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\(\mathrm{A} \cdot 540 \mathrm{~W} "\) electric heater is designed to operate from \(120 \mathrm{~V}\) lines. (a) What is its operating 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) The operating resistance is \(26.67 \Omega\) b) It draws a current of \(4.5A\) c) The heater takes a power of \(452.05W\) when voltage drops to \(110V\) d) The electrical power consumed by the heater will be larger with decreasing resistance due to decreasing temperature.

Step by step solution

01

Calculate Resistance

From the given power P and voltage V, we can calculate the resistance (R) of the heater using the formula \(R = V^{2} / P\). In this case, \(P = 540W\) and \(V = 120V\), resulting in \(R = (120)^{2} / 540 = 26.67 \Omega\)
02

Calculate Current

Ohm's law states that current (I) equals voltage (V) divided by resistance (R). Substituting the known values, \(I = V / R = 120 / 26.67 = 4.5 A\). Our heater thus draws 4.5 Amperes.
03

Calculate Power at Reduced Voltage

If the line voltage drops to \( 110V \), we need to re-calculate the power. We can use the power formula \(P = V^{2} / R\). Using the new \( V = 110V \) and the resistance calculated in Step 1 (\( R = 26.67 \Omega\)). Hence, \(P = (110)^{2} / 26.67 = 452.05W \)
04

Determine Change of Power Consumption

The resistance of the heater decreases with decreasing temperature. For a given voltage, if the resistance decreases, the current, \(I = V / R\), will increase. Consequently, the electrical power consumed, \(P = I^{2}R\), will increase. Thus, with the decrease in resistance taken into account, the electrical power consumed by the heater will be larger than what was calculated in part (c)

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Key Concepts

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

Ohm's Law
Ohm's Law is a foundational principle in electrical engineering, forming the basis for analyzing electric circuits. The law states that the current flowing through a conductor between two points is directly proportional to the voltage across the two points, provided the temperature remains constant. This can be expressed with the formula:
  • \[ I = \frac{V}{R} \]
  • Where:
    • \(I\) is the current in amperes (A),
    • \(V\) is the voltage in volts (V),
    • \(R\) is the resistance in ohms (Ω).
Understanding Ohm's Law allows us to compute the current through any component given its voltage and resistance. In our exercise, we used this principle to determine that the heater draws a current of 4.5 amperes when supplied with 120 volts.
Electrical Resistance
Electrical resistance is a measure of how much an object opposes the flow of electric current. The unit of resistance is the ohm, represented by the symbol Ω. In simple terms, resistance determines how easily electricity can pass through a material.
Foil-like materials or wires have low resistance and allow the current to flow easily, whereas materials like rubber have high resistance. The resistance of an electric heater is a crucial factor in determining how much current it draws. We calculated the resistance of the heater at 120V and 540W using the formula:
  • \[ R = \frac{V^2}{P} \]
  • Where \(P\) is the power in watts (W) and \(V\) is the voltage.
In our problem, this calculation resulted in a resistance of 26.67 Ω for the heater. This understanding helps in predicting the heater's behavior under different voltages.
Power Consumption
Power consumption in electrical devices like heaters is a measure of how much electricity they use over time, typically expressed in watts (W). It's crucial to calculate power consumption to understand how much energy an appliance uses.
The power consumed by an electric heater can be determined using the formula:
  • \[ P = \frac{V^2}{R} \]
  • Where \(P\) is power, \(V\) is voltage, and \(R\) is resistance.
In our exercise, when the voltage dropped to 110V, recalculating the power consumption using the constant resistance resulted in a new power consumption of approximately 452.05W. This illustrates how device performance varies with input voltage changes, which is essential for efficiency and operational cost assessments.
Temperature Effect on Resistance
Electrical resistance is sensitive to temperature changes, especially in metals. As the temperature of a metal increases, its resistance generally increases as well. Conversely, a drop in temperature often results in decreased resistance.
This relationship becomes significant in devices like heaters where coil temperature can vary substantially during operation. When considering a drop in line voltage to 110V, if the temperature effect on resistance is acknowledged, one can expect a reduction in resistance. This reduced resistance at lower temperatures increases the current through the relation \(I = V/R\), leading to increased power consumption as per \(P = I^2 R\). Therefore, accounting for temperature effects means predicting a higher power consumption than initially calculated, reinforcing the importance of understanding the dynamic nature of real-world electrical devices.

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Most popular questions from this chapter

A very long, straight solenoid with a cross-sectional area of \(2.00 \mathrm{~cm}^{2}\) is wound with 90.0 turns of wire per centimeter. Starting at \(t=0\) the current in the solenoid is increasing according to \(i(t)=\left(0.160 \mathrm{~A} / \mathrm{s}^{2}\right) t^{2}\). A secondary winding of 5 turns encircles the solenoid at its center, such that the secondary winding has the same cross-sectional area as the solenoid. What is the magnitude of the emf induced in the secondary winding at the instant that the current in the solenoid is \(3.20 \mathrm{~A}\) ?

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