Problem 66 - Fundamentals of Physics

Question

The headlights of a moving car require about \(10 \mathrm{~A}\) from the \(12 \mathrm{~V}\) alternator, which is driven by the engine. Assume the alternator is \(80 \%\) efficient (its output electrical power is \(80 \%\) of its input mechanical power), and calculate the horsepower the engine must supply to run the lights.

Step-by-Step Solution

Answer

The engine must supply approximately 0.201 horsepower to run the lights.

Step 1: Calculate Electrical Power Output

To find the electrical power output, use the formula for electrical power: \( P = IV \) where \( I = 10 \text{ A} \) (current) and \( V = 12 \text{ V} \) (voltage). Thus, \( P = 10 \times 12 = 120 \text{ W} \).

Step 2: Determine Mechanical Power Input

Since the alternator is 80% efficient, the mechanical power input must be higher than the electrical power output. Use the efficiency formula: \[ \text{Efficiency} = \frac{\text{Electrical Power Output}}{\text{Mechanical Power Input}} \times 100 \] Given that efficiency is 80%, \[ 0.80 = \frac{120}{\text{Mechanical Power Input}} \] Thus, \( \text{Mechanical Power Input} = \frac{120}{0.80} = 150 \text{ W} \).

Step 3: Convert Watts to Horsepower

1 horsepower is equivalent to 746 watts. To convert the mechanical power required in watts to horsepower:\[ \text{Horsepower} = \frac{\text{Mechanical Power in Watts}}{746} = \frac{150}{746} \approx 0.201 \text{ hp} \].

Key Concepts

Electrical Power

Understanding electrical power is crucial when working with electrical systems, particularly in physics problem-solving. Electrical power is the rate at which electrical energy is consumed by a circuit, and it is measured in watts (W). The formula to calculate electrical power is given by the equation \( P = IV \), where \( P \) represents power, \( I \) is the current in amperes (A), and \( V \) is the voltage in volts (V).

Let's break down the formula:

• Current (I): This is the flow of electric charge, typically measured in amperes.
• Voltage (V): This represents the potential difference that drives the electric charge through the circuit.

In the provided example, the headlights of a car draw 10 amperes at 12 volts, leading to an electrical power consumption of 120 watts. This calculation is straightforward and serves as the foundation for more complex analyses involving efficiency and conversions.

Efficiency Calculation

Efficiency plays a vital role in determining how well an electrical system converts input energy into output work. It is often expressed as a percentage that indicates how much of the input energy is successfully converted into the desired output. The general formula to determine efficiency is:

\[\text{Efficiency} = \left( \frac{\text{Output Power}}{\text{Input Power}} \right) \times 100\]

In mechanical systems like alternators, efficiency tells us how well mechanical energy is converted into electrical energy. For instance, an alternator that is 80% efficient only converts 80% of the input mechanical power into electrical power. Calculating the input required for a certain output involves understanding this efficiency:

• In the problem, the alternator's output power was calculated to be 120 watts.
• The mechanical power input then becomes \( \frac{120}{0.80} = 150 \) watts.

This means the engine needs to supply more power than the electrical output demands, due to the inefficiencies in conversion.

Horsepower Conversion

Converting between different units of power, like converting watts to horsepower, is commonplace in engineering and physics. Understanding these conversions is essential when comparing mechanical and electrical systems. Horsepower is a unit of power often used to describe engine power, and it is equivalent to 746 watts.

To convert watts to horsepower, the formula used is:
\[ \text{Horsepower} = \frac{\text{Power in Watts}}{746}\]

In the exercise example, the mechanical power needed by the engine is 150 watts. To find the equivalent in horsepower, you perform the conversion:

• \( \text{Horsepower} = \frac{150}{746} \approx 0.201 \) horsepower.

This conversion helps translate technical specifications into more familiar terms when discussing engine performance, providing a clearer understanding of the power required.