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Suppose that a voltage surge produces \(140 \mathrm{V}\) for a moment. By what percentage does the power output of a \(120-\mathrm{V}, 100-\mathrm{W}\) lightbulb increase? Assume that its resistance does not change.

Short Answer

Expert verified
The percentage increase in the power output of the lightbulb after the voltage surge is calculated by substituting the given values into the formula obtained in Step 4.

Step by step solution

01

Understand the power formula and Ohms's law

The power \( P \) in an electric circuit is given by the formula \( P = V^2 / R \), where \( V \) is the voltage and \( R \) is the resistance. Ohm's law states that the resistance \( R \) in an electric circuit is constant and equal to the voltage \( V \) divided by the current \( I \) (R = V/I). In this case, we are told that the resistance does not change.
02

Calculate the initial power output

Given that the voltage is \(120V\) and the power is \(100W\), we can use the power formula to confirm the initial power output. Substituting the given values into the formula, we get \( P = 120^2 / R \). From this, we can find the resistance as \( R = 120^2 / 100 \).
03

Calculate the power output after the surge

After the voltage surge, the voltage increases to \(140 V\). Since the resistance doesn't change, we can substitute the new voltage and the resistance obtained from the previous step into the power formula to get the new power output as \( P' = 140^2 / R \).
04

Calculate the percentage increase in power output

The percentage increase in power is given by \((P' - P) / P * 100\%. Substituting the values of \(P'\) and \(P\) we get the percentage increase in power.

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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 fundamental principle in electronics that describes the relationship between voltage, current, and resistance in an electrical circuit. It is expressed with the formula: \( V = IR \), where \( V \) is the voltage across the resistance (in volts), \( I \) is the current flowing through the resistance (in amperes), and \( R \) is the resistance itself (in ohms).
This law states that the current flowing through a conductor between two points is directly proportional to the voltage across the two points and inversely proportional to the resistance. In simpler terms, increasing the voltage increases the current, while increasing the resistance decreases the current, provided all other variables are constant.
In the context of an unchanged resistance scenario, like a lightbulb in the original exercise, Ohm's Law helps us understand that when there's a change in voltage, we can expect a direct impact on the current and consequently on the power, assuming the resistance remains constant.
Voltage Surge
A voltage surge is a sudden increase in electrical voltage that occurs in electrical circuits, often due to external factors such as lightning strikes, power outages, or malfunctions in the power grid. These surges can significantly alter the performance of electronic devices.
It's crucial to be aware of voltage surges because they can lead to increased electrical stress on circuit components. In our example, a voltage surge elevates the bulb's operating voltage from 120V to 140V momentarily. While this might not sound drastic, it is a considerable 16.67% increase in voltage, and it highlights the need to calculate its influence on the bulb's performance using the power formula.
To manage unexpected increases in voltage and avoid damage, many devices utilize surge protectors, which help stabilize voltage levels.
Resistance
Resistance is a measure of the opposition to the flow of electric current in a circuit. The unit of resistance is the ohm, symbolized by \( \Omega \). Resistance primarily depends on the material, cross-sectional area, and the length of the conductor.
In the original exercise, the resistance of the lightbulb is calculated using the initial conditions given: \( R = \frac{120^2}{100} \). Since resistance is defined as a constant in this context, it implies that any changes in voltage will directly affect the power output according to the power formula \( P = \frac{V^2}{R} \).
Understanding resistance is crucial for predicting how changes in voltage or current will affect power output, allowing for better circuit design and management.
Percentage Increase
Percentage increase is a mathematical concept used to determine how much a value has gone up in comparison to the original value. It's calculated using the formula: \((\text{New Value} - \text{Old Value}) / \text{Old Value} * 100\%\).
In the context of our lightbulb example, to find the increase in power output due to the voltage surge, you determine the old power using the original voltage and the constant resistance, and the new power using the surged voltage. Afterward, you apply the percentage increase formula.
This concept allows us to quantify changes in terms of percentages, making it easier to understand the relative significance of the change—a useful tool in evaluating performance variations in electronic devices under different conditions.

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Most popular questions from this chapter

A \(500-\) W heating coil designed to operate from \(110 \mathrm{V}\) is made of Nichrome wire \(0.500 \mathrm{mm}\) in diameter. (a) Assuming that the resistivity of the Nichrome remains constant at its \(20.0^{\circ} \mathrm{C}\) value, find the length of wire used. (b) What If? Now consider the variation of resistivity with temperature. What power will the coil of part (a) actually deliver when it is heated to \(1200^{\circ} \mathrm{C} ?\)

A straight cylindrical wire lying along the \(x\) axis has a length of \(0.500 \mathrm{m}\) and a diameter of \(0.200 \mathrm{mm} .\) It is made of a material that obeys Ohm's law with a resistivity of \(\rho=\) \(4.00 \times 10^{-8} \Omega \cdot \mathrm{m} .\) Assume that a potential of \(4.00 \mathrm{V}\) is maintained at \(x=0,\) and that \(V=0\) at \(x=0.500 \mathrm{m} .\) Find (a) the electric ficld \(\mathbf{E}\) in the wire, (b) the resistance of the wire, (c) the electric current in the wire, and (d) the current density \(J\) in the wire. Express vectors in vector notation. (e) Show that \(\mathbf{E}=\rho \mathbf{J}\).

The resistance of a platinum wire is to be calibrated for low-temperature measurements. A platinum wire with resistance \(1.00 \Omega\) at \(20.0^{\circ} \mathrm{C}\) is immersed in liquid nitrogen at \(77 \mathrm{K}\left(-196^{\circ} \mathrm{C}\right) .\) If the temperature response of the platinum wire is linear, what is the expected resistance of the platinum wire at \(-196^{\circ} \mathrm{C} ? \quad\left(\alpha_{\text {platinum }}=3.92 \times 10^{-3} /^{\circ} \mathrm{C}\right)\)

Compute the cost per day of operating a lamp that draws a current of \(1.70 \mathrm{A}\) from a \(110-\mathrm{V}\) line. Assume the cost of energy from the power company is \(\$ 0.0600 / \mathrm{kWh}\).

An electric car is designed to run off a bank of \(12.0-\mathrm{V}\) batteries with total energy storage of \(2.00 \times 10^{7} \mathrm{J} .\) (a) If the electric motor draws \(8.00 \mathrm{kW},\) what is the current delivered to the motor? (b) If the electric motor draws \(8.00 \mathrm{kW}\) as the car moves at a steady speed of \(20.0 \mathrm{m} / \mathrm{s}\), how far will the car travel before it is "out of juice"?

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