Problem 67 - Fundamentals of Physics

Question

A \(500 \mathrm{~W}\) heating unit is designed to operate with an applied potential difference of \(115 \mathrm{~V}\). (a) By what percentage will its heat output drop if the applied potential difference drops to \(110 \mathrm{~V} ?\) Assume no change in resistance. (b) If you took the variation of resistance with temperature into account, would the actual drop in heat output be larger or smaller than that calculated in (a)?

Step-by-Step Solution

Answer

(a) Heat output drops by 8.49%. (b) The actual drop would be larger.

Step 1: Understand the Relationship Between Power, Voltage, and Resistance

The power dissipated by a heating unit in terms of voltage and resistance is given by the formula:\[ P = \frac{V^2}{R} \]where \(P\) is the power, \(V\) is the potential difference, and \(R\) is the resistance.

Step 2: Calculate Initial Resistance Using Original Voltage

Using the original power \(P = 500\, \text{W}\) and voltage \(V_1 = 115\, \text{V}\), calculate the resistance:\[ R = \frac{V_1^2}{P} = \frac{115^2}{500} \,= 26.45 \Omega \]

Step 3: Calculate New Power Output with Reduced Voltage

With the reduced voltage \(V_2 = 110\, \text{V}\), calculate the new power output using the same resistance:\[ P_2 = \frac{V_2^2}{R} = \frac{110^2}{26.45} \] Solve to find:\[ P_2 = \frac{12100}{26.45} \approx 457.55\, \text{W} \]

Step 4: Determine the Percentage Drop in Power Output

Calculate the percentage drop in power from the original power to the new power:\[ \text{Percentage Drop} = \left( \frac{P_1 - P_2}{P_1} \right) \times 100 = \left( \frac{500 - 457.55}{500} \right) \times 100 \]\[ \approx 8.49\% \]

Step 5: Consider the Effect of Resistance Variation on Power Output

If resistance varies with temperature, increasing resistance would further decrease the power output due to a quadratic dependence on voltage and an increase in resistance. Thus, the actual drop in heat output would be larger.

Key Concepts

Ohm's Law

Ohm's Law is a fundamental principle used to understand how electricity flows in a circuit. It states that the current through a conductor between two points is directly proportional to the voltage across the two points and inversely proportional to the resistance. Mathematically, it is expressed as:

• \( V = IR \)

Here, \( V \) is the voltage (in volts), \( I \) is the current (in amperes), and \( R \) is the resistance (in ohms).
Ohm’s Law helps us in calculating one of these values if the other two are known. It is crucial for understanding more complex electrical concepts such as the behavior of circuits and components like resistors.
In the context of power calculations, knowing Ohm's Law means we can relate voltage and resistance to power, and determine how changes in one will affect the whole circuit.

Resistance

Resistance is a measure of the opposition to the flow of electric current in a material. It is represented by the symbol \( R \) and is measured in ohms (\( \Omega \)). Different materials have different resistances based on their properties. For example, copper wire has low resistance and is thus a good conductor of electricity.

• Resistance is calculated using the formula: \( R = \frac{V}{I} \)

When analyzing electrical circuits, resistance tells us how much the material will resist the electrical flow. In our exercise, the resistance was calculated using the initial power and voltage values. It remained constant when recalculating for different voltages, assuming that the resistive material does not change with temperature.
This assumption simplifies calculations but does not account for realistic scenarios where resistance may increase as the temperature of the device rises.

Voltage Drop

Voltage drop refers to the reduction in voltage as the electric current flows through the circuit due to the resistance it encounters.
This is a common issue in electrical systems where long cables or components with high resistance are used.

• Voltage drop can be calculated by multiplying the current by the total resistance in the circuit path.

In our heating unit example, the voltage drop was the difference between the potential differences of \(115 \mathrm{~V} \) and \(110 \mathrm{~V} \), given no change in resistance.
The concept of voltage drop helps us understand why devices might receive lower than expected power, especially in circuits with high resistance or over long distances. Addressing voltage drop is essential for maintaining efficient power delivery.

Power Loss Calculation

Power loss calculation is essential for understanding how much energy is dissipated as heat in an electrical circuit. This is crucial for designing circuits that manage energy efficiently and avoid overheating.
In the exercise, power was originally calculated using:

• \( P = \frac{V^2}{R} \)

This formula shows that power depends on the square of the voltage divided by the resistance.
When the voltage drops, as in our example, the power output decreases, resulting in a lower heating effect from the unit. Since power is proportional to the square of the voltage, even a small drop in voltage can lead to a significant decrease in power output.
The percentage change in power (calculated as around 8.49%) helps to assess how much performance is affected. If resistance changes with temperature, power loss would be greater, showing the impact of resistance on circuit performance.