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Chapter 19: Problem 33

A resistor with a 15.0 \(\mathrm{V}\) potential difference across its ends develops thermal energy at a rate of 327 \(\mathrm{W}\) . (a) What is the current in the resistor? (b) What is its resistance?

Short Answer

Expert verified
The current is 21.8 A and the resistance is approximately 0.688 惟.

Step by step solution

01

Understand Power Formula

The power (\(P\)) dissipated by a resistor in an electrical circuit is given by the formula: \( P = V \cdot I \), where \( V \) is the potential difference across the resistor and \( I \) is the current flowing through it. Rearrange this formula to solve for the current \(I\).
02

Calculate Current

Given: \( V = 15.0 \mathrm{V} \) and \( P = 327 \mathrm{W} \). Use the formula \( P = V \cdot I \) and substitute the given values: \( 327 = 15.0 \times I \). Rearrange to solve for \( I \):\[ I = \frac{327}{15.0} \]Calculate \( I \):\[ I = 21.8 \mathrm{A} \] (to one decimal place).
03

Understand Ohm's Law

Ohm's Law states \( V = I \cdot R \), where \( R \) is the resistance. Rearrange this formula to solve for the resistance \( R \):\[ R = \frac{V}{I} \].
04

Calculate Resistance

Use the Ohm's Law formula for resistance. Substitute the known values of \( V = 15.0 \mathrm{V} \) and \( I = 21.8 \mathrm{A} \):\[ R = \frac{15.0}{21.8} \]Calculate \( R \):\[ R \approx 0.688 \Omega \] (to three decimal places).

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Key Concepts

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

Electric Current
Electric current refers to the flow of electric charge through a conductor. It is analogous to the flow of water in a pipeline. Just as water flows from a region of high pressure to low pressure, electric current flows from a region of higher potential (voltage) to a region of lower potential.

Electric current is measured in amperes (A), named after Andr茅-Marie Amp猫re, a pioneer in electromagnetism. One ampere corresponds to a charge flow of one coulomb per second.
  • The formula that relates electric current with power and voltage is: \[ I = \frac{P}{V} \]
  • Where, \( I \) is the current in amperes, \( P \) is the power in watts, and \( V \) is the voltage in volts.
In the context of our exercise, the current through the resistor is calculated using the given power and voltage, resulting in a current of 21.8 A.
Power Dissipation
Power dissipation in a resistor refers to the process by which an electrical component converts electric energy into heat. This occurs due to the resistance offered to the flow of current.

According to the formula \( P = V \cdot I \), power is the product of the potential difference (voltage) across the resistor and the current flowing through it. This power loss manifests as heat, a phenomenon harnessed in devices such as electric heaters and incandescent bulbs.
  • To calculate power dissipation: \[ P = V \cdot I \]Where \( P \) is in watts (W), \( V \) is in volts (V), and \( I \) is in amperes (A).
  • Another useful formula involves resistance: \[ P = I^2 \cdot R \]Where \( R \) is resistance in ohms.
In the given problem, 327 watts of power is dissipated across the resistor, causing it to convert this energy into heat.
Resistor
A resistor is a fundamental component in electrical circuits, designed to limit the flow of electric current. Think of it like a narrow section of a pipe that slows the flow of water. The resistor controls the flow of current and divides voltage within the circuit.

Ohm's Law, \( V = I \cdot R \), is the primary relationship used to understand the behavior of resistors in circuits.
  • Here, \( R \) denotes the resistance, measured in ohms (惟). The unit honors George Ohm, who explored the relationship between voltage, current, and resistance.
  • Resistance is calculated using the rearranged Ohm's Law: \[ R = \frac{V}{I} \]
In the provided exercise, the resistance is calculated to be approximately 0.688 ohms, illustrating the resistor's opposition to the current flow. Resistors are key in both managing current levels, and safely distributing voltage in an electrical circuit.

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Most popular questions from this chapter

A \(540-\mathrm{W}\) electric heater is designed to operate from 120 \(\mathrm{V}\) lines. (a) What is its resistance? (b) What current does it draw? (c) If the line voltage drops to \(110 \mathrm{V},\) what power does the heater take? (Assume that the resistance is constant. Actually, it will change because of the change in temperature.) (d) The heater coils are metallic, so that the resistance of the heater decreases with decreasing temperature. If the change of resistance with temperature is taken into account, will the electrical power consumed by the heater be larger or smaller than what you calculated in part (c)? Explain.

A piece of wire has a resistance \(R\) . It is cut into three pieces of equal length, and the pieces are twisted together parallel to each other. What is the resistance of the resulting wire in terms of \(R ?\)

Lightbulbs. The wattage rating of a lightbulb is the power it consumes when it is connected across a 120 potential difference. For example, a 60 W lightbulb consumes 60.0 \(\mathrm{W}\) of electrical power only when it is connected across a 120 \(\mathrm{V}\) potential difference. (a) What is the resistance of a 60 \(\mathrm{W}\) lightbulb? (b) Without doing any calculations, would you expect a 100 \(\mathrm{W}\) bulb to have more or less resistance than a 60 \(\mathrm{W}\) bulb? Calculate and find out.

A capacitor is charged to a potential of 12.0 \(\mathrm{V}\) and is then connected to a voltmeter having an internal resistance of 3.40 \(\mathrm{M\Omega} .\) After a time of 4.00 \(\mathrm{s}\) the voltmeter reads 3.0 \(\mathrm{V}\) What are (a) the capacitance and (b) the time constant of the circuit?

Transmission of nerve impulses. Nerve cells transmit electric signals through their long tubular axons. These signals propagate due to a sudden rush of \(\mathrm{Na}^{+}\) ions, each with charge \(+e,\) into the axon. Measurements have revealed that typically about \(5.6 \times 10^{11} \mathrm{Na}^{+}\) ions enter each meter of the axon during a time of 10 \(\mathrm{ms}\) . What is the current during this inflow of charge in a meter of axon?
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