Rearrange to solve for \( V \): \(V = \sqrt{PR}\). Substitute \( P = 5 \) W and \( R = 15000 \ \Omega\). Calculate: \[V = \sqrt{5 \times 15000} = \sqrt{75000} \approx 273.9 \, \text{V}.\]

Step 3: Calculate Power Rating for Part (b)

Use the power formula, \(P = \frac{V^2}{R}\), to determine the power rating required for a 9 k\(\Omega\) resistor connected to a 120 V potential difference. Substitute \(V = 120\) V and \(R = 9000\ \Omega\). Calculate: \[P = \frac{(120)^2}{9000} = \frac{14400}{9000} = 1.6 \text{ W}.\]

Key Concepts

Ohm's Law

Ohm's Law is the fundamental principle used to describe the behavior of electricity through resistive circuits. It states that the current flowing through a conductor between two points is directly proportional to the voltage across the two points, and inversely proportional to the resistance between them.

Mathematically, Ohm's Law is expressed as:

• \( V = I imes R \)

where:

• \( V \) is the voltage (in volts),
• \( I \) is the current (in amperes),
• \( R \) is the resistance (in ohms).

Ohm's Law helps us to calculate any of these three parameters if the other two are known. This formula is pivotal in understanding circuits, allowing you to determine how much current flows through a resistor when a certain voltage is applied.

Electrical Resistance

Electrical resistance is a measure of the opposition that a material offers to the flow of electric current. In other words, it determines how much the material resists the flow of electrons.

Resistance is measured in ohms (\( \Omega \)) and is often represented by the symbol \( R \). Materials with low resistance allow current to flow more freely, while materials with high resistance limit the flow of current.

• In the problem, a 15 k\( \Omega \) resistor is mentioned, highlighting its ability to resist current flow.
• Resistance for such a component helps in deciding its application in a circuit, affecting both the current and potential difference.

Understanding resistance is crucial when designing circuits to ensure that the components work safely under their specified limits.

Power Dissipation

Power dissipation in a resistor refers to the conversion of electrical energy into heat. It is a critical concept, as excessive power dissipation can cause a resistor to overheat and fail. The power dissipated by a resistor is given by the formula:

• \( P = I^2 R \)

Alternatively:

• \( P = \frac{V^2}{R} \)

where \( P \) is power (in watts), \( I \) is current (in amperes), \( R \) is resistance (in ohms), and \( V \) is voltage (in volts).

The exercise speaks of a power rating, which is the maximum power a resistor can dissipate without damage. Exceeding this rating can lead to potential overheating and failure.

Maximum Current

The maximum current a resistor can handle without damage is determined by its power rating and resistance. It can be calculated using the power dissipation formula:

• \( I = \sqrt{\frac{P}{R}} \)

where \( P \) is the power rating and \( R \) is the resistance.

In practical terms, exceeding this maximum current results in increased heat generation, which can damage the resistor. As seen in the exercise, the maximum current for a 15 k\( \Omega \) resistor with a 5 W power rating is approximately 18.3 mA.
Ensuring that the current remains within safe limits is vital for the longevity and safety of electronic circuits.

Potential Difference

Potential difference, commonly referred to as voltage, is the driving force that pushes electric charges through a circuit. It is a measure of the energy provided per charge to move them between two points.

The maximum potential difference that a resistor can handle is related to its power rating and resistance, calculated as:

• \( V = \sqrt{P \times R} \)

In the step-by-step solution, the greatest allowable potential difference for the 15 k\( \Omega \) resistor with a 5 W power rating comes out to approximately 273.9 V.
Understanding potential difference is essential for ensuring that resistors are not exposed to voltages beyond their safe operating range, thereby preventing damage and ensuring circuit integrity.