91Ó°ÊÓ

Flashlight batteries. A typical small flashlight contains two batteries, each having an emf of 1.5 \(\mathrm{V}\) , connected in series with a bulb having resistance 17\(\Omega .\) (a) If the internal resistance of the batteries is negligible, what power is delivered to the bulb? (b) If the batteries last for 5.0 h, what is the total energy delivered to the bulb? (c) The resistance of real batteries increases as they run down. If the initial internal resistance is negligible, what is the combined internal resistance of both batteries when the power to the bulb has decreased to half its initial value? (Assume that the resistance of the bulb is constant. Actually, it will change somewhat when the current through the filament changes, because this changes the temperature of the filament and hence the resistivity of the filament wire.)

Step by step solution

01

Calculate Total Voltage

The flashlight has two batteries, each with an emf of 1.5 V. Since they are in series, the total voltage provided is the sum of the individual voltages.\[ V = 1.5 \, \text{V} + 1.5 \, \text{V} = 3 \, \text{V} \]

02

Determine Power Delivered to Bulb

Using Ohm's Law, and knowing that the internal resistance of the batteries is negligible, the power delivered to the bulb can be calculated using the formula for power:\[ P = \frac{V^2}{R} \]Substituting the known values:\[ P = \frac{(3 \, \text{V})^2}{17 \, \Omega} = \frac{9 \, \text{V}^2}{17 \, \Omega} \approx 0.529 \, \text{W} \]

03

Calculate Total Energy Delivered

The energy delivered to the bulb can be calculated using the power delivered and the time the batteries last. This is given by the formula:\[ E = P \times t \]Converting 5 hours to seconds (5 hours * 3600 seconds/hour):\[ t = 5 \, \text{h} \times 3600 \, \text{s/h} = 18000 \, \text{s} \]Thus, the total energy delivered is:\[ E = 0.529 \, \text{W} \times 18000 \, \text{s} = 9518 \, \text{J} \]

04

Calculate Combined Internal Resistance for Half Power

For the bulb's power to be half of its initial value, we use the modified power formula considering internal resistance:\[ P_{new} = \frac{V^2}{R + R_{int}} \]Given \( P_{new} = 0.529 \, \text{W} / 2 \), substitute and solve for \(R_{int}\):\[ 0.529/2 = \frac{9}{17 + R_{int}} \]\[ R_{int} = \frac{9}{0.2645} - 17 \approx 17 \, \Omega \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Ohm's Law

Ohm's Law is a fundamental concept in physics and electrical engineering that explains the relationship between voltage, current, and resistance in an electrical circuit. The law states that the voltage (V) across a conductor is directly proportional to the current (I) flowing through it, with the proportionality constant being the resistance (R) of the conductor. Mathematically, this relationship is expressed as:

• \( V = I \times R \)

This means that if you know any two of these variables, you can calculate the third. In simple terms, Ohm's Law tells us that to increase the current through a component, we can either increase the voltage applied or decrease the resistance of the component. This law becomes especially useful when analyzing series circuits, where many elements, such as flashlight bulbs and batteries, are connected together.

Series Circuit

A series circuit is a type of electrical circuit in which components are connected one after the other in a single path. In such circuits, the current remains constant across all components, but the voltage drops across each one adds up to the total supplied voltage. This arrangement is important when analyzing the performance of devices powered by multiple sources, like batteries in a flashlight.

• The total resistance in a series circuit is the sum of the resistances of each component. If one component fails, the circuit is broken, and all devices stop working.
• In a series circuit with two batteries, like the flashlight example, the total voltage is the sum of the emfs of the individual batteries. Therefore, each battery contributes its emf to the overall voltage.
• This setup can efficiently increase the overall voltage available to a device, which is perfect for small applications needing just a bit more power.

Electric Power

Electric power measures how quickly energy is transferred in an electrical circuit. It is a key concept needed to understand how much work or energy is provided over time by electrical components such as batteries and bulbs. The unit of power is the watt (W).In electrical terms, power is calculated using the formula:

• \( P = \, V \times I \)
• It can also be expressed in terms of resistance: \( P = \frac{V^2}{R} \)

For example, in the flashlight scenario, the power delivered to the bulb can be found by using the second formula because the voltage and resistance are known.Understanding electric power helps in determining how much total energy is consumed over time, which is crucial for knowing the efficiency of electronic devices. The formula \( E = P \times t \) helps time these calculations, allowing us to understand energy transfer in terms of joules (J), given specific periods.

Internal Resistance

Internal resistance is an inherent resistance within batteries that affects their performance. It is a critical factor to consider when evaluating real-world battery behavior because as a battery depletes, its internal resistance tends to increase.This increase in resistance can affect the power delivered to connected loads, such as a bulb in a series circuit. Over time, as internal resistance rises, it leads to a decrease in available current and thus less power output from the battery.

• The formula \( P = \frac{V^2}{R + R_{int}} \) combines the internal resistance with load resistance to show how it affects power delivery.
• When the power is halved from its initial value, internal resistance has doubled the effective resistance witnessed across the circuit, significantly impacting deliverable power.

Understanding internal resistance helps in designing circuits and choosing appropriate batteries for specific applications where consistent power delivery is essential as the battery discharges. This ensures that devices like flashlights function adequately throughout the expected usage duration.