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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.)

Short Answer

Expert verified
There are 6 different resistance values possible using all three resistors.

Step by step solution

01

Understanding the Problem

We have three unequal resistors, labeled as \(R_1\), \(R_2\), and \(R_3\). We need to calculate the number of different resistance values that can be created by using all three resistors. When combining resistors, they can be arranged in series or parallel to achieve different resistance values.
02

Calculate Resistance in Series

The total resistance when the resistors \(R_1\), \(R_2\), and \(R_3\) are in series is the sum of the individual resistances. The formula for resistors in series is:\[ R_{series} = R_1 + R_2 + R_3 \]
03

Calculate Resistance in Parallel

For resistors in parallel, the total resistance is given by the reciprocal of the sum of the reciprocals of each resistor's resistance. The formula is:\[ \frac{1}{R_{parallel}} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} \]The equivalent resistance is then:\[ R_{parallel} = \frac{1}{\left( \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} \right)} \]
04

Series-Parallel Combinations

We can explore mixed series-parallel combinations. For example:1. Two resistors in parallel and then in series with the third. - \[ R_1 \text{ and } R_2 \text{ parallel, then series with } R_3 \] gives \[ R = \left( \frac{1}{R_1} + \frac{1}{R_2} \right)^{-1} + R_3 \] - \[ R_2 \text{ and } R_3 \text{ parallel, then series with } R_1 \] gives \[ R = \left( \frac{1}{R_2} + \frac{1}{R_3} \right)^{-1} + R_1 \] - \[ R_1 \text{ and } R_3 \text{ parallel, then series with } R_2 \] gives \[ R = \left( \frac{1}{R_1} + \frac{1}{R_3} \right)^{-1} + R_2 \]
05

Three Combinations Total

Calculating all possible configurations, we obtain the following different combinations:1. All three in series: \[ R_{series} \]2. All three in parallel: \[ R_{parallel} \]3. Two resistors in parallel and then in series with the third, in three different configurations.
06

Count the Unique Combinations

Thus, the total number of different, unique resistance values from combining three unequal resistors in all possible ways (series, parallel, and mix of series-parallel) is 6. Each configuration results in a unique resistance value given unequal resistors.

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Key Concepts

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

Understanding Series Resistance
When resistors are connected in series, their total resistance is simply the sum of the individual resistances. This is because the current that flows through these resistors has only one path to take, which means it moves through each resistor in turn.
This arrangement increases the overall resistance because the current experiences resistance at each step, like hurdles on a running track.

The formula to calculate the total resistance in a series is:
  • \( R_{series} = R_1 + R_2 + R_3 \)
Here, each resistor adds its value to the total. The equivalent resistance is higher than any individual resistor.
Think of series circuits like a convoy of cars on a single-lane road; if the road is longer, it takes longer for all cars to pass any given point.
Understanding Parallel Resistance
In a parallel circuit, resistors are connected across the same two points, providing multiple paths for the current.
Each path offers a separate route for the flow of electricity. Unlike series circuits, adding more paths actually decreases the total resistance.
This is because more paths allow more current to flow, reducing the overall resistance, akin to adding extra lanes to a highway to reduce traffic.

The calculation for total resistance in a parallel circuit looks a bit more complex, but follows this formula:
  • \[ \frac{1}{R_{parallel}} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} \]
After determining the sum of the reciprocals, invert that sum to find the total parallel resistance:
  • \[ R_{parallel} = \frac{1}{\left( \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} \right)} \]
This method ensures that the total resistance in a parallel arrangement is always lower than the smallest individual resistor. Think of parallel circuits like multiple tunnels under a mountain; each tunnel reduces congestion and makes the total travel time shorter by providing more options.
Exploring Mixed Series-Parallel Circuits
Mixed series-parallel circuits combine both series and parallel components, leading to a variety of possible configurations and resistance values. This flexibility allows for more control over the circuit’s resistance.
By understanding the behavior of series and parallel resistors, you can apply this knowledge to mix them creatively.

Here are a few ways resistors can be combined using mixed configurations:
  • Two resistors in parallel followed by one in series: Calculate the parallel resistance first for two resistors, then add the third in series.
  • Two resistors in series followed by one in parallel: Add the series resistances first, then calculate the parallel resistance with the remaining resistor.
To calculate resistance for these types of combinations, use the applicable formulas for series and parallel connections.
For example:
  • For two resistors \( R_1 \) and \( R_2 \) in parallel, then in series with \( R_3 \):
  • \[ R = \left( \frac{1}{R_1} + \frac{1}{R_2} \right)^{-1} + R_3 \]
Each mix provides unique resistance values, enhancing the functionality of the circuit.
Much like a water park with a mix of slides and lazy rivers, mixed circuits offer diverse options for routing the current, resulting in a variety of total resistances.

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Most popular questions from this chapter

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.

A \(1.0 \Omega\) toaster and a \(2.0 \Omega\) lamp are connected in parallel with the 110-V supply of your house. (Ignore the fact that the voltage is \(\mathrm{AC}\) rather than \(\mathrm{DC.}\).) (a) Draw a schematic of the circuit. (b) For each of the three components in the circuit, find the current passing through it and the voltage drop across it. (answer check available at lightandmatter.com) (c) Suppose they were instead hooked up in series. Draw a schematic and calculate the same things.(answer check available at lightandmatter.com)

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.

Wire is sold in a series of standard diameters, called "gauges." The difference in diameter between one gauge and the next in the series is about \(20 \%\). How would the resistance of a given length of wire compare with the resistance of the same length of wire in the next gauge in the series? (answer check available at lightandmatter.com)

A silk thread is uniformly charged by rubbing it with llama fur. The thread is then dangled vertically above a metal plate and released. As each part of the thread makes contact with the conducting plate, its charge is deposited onto the plate. Since the thread is accelerating due to gravity, the rate of charge deposition increases with time, and by time \(t\) the cumulative amount of charge is \(q=c t^{2}\), where \(c\) is a constant. (a) Find the current flowing onto the plate.(answer check available at lightandmatter.com) (b) Suppose that the charge is immediately carried away through a resistance \(R .\) Find the power dissipated as heat.(answer check available at lightandmatter.com)

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