Problem 6 - Physics

Question

A resistor is connected across the terminals of a \(9.0\) -V battery, which delivers \(1.1 \times 10^{5} \mathrm{~J}\) of energy to the resistor in six hours. What is the resistance of the resistor?

Step-by-Step Solution

Answer

The resistance of the resistor is approximately 15.91 ohms.

Step 1: Identify known quantities and formulas

We know the voltage (V) across the resistor is \(9.0\, \text{V}\), the energy (E) delivered is \(1.1 \times 10^{5} \text{ J}\), and the time (t) over which the energy is delivered is \(6 \text{ hours}\). We need the resistance (R) of the resistor. We'll use the formula for power (P): \[ P = \frac{E}{t} \]Also, power in terms of voltage and resistance is given by:\[ P = \frac{V^2}{R} \]

Step 2: Convert time from hours to seconds

To use the formula for power, time should be in seconds. We convert \(6 \text{ hours}\) to seconds: \[ 6 \times 3600 = 21600 \text{ seconds} \]

Step 3: Calculate the power delivered

Using the formula for power, we calculate:\[ P = \frac{1.1 \times 10^{5}}{21600} = 5.09 \text{ W (watts)} \]

Step 4: Use the power and voltage to find resistance

Using the formula \( P = \frac{V^2}{R} \), we substitute \(P = 5.09 \text{ W}\) and \(V = 9.0 \text{ V}\) to solve for \(R\):\[ R = \frac{V^2}{P} = \frac{9^2}{5.09} = \frac{81}{5.09} \approx 15.91 \text{ ohms} \]

Key Concepts

electric power

Electric power is a measure of the rate at which energy is transferred or converted. It is closely related to both energy and time, as it tells us how much energy is being used in a certain amount of time.

Power is often measured in watts (W), where 1 watt is equal to 1 joule of energy per second. This means if a device is using 5 watts of power, it is converting 5 joules of energy every second.

In this exercise, we calculated the electric power of the resistor using the formula:

• \[ P = \frac{E}{t} \]

where \(E\) represents the energy in joules and \(t\) represents the time in seconds. By knowing the total energy delivered to the resistor and the total time, we can find the average power consumption.

With the values given, we converted hours into seconds to proceed with calculations correctly and determined the power to be approximately 5.09 watts.

resistance calculation

Resistance is a measure of how much a component, like a resistor, opposes the flow of electric current. It is measured in ohms (Ω). Understanding resistance is crucial for analyzing electrical circuits, as it helps in controlling how much current flows in different parts of a circuit.

To calculate the resistance of the resistor, we utilized Ohm's Law related formulas and rearranged them to be useful for resistance calculation:

• \[ P = \frac{V^2}{R} \]

This formula relates power \(P\), voltage \(V\), and resistance \(R\). We rearranged it to find:\

• \[ R = \frac{V^2}{P} \]

By substituting the known values of voltage (9 V) and power (5.09 W), the resistance calculated was approximately 15.91 ohms.

These equations show that higher resistance results from either higher voltage or lower power for a constant setup, reflecting the opposition to current in the circuit.

energy delivered by battery

A battery stores energy in the form of chemical energy and provides electrical energy to a circuit when connected. The energy delivered by a battery can be calculated if the power and time are known, demonstrating how energy supply happens in practical terms.

In the given exercise, the battery delivered \(1.1 \times 10^{5}\) joules of energy to the resistor over a period of 6 hours. This energy allows the resistor to do its work, such as producing heat in this case.

The formula linking these aspects is:

• \[ E = P \times t \]

where \(E\) is energy in joules, \(P\) is power in watts, and \(t\) is time in seconds.

This exercise showed us how important it is to correctly calculate energy requirements and conversions for a circuit to function properly. Understanding these parts helps in designing and managing electrical systems effectively in real-world applications.