/*! 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 33 A person in a rural area who has... [FREE SOLUTION] | 91Ó°ÊÓ

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A person in a rural area who has no electricity runs an extremely long extension cord to a friend's house down the road so she can run an electric light. The cord is so long that its resistance, \(x\), is not negligible. Show that the lamp's brightness is greatest if its resistance, \(y\), is equal to \(x\). Explain physically why the lamp is dim for values of \(y\) that are too small or too large.

Short Answer

Expert verified
The lamp brightness is greatest when the lamp's resistance \( y \) equals the extension cord's resistance \( x \).

Step by step solution

01

Understanding the Problem

We are given that the extension cord connecting to the lamp has a resistance of \( x \), and the lamp's resistance is \( y \). We need to find when the lamp's brightness is maximized, suggesting an optimal relationship between \( x \) and \( y \).
02

Applying Ohm's Law

Ohm's Law states that the current through a conductor between two points is directly proportional to the voltage across the two points: \( V = IR \), where \( V \) is the voltage, \( I \) is the current, and \( R \) is the total resistance. In this case, the total resistance in the circuit is \( x + y \). Therefore, the current \( I \) can be given by \( I = \frac{V}{x + y} \).
03

Power Expression for the Lamp

The power consumption by the lamp, which relates to its brightness, is given by \( P = I^2 y \). Inserting the current expression from Ohm's Law, we get \( P = \left(\frac{V}{x + y}\right)^2 y \).
04

Optimizing the Power

To find the maximum brightness, we need to differentiate \( P = \frac{V^2 y}{(x + y)^2} \) with respect to \( y \) and set the derivative to zero to find the critical points: \( \frac{dP}{dy} = \frac{V^2((x + y)^2 - 2y(x + y))}{(x + y)^4} = 0 \). Solving gives \( 2y(x + y) = (x + y)^2 \). Simplifying results in \( y = x \).
05

Analyzing Cases of Small and Large \( y \)

If \( y \) is too small, the power \( P = \frac{V^2 y}{(x+y)^2} \) approaches zero because the numerator \( y \) is very small. If \( y \) is too large, the total resistance \( x + y \) becomes large, reducing the current \( I = \frac{V}{x + y} \), which in turn reduces the power \( P \).
06

Conclusion on Brightness Conditions

Thus, for maximum brightness, the resistances of the extension cord and the lamp should match (\( y = x \)). This allows for efficient power transfer and maximizes the brightness without unnecessary losses.

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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 one of the fundamental concepts in electrical circuits. It describes the relationship between voltage, current, and resistance. The law is simply stated as: \( V = IR \), where:
  • \( V \) is the voltage across the circuit
  • \( I \) is the current flowing through the circuit
  • \( R \) is the total resistance in the circuit
This means that the voltage in a circuit is directly proportional to the current, provided that the resistance stays constant. If you increase the resistance while keeping the voltage the same, the current decreases, and vice versa. In the context of the exercise, Ohm's Law helps us understand how the resistance of the lamp, \( y \), and the resistance of the extension cord, \( x \), together affect the total current flowing to the lamp. The formula for current in this circuit becomes \( I = \frac{V}{x + y} \). The balance between resistances \( x \) and \( y \) plays a crucial role in how much current can flow, directly impacting the lamp's brightness.
Power Consumption
Power consumption in an electrical circuit is directly linked to how much energy is being used by the components of the circuit. For electrical devices, power, represented as \( P \), is calculated as: \( P = I^2 R \). For the lamp in our example, the power also represents its brightness. Indeed, the more power the lamp consumes, the brighter it will be. The formula for power in this scenario using the previously calculated current is \( P = \frac{V^2 y}{(x + y)^2} \). This formula shows us that:
  • The power consumed by the lamp is influenced by both its resistance, \( y \), and the total resistance in the circuit \( x + y \).
  • If \( y \) (lamp's resistance) is too small, the numerator of the formula—and consequently the power—also becomes small.
  • If \( y \) is too large, the increased total resistance \( x+y \) diminishes the current, thereby reducing power.
The key takeaway is that optimal power consumption, and hence brightness, is achieved when the resistances are perfectly balanced.
Resistance Matching
Resistance matching is a principle used in many areas of electronics and electrical engineering to ensure maximum efficiency. This principle stems from the need to minimize power loss and maximize energy transfer. In the case of the lamp and extension cord circuit, resistance matching occurs when the lamp's resistance, \( y \), is equal to the resistance of the extension cord, \( x \).Why is this important? When \( y = x \), the system reaches optimal conditions for power transfer, maximizing the brightness of the lamp. The mathematical explanation for this lies in the power formula \( P = \frac{V^2 y}{(x + y)^2} \), where simplifying derivatives shows that maximum power is transferred when the resistances are equal.If \( y \) is either too low or too high compared to \( x \), energy is not optimally transferred, leading to dim lighting. Thus, matching the resistance allows the electrical energy from the extension cord to flow effectively into the lamp, ensuring it shines as brightly as possible. This concept not only applies useful strategies for powering devices but also demonstrates the importance of understanding how resistance impacts electrical devices.

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Most popular questions from this chapter

(a) You take an LP record out of its sleeve, and it acquires a static charge of \(1 \mathrm{nC}\). You play it at the normal speed of \(33 \frac{1}{3}\) r.p.m. and the charge moving in a circle creates an electric current. What is the current, in amperes? (b) Although the planetary model of the atom can be made to work with any value for the radius of the electrons' orbits, more advanced models that we will study later in this course predict definite radii. If the electron is imagined as circling around the proton at a speed of \(2.2 \times 10^{6} \mathrm{~m} / \mathrm{s}\), in an orbit with a radius of \(0.05 \mathrm{~nm}\), what electric current is created? The charge of an electron is \(-e=-1.60 \times\) \(10^{-19} \mathrm{C}\)

Suppose six identical resistors, each with resistance \(R\), are connected so that they form the edges of a tetrahedron (a pyramid with three sides in addition to the base, i.e., one less side than an Egyptian pyramid). What resistance value or values can be obtained by making connections onto any two points on this arrangement?

In a wire carrying a current of \(1.0 \mathrm{pA}\), how long do you have to wait, on the average, for the next electron to pass a given point? Express your answer in units of microseconds. The charge of an electron is \(-e=-1.60 \times 10^{-19} \mathrm{C}\).

(a) Many battery-operated devices take more than one battery. If you look closely in the battery compartment, you will see that the batteries are wired in series. Consider a flashlight circuit. What does the loop rule tell you about the effect of putting several batteries in series in this way? (b) The cells of an electric eel's nervous system are not that different from ours - each cell can develop a voltage difference across it of somewhere on the order of one volt. How, then, do you think an electric eel can create voltages of thousands of volts between different parts of its body?

(a) Express the power dissipated by a resistor in terms of \(R\) and \(\Delta V\) only, eliminating \(I\). (b) Electrical receptacles in your home are mostly \(110 \mathrm{~V}\), but circuits for electric stoves, air conditioners, and washers and driers are usually \(220 \mathrm{~V}\). The two types of circuits have differently shaped receptacles. Suppose you rewire the plug of a drier so that it can be plugged in to a \(110 \mathrm{~V}\) receptacle. The resistor that forms the heating element of the drier would normally draw \(200 \mathrm{~W}\). How much power does it actually draw now?

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