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Chapter 18: Problem 110

A \(500-\) W electric heater unit is designed to operate with an applied potential difference of \(120 \mathrm{V}\). (a) If the local power company imposes a voltage reduction to lighten its load, dropping the voltage to $110 \mathrm{V}$, by what percentage does the heat output of the heater drop? (Assume the resistance does not change.) (b) If you took the variation of resistance with temperature into account, would the actual drop in heat output be larger or smaller than calculated in part (a)?

Short Answer

Expert verified
Additionally, discuss the potential impact of the variation of resistance with temperature on the actual drop in heat output. Answer: The approximate percentage drop in heat output, assuming constant resistance, is 15.9%. However, taking into account the variation of resistance with temperature, the actual drop in heat output may be smaller than 15.9% because the resistance may decrease as the temperature decreases, resulting in higher actual power output than calculated.

Step by step solution

01

Calculate the initial power output

To find the initial power output, we can use the power formula: P = V²/R. We are given the power (P = 500 W) and the voltage (V = 120 V). First, we need to find the resistance (R). Rearrange the formula to solve for R: R = V²/P. Plug in the given values: R = (120 V)² / 500 W = 28.8 Ω.
02

Calculate the power output after the voltage reduction

Now, we will find the power output when the voltage is reduced to 110 V, assuming the resistance remains constant. Use the power formula again: P = V²/R. Plug in the new voltage and the resistance: P = (110 V)² / 28.8 Ω ≈ 420.49 W.
03

Calculate the percentage drop in heat output

To find the percentage drop in heat output, we can use the formula: percentage drop = ((initial power - new power) / initial power) * 100. Plug in the values we found: percentage drop = ((500 W - 420.49 W) / 500 W) * 100 ≈ 15.9%. So, when the voltage is reduced to 110 V, the heat output drops by approximately 15.9%.
04

Discuss the effect of the variation of resistance with temperature

In part (b), we are asked to consider the effect of the variation of resistance with temperature on the actual drop in heat output. As the temperature of the heater decreases, the resistance of the heater may also decrease (as most materials have a positive temperature coefficient of resistance). This means that the actual resistance of the heater could be lower when the voltage is reduced. Since P = V²/R, if the resistance is lower than we assumed, the actual power output would be higher than we calculated in Step 2. Therefore, taking the variation of resistance with temperature into account, the actual drop in heat output would likely be smaller than the value we calculated in part (a) (15.9%).

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Most popular questions from this chapter

What is the energy stored in a small battery if it can move \(675 \mathrm{C}\) through a potential difference of \(1.20 \mathrm{V} ?\)
Chelsea inadvertently bumps into a set of batteries with an emf of $100.0 \mathrm{V}\( that can supply a maximum power of \)5.0 \mathrm{W} .$ If the resistance between the points where she contacts the batteries is $1.0 \mathrm{k} \Omega,$ how much current passes through her?
A voltmeter has a switch that enables voltages to be measured with a maximum of \(25.0 \mathrm{V}\) or \(10.0 \mathrm{V}\). For a range of voltages to $25.0 \mathrm{V},\( the switch connects a resistor of magnitude \)9850 \Omega$ in series with the galvanometer; for a range of voltages to \(10.0 \mathrm{V}\), the switch connects a resistor of magnitude \(3850 \Omega\) in series with the galvanometer. Find the coil resistance of the galvanometer and the galvanometer current that causes a full-scale deflection. [Hint: There are two unknowns, so you will need to solve two equations simultaneously.]
A \(12-\Omega\) resistor has a potential difference of \(16 \mathrm{V}\) across it. What current flows through the resistor?
A 2.00 - \(\mu\) F capacitor is charged using a \(5.00-\mathrm{V}\) battery and a 3.00 - \(\mu\) F capacitor is charged using a \(10.0-\mathrm{V}\) battery. (a) What is the total energy stored in the two capacitors? (b) The batteries are disconnected and the two capacitors are connected together \((+\text { to }+\) and \- to \(-\) ). Find the charge on each capacitor and the total energy in the two capacitors after they are connected. (c) Explain what happened to the "missing" energy. [Hint: The wires that connect the two have some resistance.]
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