/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 29 Find the power dissipated in a \... [FREE SOLUTION] | 91Ó°ÊÓ

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Find the power dissipated in a \(25-\Omega\) electric heater connected to a \(120-\mathrm{V}\) outlet.

Short Answer

Expert verified
The power dissipated is 576 watts.

Step by step solution

01

Identify the given values

We have an electric heater with a resistance of \( R = 25 \, \Omega \) and it is connected to a voltage source of \( V = 120 \, \mathrm{V} \). Our task is to find the power dissipated, denoted as \( P \).
02

Recall the formula for electrical power

The power dissipated in a resistor can be calculated using the formula: \[ P = \frac{V^2}{R} \]where \( P \) is the power in watts (W), \( V \) is the voltage across the resistor in volts (V), and \( R \) is the resistance in ohms (\( \Omega \)).
03

Substitute the given values into the formula

Substitute \( V = 120 \, \mathrm{V} \) and \( R = 25 \, \Omega \) into the power formula:\[P = \frac{(120 \, \mathrm{V})^2}{25 \, \Omega}\]
04

Calculate the power

Calculate the expression:\[P = \frac{14400}{25} = 576 \, \mathrm{W}\]Therefore, the power dissipated by the heater is 576 watts.

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Key Concepts

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

Electric Power
Electric power is an essential concept in understanding how electrical systems work. It is defined as the rate at which electrical energy is converted into another form, such as heat, light, or motion, through a device such as a resistor.The formula to calculate electric power is given by:
  • \( P = \frac{V^2}{R} \)
  • \( P = V \cdot I \)
Here, \( P \) stands for power in watts (W), \( V \) is the voltage in volts (V), \( I \) is the current in amperes (A), and \( R \) is resistance in ohms (\( \Omega \)).Using the formula \( P = \frac{V^2}{R} \), we calculate power when only voltage and resistance values are known. The formula \( P = V \cdot I \) is used when both voltage and current are given. Understanding how to interchangeably use these formulas can help you solve a variety of power-related problems in electronics. The unit of power, the watt, signifies how much energy the device is using per second. The higher the power rating, the more efficient the device is at converting electrical energy into work. In this context, a heater converts electrical energy primarily into heat.
Resistance
Resistance is a measure of how much an object opposes the passage of an electric current. When dealing with resistors, it's crucial to understand that they convert electrical energy into heat, thereby affecting the overall power output of an electrical appliance.The resistance of an object is measured in ohms (\( \Omega \)) and can be impacted by various factors, like:
  • The material of the resistor
  • Its temperature
  • The physical dimensions of the resistor
In a simple resistor circuit, Ohm's Law, which is \( V = I \cdot R \), relates voltage \( V \), current \( I \), and resistance \( R \). This fundamental principle allows us to calculate one of these quantities if the other two are known. For example, knowing the resistance and current allows you to determine the voltage.In our exercise, the electric heater has a resistance of \(25 \Omega\). This determines how much current will flow through the heater when connected to a voltage source, impacting how much electric power is dissipated as heat.
Voltage
Voltage, often referred to as electric potential difference, is a measure of the energy difference per charge between two points in an electrical circuit. It's essentially the "push" that makes electric charges move in a wire, driving the current through a circuit.Measured in volts (V), voltage can be thought of as the force that causes electrons to flow through a conductor. In a practical sense, increasing the voltage increases the power output of an electrical device if resistance stays the same (as shown by \( P = \frac{V^2}{R} \)).There are some key points to remember about voltage:
  • It is measured across components, such as a battery or resistor.
  • Higher voltage means more potential energy is available.
  • In the context of alternating current (AC) supply, it is often the root mean square (RMS) value that is used.
In the exercise, the heater is connected to a \(120 \mathrm{V}\) outlet, which provides the necessary electric pressure to drive the current through the heater's resistance, converting electrical energy into heat. Understanding voltage helps in grasping how various electrical components interact within a circuit.

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Most popular questions from this chapter

A circuit consists of three resistors, \(R_{1}

An \(89-\Omega\) resistor has a current of \(0.72 \mathrm{A}\) and is connected in series with a \(130-\Omega\) resistor. What is the emf of the battery to which the resistors are connected?

Three resistors \(\left(R, \frac{1}{2} R, 2 R\right)\) are connected to a battery. (a) If the resistors are connected in series, which one has the greatest rate of energy dissipation? (b) Repeat part (a), this time assuming that the resistors are connected in parallel.

Predict/Explain An electric space heater has a power rating of \(500 \mathrm{W}\) when connected to a given voltage \(V\). (a) If two of these heaters are connected in series to the same voltage, is the power consumed by the two heaters greater than, less than, or equal to 1000 W? (b) Choose the best explamation from among the following: I. Each heater consumes \(500 \mathrm{W}\); therefore two of them will consume \(500 \mathrm{W}+500 \mathrm{W}=1000 \mathrm{W}\) II. The voltage is the same, but the resistance is doubled by connecting the heaters in series. Therefore, the power consumed \(\left(P=V^{2} / R\right)\) is less than \(1000 \mathrm{W}\) III. Connecting two heaters in series doubles the resistance. since power depends on the resistance squared, it follows that the power consumed is greater than \(1000 \mathrm{W}\).

IP Three resistors, \(22 \Omega, 67 \Omega\), and \(R\), are connected in parallel with a \(12.0-\mathrm{V}\) battery. The total current flow ing through the battery is \(0.88 \mathrm{A}\) (a) Find the value of resistance \(R\). (b) Find the current through each resistor. \((c)\) If the total current in the battery had been greater than \(0.88 \mathrm{A}\), would your answer to part (a) have been larger or smaller? Explain.

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