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Chapter 26: Problem 14

Thermal energy is produced in a resistor at a rate of \(90 \mathrm{~W}\) when the current is \(3.00 \mathrm{~A}\). What is the resistance?

Short Answer

Expert verified
The resistance is \( 10.0 \, \Omega \).

Step by step solution

01

Identify the Given Information

You're given the power across the resistor as \( 90 \, \text{W} \), and the current flowing through it as \( 3.00 \, \text{A} \). Your goal is to find the resistance.
02

Write the Formula for Power in a Resistor

The power \( P \) dissipated in a resistor is given by the formula:\[ P = I^2 R \]where \( P \) is the power, \( I \) is the current, and \( R \) is the resistance.
03

Rearrange the Formula to Solve for Resistance

You'll need to rearrange the formula to solve for resistance \( R \):\[ R = \frac{P}{I^2} \]
04

Substitute the Known Values into the Equation

Substitute \( P = 90 \, \text{W} \) and \( I = 3.00 \, \text{A} \) into the rearranged formula:\[ R = \frac{90}{(3.00)^2} \]
05

Perform the Calculation

Calculate the power of the current and then divide the power by this result:\( (3.00)^2 = 9.00 \).Then, \( R = \frac{90}{9} = 10.0 \, \Omega \).
06

Finalize the Answer

The resistance \( R \) of the resistor is found to be \( 10.0 \, \Omega \).

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Key Concepts

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

Resistance Calculation
Resistance is a key concept in electrical circuits, representing how much a particular component resists the flow of electric current. It's measured in ohms (\(\Omega\)). Understanding how to calculate resistance is crucial for designing circuits and ensuring they function correctly.
To find resistance when knowing the power and current is a common task. The formula we use for this is derived from Ohm's Law and the power formula:
  • Power (\(P\)) in a resistor is given by \(P = I^2 R\), where \(I\) is the current and \(R\) is the resistance.
To make resistance the subject of the formula:
  • Rearrange \(P = I^2 R\) to find \(R\): \(R = \frac{P}{I^2}\).
  • Substitute the known values of power and current into the formula.
  • Perform the arithmetic calculation to find resistance in ohms.
This process allows us to find the resistance, helping us understand how well a component inhibits current, enhancing the stability and performance of circuits.
Power in Electrical Circuits
Power in electrical circuits is a way to measure how much energy is being used. It's critical when planning how circuits operate and ensuring they don't overheat or underperform.
The power formula tells us the relationship between power, current, and resistance in a circuit:
  • The formula used is \(P = I^2 R\), showcasing that power is the product of the square of the current and the resistance.
  • This means as current increases, the power loss increases significantly, as it is proportional to the current squared.
  • If current remains constant, increasing resistance results in more power used. Therefore, choice of materials and resistance types is vital in circuit planning.
Monitoring the power also helps in designing systems to enhance energy efficiency. Such calculations enable engineers to keep devices safe and functioning optimally.
Current and Resistance Relationship
Current and resistance share an inverse relationship, described by Ohm's Law. This relationship is fundamental when understanding electric circuits and their behavior.
Ohm's Law is given by:
  • \(V = IR\), where \(V\) is voltage, \(I\) is current, and \(R\) is resistance. It shows how voltage across a component in the circuit is the product of the current flowing through it and its resistance.
  • As resistance increases, for the same voltage, the current must decrease. Conversely, with lower resistance, current flows more freely.
  • This balance governs how circuits are designed, ensuring components don't burn out from excess current.
Grasping the current-resistance relationship helps in both designing and troubleshooting circuits, ensuring they perform as expected and safely.

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Most popular questions from this chapter

The magnitude \(J(r)\) of the current density in a certain cylindrical wire is given as a function of radial distance from the center of the wire's cross section as \(J(r)=B r\), where \(r\) is in meters, \(J\) is in amperes per square meter, and \(B=2.00 \times 10^{5} \mathrm{~A} / \mathrm{m}^{3}\). This function applies out to the wire's radius of \(2.00 \mathrm{~mm}\). How much current is contained within the width of a thin ring concentric with the wire if the ring has a radial width of \(10.0 \mu \mathrm{m}\) and is at a radial distance of \(0.750 \mathrm{~mm}\) ?

A small but measurable current of \(1.2 \times 10^{-10}\) A exists in a copper wire whose diameter is \(3.0 \mathrm{~mm}\). The number of charge carriers per unit volume is \(8.49 \times 10^{28} \mathrm{~m}^{-3}\). Assuming the current is uniform, calculate the (a) current density and (b) electron drift speed.

A \(890 \mathrm{~W}\) radiant heater is constructed to operate at \(115 \mathrm{~V}\). (a) What is the current in the heater when the unit is operating? (b) What is the resistance of the heating coil? (c) How much thermal energy is produced in \(5.00 \mathrm{~h} ?\)

A wire initially has length \(L_{0}\) and resistance \(5.00 \Omega\). The resistance is to be increased to \(45.0 \Omega\) by stretching the wire. Assuming that the resistivity and density of the material are unaffected by the stretching, find the ratio of the new length to \(L_{0}\).

Figure \(26-18\) shows wire section 1 of diameter \(D_{1}=4.00 R\) and wire section 2 of diameter \(D_{2}=1.75 R\), connected by a tapered section. The wire is copper and carries a current. Assume that the current is uniformly distributed across any cross-sectional area through the wire's width. The electric potential change \(V\) along the length \(L=2.00 \mathrm{~m}\) shown in section 2 is \(10.0 \mu \mathrm{V}\). The number of charge carriers per unit volume is \(8.49 \times 10^{28} \mathrm{~m}^{-3}\). What is the drift speed of the conduction electrons in section \(1 ?\)
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