Question

An idealized voltmeter is connected across the terminals of a 15.0-V battery, and a 75.0-\(\Omega\) appliance is also connected across its terminals. If the voltmeter reads 11.9 V, (a) how much power is being dissipated by the appliance, and (b) what is the internal resistance of the battery?

Short answer

Power dissipated is approximately 1.89 W, and internal resistance is around 19.53 Ω.

Step-by-step solution

Analyze the Circuit

Given a 15.0 V battery and a 75.0 Ω appliance, the voltmeter reads 11.9 V across the appliance. This means there's a voltage drop due to internal resistance of the battery.

Determine Voltage Drop due to Internal Resistance

The difference in voltage (15.0 V - 11.9 V) equals the voltage drop due to the internal resistance of the battery, which is 3.1 V.

Calculate Current through the Appliance

Using Ohm's Law: \( V = IR \), where \( V = 11.9 \text{ V} \) and \( R = 75.0 \Omega \), the current \( I \) can be calculated as: \( I = \frac{V}{R} = \frac{11.9}{75.0} \approx 0.1587 \text{ A} \).

Calculate Power Dissipated by the Appliance

Using the formula for power: \( P = IV \), where \( I = 0.1587 \text{ A} \) and \( V = 11.9 \text{ V} \). Therefore, \( P = 0.1587 \times 11.9 \approx 1.886 \text{ W} \).

Calculate Internal Resistance of the Battery

The voltage drop due to internal resistance is 3.1 V and the current is 0.1587 A. Using \( V = IR \), the internal resistance \( r \) is \( r = \frac{3.1}{0.1587} \approx 19.53 \Omega \).

Key concepts

Ohm's Law

Ohm's Law is a fundamental principle in electronics that relates the voltage across a conductor to the current flowing through it and its resistance. The formula is:

• \( V = IR \)

where:

• \( V \) is the voltage (volts),
• \( I \) is the current (amperes), and
• \( R \) is the resistance (ohms).

In practical terms, if you know two of these values, you can easily find the third. For instance, in the exercise, the appliance had a resistance of \( 75.0 \, \Omega \) and the voltage across it was \( 11.9 \, V \). By rearranging Ohm's Law to find the current, \( I = \frac{V}{R} \), we found the current to be approximately \( 0.1587 \, A \). Understanding and applying Ohm's Law is crucial for analyzing electrical circuits and determining how current flows across different components, affecting overall circuit behavior.

power dissipation

Power dissipation in electrical circuits refers to the conversion of electrical energy into thermal energy, which is typically seen as the "loss" of power across a component, such as a resistor or appliance. To calculate power dissipation, we use the formula:

• \( P = IV \)

where:

• \( P \) is the power (watts),
• \( I \) is the current (amperes), and
• \( V \) is the voltage (volts) across the component.

In the provided exercise, once we found the current to be \( 0.1587 \, A \) and knew the voltage across the appliance to be \( 11.9 \, V \), calculating the power dissipation was straightforward: \( P = 0.1587 \, A \times 11.9 \, V \approx 1.886 \, W \). This value represents the rate at which electrical energy is being converted into heat within the appliance, making it an important factor in circuit design and electrical engineering.

voltage drop

Voltage drop is a reduction in voltage as electrical current flows through a part of an electric circuit. This drop is often due to the internal resistance of components like batteries or resistors. In the context of the exercise, the battery's expected voltage was \( 15.0 \, V \), but only \( 11.9 \, V \) was measurable across the appliance, signifying a voltage drop due to the battery's internal resistance. The formula for voltage drop, based on Ohm's Law \( V = IR \), was used to calculate the missing \( 3.1 \, V \) as:

• \( 15.0 \, V - 11.9 \, V = 3.1 \, V \)

This drop occurs because of energy loss in the form of heat inside the battery. Recognizing and calculating voltage drops is crucial for ensuring reliable performance and efficiency in electronic circuits. Understanding where these voltage losses occur can help in diagnosing circuit issues and enhancing design techniques to minimize unwanted effects.