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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's brightness is maximized when its resistance equals the extension cord's resistance, i.e., \(y = x\).

Step by step solution

01

Setting up the Circuit

Consider the circuit where the extension cord with resistance \(x\) and the lamp with resistance \(y\) are in series. According to Ohm's law, the total resistance of the circuit is \(R = x + y\). The voltage across the whole circuit is \(V\).
02

Applying Ohm's Law

Using Ohm's Law, the current \(I\) through the circuit can be expressed as \(I = \frac{V}{R} = \frac{V}{x + y}\). This current flows through the lamp, determining its brightness.
03

Calculating Power Dissipated by the Lamp

The power \(P\) dissipated by the lamp, which correlates to its brightness, is given by \(P = I^2 y\). Substituting \(I\) from Step 2, we get \(P = \left( \frac{V}{x + y} \right)^2 y\).
04

Maximizing the Power

To find the maximum power, take the derivative of \(P\) with respect to \(y\), set it to zero, and solve for \(y\). Let’s find \( \frac{dP}{dy} \).
05

Using Calculus to Find Maximum Power

Differentiate \(P = \left( \frac{V}{x + y} \right)^2 y\) with respect to \(y\) to get \( \frac{dP}{dy} = \left( \frac{V}{x + y} \right)^2 - \frac{2Vy}{(x + y)^3} \). Set \( \frac{dP}{dy} = 0\).
06

Solving the Derivative Condition

From \( \left( \frac{V}{x + y} \right)^2 = \frac{2Vy}{(x + y)^3} \), cross-multiply and simplify to find \(x = y\). Therefore, the lamp’s brightness is maximized when its resistance \(y\) equals the resistance \(x\) of the cord.
07

Physical Explanation for Dim Light

When \(y\) is too small, the lamp's power \(P = I^2 y\) is low because \(y\) cannot dissipate much power. When \(y\) is too large, most of the voltage drop occurs across the lamp, reducing the current \(I\), thus decreasing brightness. The lamp is brightest when the power dissipation is optimal, i.e., \(x = y\).

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Key Concepts

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

Resistance
Resistance is a measure of how much a material opposes the flow of electric current. It is calculated using the formula \( R = \frac{V}{I} \), where \( R \) is resistance, \( V \) is voltage, and \( I \) is current. The unit of resistance is ohms (\( \Omega \)). The higher the resistance, the more difficult it is for current to flow through the wire.

In our context, the extension cord's resistance, \( x \), and the lamp's resistance, \( y \), affect how much electrical power reaches the lamp. With higher resistance, less current passes through, dimming the light. Conversely, lower resistance increases current flow. However, too low a resistance can lead to wastage and inefficiencies.
Electric Circuit
An electric circuit is a closed pathway that allows electric current to flow. It consists of various components like resistors (where resistance plays a crucial role), power sources, and conductive paths (such as wires).

In our scenario, the circuit includes the resistive extension cord and the lamp connected in a series. This setup ensures that the same current flows through both components, influencing the brightness of the lamp as a function of the total resistance \( R = x + y \). Proper circuit design is essential for efficient energy use and performance.
Power Dissipation
Power dissipation refers to the process in which an electric component taps energy from an electric current and converts it into heat or light, like a lamp lighting up. It is calculated using the formula \( P = I^2 R \), where \( P \) is power, \( I \) is current, and \( R \) is resistance. Power is measured in watts.

For the lamp and extension cord set up, the power dissipated by the lamp, \( P = \left( \frac{V}{x + y} \right)^2 y \), is what determines its brightness. For the lamp to be at its brightest, its resistance \( y \) must be equal to \( x \), minimizing power loss across the circuit.
Current
Current is the flow of electric charge through a conductor. It's determined by the potential difference and the resistance in the circuit as per Ohm's Law: \( I = \frac{V}{R} \), where \( I \) is current, \( V \) is voltage, and \( R \) is total resistance in the circuit. The unit of current is amperes (A).

With our circuit, the amount of current flowing through will dictate how much energy is available to light the lamp. The current is limited by the combined resistance of both the extension cord and the lamp, \( x + y \). Adjusting the resistances within the circuit can thus optimize the current for the desired level of brightness.
Series Circuit
In a series circuit, all components are connected end-to-end, forming a single path for current flow. This contrasts with parallel circuits, where components are connected across multiple paths.

For the exercise, the lamp and extension cord form a series, meaning the same current flows through both. This makes the total resistance \( R = x + y \) a key factor since any change in either the cord's or lamp's resistance directly affects the whole circuit. The design of series circuits is critical, as the failure of one component can affect the entire circuit, and resistances accumulate, changing the overall behavior of the current and voltage in the system.

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Most popular questions from this chapter

What resistance values can be created by combining a \(1 \mathrm{k} \Omega\) resistor and a \(10 \mathrm{k} \Omega\) resistor? (solution in the pdf version of the book)

In AM (amplitude-modulated) radio, an audio signal \(f(t)\) is multiplied by a sine wave \(\sin \omega t\) in the megahertz frequency range. For simplicity, let's imagine that the transmitting antenna is a whip, and that charge goes back and forth between the top and bottom. Suppose that, during a certain time interval, the audio signal varies linearly with time, giving a charge \(q=(a+b t) \sin \omega t\) at the top of the whip and \(-q\) at the bottom. Find the current as a function of time. (answer check available at lightandmatter.com)

We have referred to resistors dissipating heat, i.e., we have assumed that \(P=I \Delta V\) is always greater than zero. Could \(I \Delta V\) come out to be negative for a resistor? If so, could one make a refrigerator by hooking up a resistor in such a way that it absorbed heat instead of dissipating it?

How many different resistance values can be created by combining three unequal resistors? (Don't count possibilities in which not all the resistors are used, i.e., ones in which there is zero current in one or more of them.)

A resistor has a voltage difference \(\Delta V\) across it, causing a current \(I\) to flow. (a) Find an equation for the power it dissipates as heat in terms of the variables \(I\) and \(R\) only, eliminating \(\Delta V\). (answer check available at lightandmatter.com) (b) If an electrical line coming to your house is to carry a given amount of current, interpret your equation from part a to explain whether the wire's resistance should be small, or large.

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