Problem 6 - Applied Physics

Question

What current does a \(75.0-\Omega\) resistance draw on \(115 \mathrm{~V} ?\)

Step-by-Step Solution

Answer

The current is approximately 1.53 A.

Step 1: Understand Ohm's Law

Ohm's Law states that the current (I) flowing through a conductor between two points is directly proportional to the voltage (V) across the two points and inversely proportional to the resistance (R) of the conductor. The law is mathematically expressed as: \( I = \frac{V}{R} \).

Step 2: Identify the Given Values

From the problem, the given resistance \( R \) is \( 75.0 \, \Omega \) and the voltage \( V \) is \( 115 \, \text{V} \).

Step 3: Insert Values into Ohm's Law Equation

Using the formula from Ohm's Law \( I = \frac{V}{R} \), substitute the given values: \( I = \frac{115}{75.0} \).

Step 4: Perform the Calculation

Divide 115 by 75.0 to find the current \( I \): \( I = 1.5333 \ldots \approx 1.53 \, \text{A} \). We round the final value to two decimal places as is standard in physics for precision.

Key Concepts

Understanding Current Calculation

Current calculation is a fundamental concept in electrical physics. This involves determining the amount of charge flow per second through a circuit.
According to Ohm's Law, the current (denoted by \( I \)) can be calculated using the formula:

• \( I = \frac{V}{R} \)

where \( V \) stands for voltage, and \( R \) represents resistance.
This formula shows that the current is directly proportional to the voltage and inversely proportional to the resistance.
To find the current, divide the voltage by the resistance.
In practical terms, this tells us how many amperes flow through the circuit when a certain voltage is applied to a given resistance. Understanding how to calculate current helps in designing and analyzing electrical circuits, ensuring efficiency and safety.

Concept of Resistance

Resistance is a measure of how much a material opposes the flow of electric current through it. It's like the electrical friction that slows down the current.
Resistance is measured in ohms (\( \Omega \)), named after Georg Simon Ohm, who formulated Ohm’s Law.
The higher the resistance, the lower the current for a given voltage, as the current and resistance are inversely proportional.

• Factors affecting resistance include the material’s nature, its length, cross-sectional area, and temperature.
• Materials with low resistance, like copper, are called conductors, while those with high resistance, like rubber, are called insulators.

Understanding resistance helps us choose the right materials for building circuits and ensures they function properly by allowing just the right amount of current to pass through.

Exploring Voltage

Voltage, also known as electric potential difference, is the force that pushes electric charges through a conductor. It’s like the pressure forcing water through a pipe.
Measured in volts (V), the voltage indicates the energy difference that drives the current from one point to another in a circuit.

• Higher voltage means greater force, pushing more current through the circuit given a constant resistance.
• This is why devices have voltage ratings; connecting them to voltages beyond their capacity can cause damage due to excessive current.

In circuits, voltage sources like batteries or power supplies provide the energy needed to move the charges. Understanding voltage is crucial in the design of safe and effective electrical systems.

Physics Problem-Solving

Physics problem-solving often involves applying concepts like Ohm's Law to real-world scenarios.
It requires careful analysis of the given data and insertion of these values into appropriate formulas to find solutions.
For instance, solving for current when given resistance and voltage involves:

• Identifying known quantities (resistance \( R \) and voltage \( V \)).
• Substitute them into \( I = \frac{V}{R} \).
• Performing mathematical operations to solve for the unknown (current \( I \)).

Effective problem-solving means understanding the relationship between physical quantities, applying the equations correctly, and interpreting the results in the context of the problem. Practice and familiarity with these concepts enhance one’s ability to tackle complex electrical problems with confidence.