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The heating element of an electric dryer is rated at 4.1 kW when connected to a 240-V line. (a) What is the current in the heating element? Is 12-gauge wire large enough to supply this current? (b) What is the resistance of the dryer's heating element at its operating temperature? (c) At 11 cents per kWh, how much does it cost per hour to operate the dryer?

Short Answer

Expert verified
(a) 17.08 A; 12-gauge wire is sufficient. (b) Resistance is 14.05 ohms. (c) Cost is \$0.451 per hour.

Step by step solution

01

Calculate the Current

To find the current in the heating element, use the power formula \( P = VI \), where \( P \) is the power in watts (W), \( V \) is the voltage in volts (V), and \( I \) is the current in amperes (A). Rearrange the formula to solve for \( I \): \( I = \frac{P}{V} \).Given \( P = 4.1 \) kW or \( 4100 \) watts, and \( V = 240 \) volts:\[ I = \frac{4100}{240} \approx 17.08 \text{ A} \]Using a 12-gauge wire, which can typically carry up to 20 A safely, it is large enough to supply this current (17.08 A).
02

Calculate the Resistance

Use Ohm's Law, which states \( V = IR \), to find the resistance \( R \) of the heating element at its operating temperature.Rearrange the formula to solve for \( R \): \( R = \frac{V}{I} \).Using \( V = 240 \) volts and \( I = 17.08 \) amperes:\[ R = \frac{240}{17.08} \approx 14.05 \text{ ohms} \]
03

Calculate the Operating Cost

To calculate the cost per hour to operate the dryer, multiply the power in kilowatts by the cost per kWh.Given that the power is \( 4.1 \) kW and the cost is \\(0.11 per kWh:\[ \text{Cost per hour} = 4.1 \times 0.11 = \\)0.451 \text{ per hour} \]

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Key Concepts

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

Power Calculation
When we need to compute power in electrical circuits, we typically use the formula:
  • Power (\(P\)) is the product of voltage (\(V\)) and current (\(I\)): \(P = VI\).

In this specific example, a drying element with a power rating of 4.1 kilowatts (kW) is connected to a 240-volt (V) electrical line.
If the power is given in kilowatts, like 4.1 kW, you should convert it to watts (W) by multiplying by 1,000.
So, \(4.1 ext{ kW} = 4100 ext{ W}\).When calculating the amount of electrical power, simply plug the values into the formula.
  • To derive the current (\(I\)), rearrange the formula: \(I = \frac{P}{V}\).

Here, plug in power (\(P = 4100 ext{ W}\)) and voltage (\(V = 240 ext{ V}\)) to determine the current: \(I = \frac{4100}{240} \approx 17.08 ext{ amperes}\).
This method allows us to understand how different elements like voltage and power are interconnected in electrical circuits.It's crucial to appreciate that power calculation is fundamental for determining whether components, such as wires, are suitable for handling specific electrical loads, ensuring safe and efficient operation.
Ohm's Law
Ohm's Law is an essential concept in electrical circuits, providing a direct relationship between voltage, current, and resistance.
  • It defines that \(V = IR\), where voltage (\(V\)) is the product of current (\(I\)) and resistance (\(R\)).

In our earlier situation, to find the resistance of the dryer's heating element, we need this foundational equation.
Having already calculated the current at approximately \(17.08 ext{ A}\) from the previous section, and knowing the voltage (\(240 ext{ V}\)), we can rearrange Ohm's Law to solve for resistance.
  • Rearrange it to \(R = \frac{V}{I}\).

Inserting the known values: \(R = \frac{240}{17.08} \approx 14.05 ext{ ohms}\).Understanding this concept not only enables one to find resistance but also underpins all calculations involving electric currents, helping one gauge the impact of different resistive loads in a circuit.
Electrical Resistance
Electrical resistance is a property that quantifies how strongly a material opposes the flow of electric current. It's measured in ohms (Ω) and defined by Ohm's Law.
  • Higher resistance means less current flow at a given voltage.
  • In practical terms, resistance is similar to friction in a mechanical system.

In the dryer example, the calculated resistance was found to be \(14.05 ext{ ohms}\).Materials used in electrical devices are chosen based on their resistance characteristics.
  • This affects how much energy is consumed and converted into heat.Electrical resistance helps determine the efficiency of electrical components.

For example, high resistance could denote poor efficiency and potential overheating.
In our scenario, knowing the resistance enabled us to assess its efficiency and ensure that the wiring and the heating element are correctly rated for safe operation. Understanding resistance is vital for designing circuits that use the right materials to safely manage voltage and current, ensuring reliability.

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Most popular questions from this chapter

A 4.60-\(\mu\)F capacitor that is initially uncharged is connected in series with a 7.50-k\(\Omega\) resistor and an emf source with \(\varepsilon =\) 245 V and negligible internal resistance. Just after the circuit is completed, what are (a) the voltage drop across the capacitor; (b) the voltage drop across the resistor; (c) the charge on the capacitor; (d) the current through the resistor? (e) A long time after the circuit is completed (after many time constants) what are the values of the quantities in parts (a)-(d)?

A resistor \(R_1\) consumes electrical power \(P_1\) when connected to an emf \(\varepsilon\). When resistor \(R_2\) is connected to the same emf, it consumes electrical power \(P_2\). In terms of \(P_1\) and \(P_2\), what is the total electrical power consumed when they are both connected to this emf source (a) in parallel and (b) in series?

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An emf source with \(\varepsilon =\) 120 V, a resistor with \(R =\) 80.0 \(\Omega\), and a capacitor with \(C =\) 4.00 \(\mu\)F are connected in series. As the capacitor charges, when the current in the resistor is 0.900 A, what is the magnitude of the charge on each plate of the capacitor?

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