/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 78 When a "dry-cell" flashlight bat... [FREE SOLUTION] | 91影视

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When a "dry-cell" flashlight battery with an internal resistance of 0.33\(\Omega\) is connected to a \(1.50-\Omega\) light bulb, the bulb shines dimly. However, when a lead-acid "wet-cell" battery with an internal resistance of 0.050\(\Omega\) is connected, the bulb is noticeably brighter. Both batteries have the same emf. Find the ratio \(P_{\text { wet }} / P_{\text { dry }}\) of the power delivered to the bulb by the wet-cell battery to the power delivered by the dry-cell battery.

Short Answer

Expert verified
The ratio of powers is approximately 1.393.

Step by step solution

01

Calculate Total Resistance with Dry-cell Battery

Calculate the total resistance when the dry-cell battery is used. The internal resistance is given as \(r_{\text{dry}} = 0.33 \Omega\) and resistance of the bulb is \(R = 1.50 \Omega\). The total resistance \(R_{\text{total, dry}}\) is the sum of these two resistances:\[R_{\text{total, dry}} = R + r_{\text{dry}} = 1.50 \Omega + 0.33 \Omega = 1.83 \Omega\]
02

Calculate Power Delivered with Dry-cell Battery

Use the formula for power delivered to the bulb, \(P = \frac{V^2}{R_{\text{total}}} \cdot R\), where \(V\) is the emf of the battery. Assuming emf to be \(E\), the power delivered to the bulb by the dry-cell battery is:\[P_{\text{dry}} = \frac{E^2}{R_{\text{total, dry}}^2} \cdot R = \frac{E^2}{1.83^2} \cdot 1.5\]
03

Calculate Total Resistance with Wet-cell Battery

Calculate the total resistance when the wet-cell battery is used. The internal resistance is \(r_{\text{wet}} = 0.050 \Omega\). Hence, the total resistance \(R_{\text{total, wet}}\) is:\[R_{\text{total, wet}} = R + r_{\text{wet}} = 1.50 \Omega + 0.050 \Omega = 1.55 \Omega\]
04

Calculate Power Delivered with Wet-cell Battery

Similarly, calculate the power delivered to the bulb by the wet-cell battery:\[P_{\text{wet}} = \frac{E^2}{R_{\text{total, wet}}^2} \cdot R = \frac{E^2}{1.55^2} \cdot 1.5\]
05

Calculate the Ratio of Powers

To find the ratio of the power delivered to the bulb by the wet-cell to that by the dry-cell battery, divide \(P_{\text{wet}}\) by \(P_{\text{dry}}\):\[\frac{P_{\text{wet}}}{P_{\text{dry}}} = \left(\frac{1.83}{1.55}\right)^2\]Simplifying, we get:\[\frac{P_{\text{wet}}}{P_{\text{dry}}} = \left(\frac{1.83}{1.55} \right)^2 = 1.393\]

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Key Concepts

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

Electrical Resistance
Electrical resistance is a fundamental concept that represents the opposition to the flow of electric current in a conductor. It is measured in ohms (\( \Omega \)) and varies depending on the material and dimensions of the conductor. The opposition occurs due to collisions between the free electrons and the atoms within the conductor.
A higher resistance means that the conductor allows less current to flow, leading to energy being dissipated as heat. In our example, the flashlight battery with an internal resistance of 0.33\( \Omega \) demonstrates this opposition more than the wet-cell battery with only 0.050\( \Omega \). This difference in internal resistance explains why the light bulb is dimmer with the dry-cell battery while it's brighter with the wet-cell battery.
Remember, electrical resistance not only depends on the element through which the electricity flows, but it also includes any internal elements of devices such as batteries or bulbs connected in the circuit. This internal resistance becomes crucial when calculating the total resistance in a circuit.
Power Calculation
Calculating power is essential in understanding how much energy is being used or delivered by electrical devices. Power, measured in watts (W), is described by the equation \(P = \frac{V^2}{R_{\text{total}}} \cdot R\), where \(V\) is the electromotive force (emf) or potential difference, and \(R\) represents the resistance across which power is being calculated.
Power calculation helps determine the efficiency of a circuit in delivering energy to a device. For instance, when calculating the power delivered by a battery to a light bulb, it's essential to consider the total resistance, which includes both the internal resistance of the battery and the resistance of the bulb.
The exercise highlights how different internal resistances of batteries affect power delivery to the connected bulb, consequently affecting its brightness. A smaller internal resistance leads to lower total resistance, resulting in higher power delivered to the bulb. This principle is central to understanding why a wet-cell battery provides more power to the light bulb than a dry-cell battery.
Internal Resistance
Internal resistance is the resistance within the battery itself. It is a property that impedes the flow of charge and dissipates energy as heat inside the battery. The lower the internal resistance, the more current the battery can deliver to external circuits.
This concept is vital in determining the efficiency and performance of batteries. In the given problem, the internal resistance is a significant factor in comparing the performance of the dry-cell and wet-cell batteries.
The dry-cell battery has a higher internal resistance of 0.33\( \Omega \) compared to the 0.050\( \Omega \) of the wet-cell battery. This difference dramatically impacts how much power is delivered to the bulb. By reducing internal resistance, more energy is available for external use, leading to brighter bulb illumination when using the wet-cell battery.
Ohm's Law
Ohm's Law is a foundational principle in electrical engineering that relates voltage (V), current (I), and resistance (R) in an electrical circuit. It states that \(V = I \cdot R\), meaning the voltage across a resistor is directly proportional to the current flowing through it and its resistance.
This law is crucial for calculating various parameters within a circuit, such as current, voltage, and resistance. In this context, while the primary focus is on power calculations, Ohm's law is indirectly applied to understand relationships in the circuit's electrical characteristics.
For instance, as the total resistance changes due to different internal resistances from batteries, the current flowing through the circuit is affected. This change alters how much power is delivered to the light bulb, as seen in the exercise. Thus, Ohm's Law is implicitly used in calculating how different resistances impact the overall functionality of electrical devices.

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Most popular questions from this chapter

A defibrillator is used during a heart attack to restore the heart to its normal beating pattern (see Section 19.5). A defibrillator passes 18 A of current through the torso of a person in 2.0 ms. (a) How much charge moves during this time? (b) How many electrons pass through the wires connected to the patient?

The total current delivered to a number of devices connected in parallel is the sum of the individual currents in each device. Circuit breakers are resettable automatic switches that protect against a dangerously large total current by 鈥渙pening鈥 to stop the current at a specified safe value. A 1650-W toaster, a 1090-W iron, and a 1250-W microwave oven are turned on in a kitchen. As the drawing shows, they are all connected through a 20-A circuit breaker (which has negligible resistance) to an ac voltage of 120 V. (a) Find the equivalent resistance of the three devices. (b) Obtain the total current delivered by the source and determine whether the breaker will 鈥渙pen鈥 to prevent an accident.

The coil of wire in a galvanometer has a resistance of \(R_{C}=60.0 \Omega\) . The galvanometer exhibits a full-scale deflection when the current through it is 0.400 \(\mathrm{mA}\) . A resistor is connected in series with this combination so as to produce a nondigital voltmeter. The voltmeter is to have a full-scale deflection when it measures a potential difference of 10.0 \(\mathrm{V}\) . What is the resistance of this resistor?

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For the three-way bulb (50 W, 100 W, 150 W) discussed in Conceptual Example 11, find the resistance of each of the two filaments. Assume that the wattage ratings are not limited by significant figures, and ignore any heating effects on the resistances.

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