Question

A voltage spike causes the line voltage in a home to jump rapidly from \(110 . V\) to \(150 . V\). What is the percentage increase in the power output of a \(100 .-\mathrm{W}\) tungsten-filament incandescent light bulb during this spike, assuming that the bulb's resistance remains constant?

Short answer

Answer: The percentage increase in the power output of the light bulb during the voltage spike is approximately 85.95%.

Step-by-step solution

Calculate the initial power output at 110V

The given power of the light bulb is 100 W at 110V. We can calculate the resistance using the formula P = V^2/R, where P = 100 W, V = 110 V. Rearranging the formula, we can get R = V^2/P. R = (110)^2/100 R = 12100/100 R = 121 ohms The resistance of the tungsten filament is 121 ohms.

Calculate the power output at 150V

We now have the resistance of the tungsten filament. With the increased voltage to 150V, we can calculate the new power output (P_new) using the formula P_new = V^2/R, where V = 150 V and R = 121 ohms. P_new = (150)^2/121 P_new = 22500/121 P_new ≈ 185.95 W The new power output at 150V is approximately 185.95 W.

Calculate the percentage increase in power output

Now, we will find the percentage increase in the power output by comparing the initial power output (P_initial = 100 W) and the new power output (P_new ≈ 185.95 W). Percentage increase = ((P_new - P_initial) / P_initial) * 100 Percentage increase = ((185.95 - 100) / 100) * 100 Percentage increase ≈ 85.95% The percentage increase in the power output of the light bulb during the voltage spike is approximately 85.95%.

Key concepts

Ohm's Law

Understanding Ohm's Law is foundational in analyzing the behavior of electrical circuits. Essentially, Ohm's Law reveals how voltage, current, and resistance interact. The formula, often written as \( V = IR \), tells us that voltage (\( V \)) across a component is the product of the current (\( I \)) flowing through it and its resistance (\( R \)).

This relationship becomes especially useful when considering electronic devices like the tungsten-filament incandescent light bulb in our exercise. When the voltage spike occurs, Ohm's Law explains how the unchanged resistance of the bulb's filament influences the current and power output. If the resistance remains constant, and the voltage increases, the current must also increase proportionally. Subsequently, this increased current leads to a higher power output, which is what we are investigating in the calculation.

Voltage Spike Impact

A voltage spike is a sudden and sharp increase in voltage, which can dramatically impact electrical devices. The most immediate effect is an increase in the power consumed by devices, if their resistance remains constant. Using Ohm's Law and the power formula \( P = V^2/R \), we can see that power is directly proportional to the square of the voltage.

For example, in a light bulb, a voltage spike means that for a moment, the filament is subjected to a higher potential difference. Given the bulb's resistance doesn't change quickly, this leads to an exponential increase in power. To illustrate, when the voltage in our exercise goes from 110V to 150V, using the formula mentioned, the calculation shows a significant rise in power – well beyond what the bulb is rated for. This not only causes a brighter light output but could also shorten the lifespan of the bulb due to overheating.

Power Output Percentage Increase

Now, to gauge how a voltage spike impacts an electrical device's performance, determining the power output percentage increase is key. This measure gives us a clear indication of how much more power the device is using compared to its standard operation.

In our example, the original power output was 100 W, and after the voltage spike, it increased to approximately 185.95 W. To find the percentage increase, we take the new power (\( P_{new} \)), subtract the original power (\( P_{initial} \)), divide by the original power, and then multiply by 100. The resulting figure, in this case, approximately 85.95%, reflects the surge in power output. It is crucial to note that this sharp increase can overburden the circuit and potentially cause damage to electrical components.