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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?

Short Answer

Expert verified
(a) Max current: ~0.018 A; Max voltage: ~273.86 V. (b) Required power rating: 1.6 W.

Step by step solution

01

Identify given values and equations for (a)

First, we need to identify the given values for the first part (a) of the problem: a resistor of resistance \( R = 15\, \text{k}\Omega = 15,000 \, \Omega \) and a power rating \( P = 5.0 \, \text{W} \). We are to find the maximum current \( I \) and potential difference \( V \) using the power formula \( P = I^2 R = \frac{V^2}{R} \).
02

Solve for maximum current (a)

The formula for power in terms of current and resistance is \( P = I^2 R \). Rearranging for current gives us \( I = \sqrt{\frac{P}{R}} \). Substituting the values: \[ I = \sqrt{\frac{5.0}{15,000}} \approx 0.018 \text{ A} \]
03

Solve for maximum voltage (a)

To find the potential difference, use the power formula in terms of voltage: \( P = \frac{V^2}{R} \). Solving for \( V \) gives us \( V = \sqrt{P \times R} \). Substituting the values:\[ V = \sqrt{5.0 \times 15,000} \approx 273.86 \text{ V} \]
04

Identify given values and equations for (b)

For part (b), we are given a resistor \( R = 9,000 \, \Omega \) and a potential difference \( V = 120 \, \text{V} \). We need to determine the power rating \( P \). Use the power equation: \( P = \frac{V^2}{R} \).
05

Solve for required power rating (b)

Using the formula \( P = \frac{V^2}{R} \) and substituting the given values:\[ P = \frac{120^2}{9,000} = \frac{14400}{9,000} = 1.6 \, \text{W} \] The resistor needs a power rating of at least 1.6 W.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Ohm's Law
Ohm's Law is a fundamental concept in physics and electronics. It describes the relationship between voltage, current, and resistance.
Understanding this law is essential for solving many electrical problems. The law is usually stated as:
  • Voltage ( V ) is equal to the product of current ( I ) and resistance ( R )
The equation can be written as:\[ V = I imes R \]
Ohm's Law helps us understand how electrical circuits operate under different conditions. By knowing two of the variables, we can calculate the third. For example, if we know the current and resistance, we can find the voltage.To gain mastery over Ohm's Law, students should:- Practice solving problems with different given values.- Ensure a strong conceptual understanding of how increasing resistance affects current and voltage.Understanding Ohm's Law can aid in solving complex problems, such as calculating voltages or currents across resistors in circuits.
Electric Power
Electric power is the rate at which electrical energy is generated or consumed in a circuit. It's crucial for determining how much energy a device uses or dissipates as heat when operating. Power is measured in watts (W), and it can be calculated using the formulas:
  • \[ P = V imes I \] (where \( V \) is voltage and \( I \) is current)
  • \[ P = I^2 imes R \] (where \( I \) is current through and \( R \) is resistance of the device)
  • \[ P = \frac{V^2}{R} \], expressing power in terms of voltage and resistance

Understanding how to calculate electric power helps in designing circuits to ensure that components function safely within their power limits.
This knowledge can prevent damage due to overheating, as components have power ratings that indicate the maximum power they can handle.
Resistance Calculation
Resistance calculation is a vital concept when designing or analyzing electrical circuits. Resistance, measured in ohms (Ω), determines how much a resistor impedes the flow of current. There are various formulas used to calculate resistance based on different scenarios, such as current and power or voltage and power:
  • If you have voltage and current, resistance can be derived from Ohm's Law: \[ R = \frac{V}{I} \]
  • If power and voltage are known, resistance can be calculated: \[ R = \frac{V^2}{P} \]
  • If power and current are known, use: \[ R = \frac{P}{I^2} \]

By calculating resistance, you can ensure that the components are correctly selected for the desired current and voltage levels.
Accurate calculation is key to building efficient and safe electrical systems, and understanding it helps to troubleshoot circuits when things aren't working as expected.

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Most popular questions from this chapter

A 5.0 A current flows through the leads of an electrical appliance. (a) Calculate how much charge passes through the appliance in \(1.0 \mathrm{~min} .\) (b) How many electrons are in that amount of charge?

A typical small flashlight contains two batteries, each having an emf of \(1.5 \mathrm{~V},\) connected in series with a bulb having resistance \(17 \Omega\). (a) If the internal resistance of the batteries is negligible, what power is delivered to the bulb? (b) If the batteries last for \(5.0 \mathrm{~h}\), what is the total energy delivered to the bulb? (c) The resistance of real batteries increases as they run down. If the initial internal resistance is negligible, what is the combined internal resistance of both batteries when the power to the bulb has decreased to half its initial value? (Assume that the resistance of the bulb is constant. Actually, it will change somewhat when the current through the filament changes because this changes the temperature of the filament and hence the resistivity of the filament wire.

Calculate the (a) maximum and (b) minimum values of resistance that can be obtained by combining resistors of \(36 \Omega, 47 \Omega\), and \(51 \Omega\).

You want to precut a set of \(1.00 \Omega\) strips of 14 gauge copper wire (of diameter \(1.628 \mathrm{~mm}\) ). How long should each strip be?

Some types of spiders build webs that consist of threads made of dry silk coated with a solution of a variety of compounds. This coating leaves the threads, which are used to capture prey, hygroscopic-that is, they attract water from the atmosphere. It has been hypothesized that this aqueous coating makes the threads good electrical conductors. To test the electrical properties of coated thread, researchers placed a \(5 \mathrm{~mm}\) length of thread between two electrical contacts. The researchers stretched the thread in \(1 \mathrm{~mm}\) increments to more than twice its original length, and then allowed it to return to its original length, again in \(1 \mathrm{~mm}\) increments. Some of the resistance measurements are given in the table: $$\begin{array}{l|cccccccc}\begin{array}{l}\text { Resistance of } \\\\\text { thread }\left(10^{9} \Omega\right)\end{array} & 9 & 19 & 41 & 63 & 102 & 76 & 50 & 24 \\ \hline \begin{array}{l}\text { Length of } \\\\\text { thread }(\mathrm{mm})\end{array} & 5 & 7 & 9 & 11 & 13 & 9 & 7 & 5\end{array}$$ What is the maximum current that flows in the thread during this experiment if the voltage source is a \(9 \mathrm{~V}\) battery? A. about \(1 A\) B. about \(0.1 A\) C. about \(1 \mu A\) D. about 1 nA

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