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Chapter 19: Problem 43

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.

Short Answer

Expert verified
(a) 26.67 惟, (b) 4.5 A, (c) 453.76 W, (d) Larger power if resistance decreases with temperature.

Step by step solution

01

Calculate Resistance (a)

To find the resistance of the heater, we use the power formula: \[ P = V^2 / R \] where \( P \) is the power (540 W) and \( V \) is the voltage (120 V). Solve for \( R \): \[ R = V^2 / P = 120^2 / 540 \].Perform the calculation: \[ R = 14400 / 540 \approx 26.67 \mathrm{\, \Omega} \]
02

Calculate Current Draw (b)

To find the current draw by the heater, use Ohm's Law: \[ I = V / R \]where \( V \) is 120 V and \( R \approx 26.67 \mathrm{\, \Omega} \). Plug in the values:\[ I = 120 / 26.67 \]This calculation results in:\[ I \approx 4.5 \mathrm{\, A} \]
03

Calculate Power at Lower Voltage (c)

When the voltage drops to 110 V, use the power formula: \[ P = V^2 / R \]where \( V = 110 \mathrm{\, V} \) and \( R \approx 26.67 \mathrm{\, \Omega} \). Substitute the values and calculate:\[ P = 110^2 / 26.67 \]This calculation results in:\[ P = 12100 / 26.67 \approx 453.76 \mathrm{\, W} \]
04

Discuss Resistance Change with Temperature (d)

Metallic resistance decreases with temperature, causing the resistance to be lower than calculated in part (c). This reduces the actual resistance and increases the current, leading to increased power consumption. Thus, the electrical power consumed will be larger than 453.76 W if the resistance change with temperature is considered.

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Key Concepts

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

Power Calculation
Power calculation in electric circuits is fundamental for understanding how devices like electric heaters work. Power (\( P \) ) in an electrical circuit is the rate at which energy is consumed or transformed. For resistive components, power can be calculated using the formula:\[ P = \frac{V^2}{R} \] Here, \( V \) represents the voltage across the device, and \( R \) stands for resistance. Knowing how power is calculated helps you determine how much energy a device will consume over time. This is essential for designing circuits that safely handle specific power levels.
Ohm's Law
Ohm's Law is a crucial principle in the study of electric circuits that relates voltage (\( V \) ), current (\( I \) ), and resistance (\( R \) ) with the formula:\[ V = I \times R \]This means the voltage across a resistor is equal to the product of its current and resistance. In the heater problem, to find the current drawn by the heater, Ohm鈥檚 Law is rearranged to:\[ I = \frac{V}{R} \]This formula allows you to calculate how much current flows through a device when you know its resistance and the voltage applied.
Resistance and Temperature
Resistance in electric circuits is not always constant and can change with varying temperatures, especially in metallic elements. Most metals increase in resistance with a rise in temperature due to increasing atomic vibrations that impede the flow of electrons. However, for some temperature ranges, resistance might slightly decrease as described in the exercise if cooling is taken into account. This changing resistance affects power and current calculations, meaning real-world scenarios can slightly deviate from ideal calculations where resistance is assumed constant.
Current Calculation
Current calculation in electric circuits is about determining the flow of electric charge. The current (\( I \) ) is usually measured in amperes (A) and can be determined using the rearranged formula from Ohm's Law:\[ I = \frac{V}{R} \]Here, \( V \) stands for voltage, and \( R \) represents resistance. Understanding current is essential since knowing the amount of current that flows ensures that circuits function correctly without overheating or causing damage. In practical applications, calculating the correct current is also important while designing electrical systems to match power requirements safely and efficiently.

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Most popular questions from this chapter

Struck by lightning. Lightning strikes can involve currents as high as \(25,000\) A that last for about 40\(\mu\) s. If a person is struck by a bolt of lightning with these properties, the current will pass through his body. We shall assume that his mass is 75 kg, that he is wet (after all, he is in a rainstorm) and therefore has a resistance of \(1.0 \mathrm{k} \Omega,\) and that his body is all water (which is reasonable for a rough, but plausible, approximation).(a) By how many degrees Celsius would this lightning bolt increase the temperature of 75 kg of water? (b) Given that the internal body temperature is about \(37^{\circ} \mathrm{C}\) , would the person's temperature actually increase that much? Why not? What would happen first?

With a 1500 \(\mathrm{M\Omega}\) resistor across its terminals, the terminal voltage of a certain battery is 2.50 \(\mathrm{V}\) . With only a 5.00\(\Omega\) ? resistor across its terminals, the terminal voltage is 1.75 \(\mathrm{V}\) . (a) Find the internal emf and the internal resistance of this battery. (b) What would be the terminal voltage if the 5.00\(\Omega\) resistor were replaced by a 7.00 \(\Omega\) resistor?

. A complete series circuit consists of a 12.0 \(\mathrm{V}\) battery, a 4.70\(\Omega\) resistor, and a switch. The internal resistance of the battery is 0.30\(\Omega .\) The switch is open. What does an ideal voltmeter read when placed (a) across the terminals of the battery, (b) across the resistor, (c) across the switch? (d) Repeat parts (a), (b), and (c) for the case when the switch is closed.

When you connect an unknown resistor across the terminals of a 1.50 \(\mathrm{V}\) AAA battery having negligible internal resistance, you measure a current of 18.0 \(\mathrm{mA}\) flowing through it. (a) What is the resistance of this resistor? (b) If you now place the resistor across the terminals of 12.6 \(\mathrm{V}\) car battery having no inter- nal resistance, how much current will flow? (c) You now put the resistor across the terminals of an unknown battery of negligible internal resistance and measure a current of 0.453 \(\mathrm{A}\) flowing through it. What is the potential difference across the terminals of the battery?

An idealized voltmeter is connected across the terminals of a 15.0 \(\mathrm{V}\) battery, and a 75.0\(\Omega\) appliance is also connected across its terminals. If the voltmeter reads \(11.3 \mathrm{V} :\) (a) how much power is being dissipated by the appliance, and (b) what is the internal resistance of the battery?
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