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Chapter 28: Problem 23

By use of one or more of the three resistors \(3.0 \Omega, 5.0 \Omega\), and \(6.0\) \(\Omega\), a total of 18 resistances can be obtained. What are they?

Short Answer

Expert verified
Possible resistances: 1.176, 1.875, 2, 2.727, 3, 5, 5.727, 6, 7, 7.875, 8, 9, 11, and 14 \(\Omega\).

Step by step solution

01

Understand Series and Parallel Combinations

Resistors can be combined in two basic ways: in series and in parallel. In a series circuit, the total resistance is the sum of individual resistances. In a parallel circuit, the total resistance is found using the formula: \( R_{total} = \frac{1}{\frac{1}{R_1} + \frac{1}{R_2} + \cdots}\).
02

Calculate Individual and Series Resistances

List all individual resistances: 3 \(\Omega\), 5 \(\Omega\), 6 \(\Omega\). Calculate total resistances for series combinations:- Series of 3 \(\Omega\) and 5 \(\Omega\): \(3 + 5 = 8\) \(\Omega\)- Series of 3 \(\Omega\) and 6 \(\Omega\): \(3 + 6 = 9\) \(\Omega\)- Series of 5 \(\Omega\) and 6 \(\Omega\): \(5 + 6 = 11\) \(\Omega\)- Series of 3 \(\Omega\), 5 \(\Omega\), and 6 \(\Omega\): \(3 + 5 + 6 = 14\) \(\Omega\)
03

Calculate Resistances for Parallel Combinations

Using the formula for parallel resistances, calculate the following combinations:- Parallel of 3 \(\Omega\) and 5 \(\Omega\): \(\frac{1}{\frac{1}{3} + \frac{1}{5}} = 1.875\) \(\Omega\)- Parallel of 3 \(\Omega\) and 6 \(\Omega\): \(\frac{1}{\frac{1}{3} + \frac{1}{6}} = 2\) \(\Omega\)- Parallel of 5 \(\Omega\) and 6 \(\Omega\): \(\frac{1}{\frac{1}{5} + \frac{1}{6}} = 2.727\) \(\Omega\)- Parallel of all three (3 \(\Omega\), 5 \(\Omega\), 6 \(\Omega\)): \(\frac{1}{\frac{1}{3} + \frac{1}{5} + \frac{1}{6}} = 1.176\) \(\Omega\)
04

Determine Other Combinations

Consider combinations of series and parallel circuits together. For example:- Series of parallel (3 \(\Omega\) and 5 \(\Omega\)) + 6 \(\Omega\): \(1.875 + 6 = 7.875\) \(\Omega\)- Series of 3 \(\Omega\) and parallel (5 \(\Omega\) and 6 \(\Omega\)): \(3 + 2.727 = 5.727\) \(\Omega\)- Series of parallel (3 \(\Omega\) and 6 \(\Omega\)) + 5 \(\Omega\): \(2 + 5 = 7\) \(\Omega\)Explore all these combinations to find other unique resistances.
05

Compile All Unique Resistances

List all unique resistances found: 1.176 \(\Omega\), 1.875 \(\Omega\), 2 \(\Omega\), 2.727 \(\Omega\), 3 \(\Omega\), 5 \(\Omega\), 5.727 \(\Omega\), 6 \(\Omega\), 7 \(\Omega\), 7.875 \(\Omega\), 8 \(\Omega\), 9 \(\Omega\), 11 \(\Omega\), 14 \(\Omega\). More unique values can be found by exploring additional combinations.

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Key Concepts

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

Series Circuits
In a series circuit, resistors are connected end-to-end, forming a single path for electric current to flow through. Here's a simple way to understand it:

- **Current Flow:** The same current passes through each resistor, unaffected by their individual resistances.
- **Total Resistance:** To find the total resistance, simply add the resistances of each resistor together. The formula is:
\[ R_{total} = R_1 + R_2 + R_3 + \ldots \]

For the given resistors in this exercise - 3 惟, 5 惟, and 6 惟 - we can find different series combinations. For example:

- A 3 惟 and 5 惟 combination gives a total of 8 惟.
- Adding another, 3 惟 + 5 惟 + 6 惟, yields 14 惟.

Understanding how series circuits combine their resistances can help you design or decode a circuit's total resistance effectively.
Parallel Circuits
In contrast to series circuits, parallel circuits provide multiple paths for current to flow. Each path, or branch, has its own resistor. Here are the essential points:

- **Voltage Consistency:** The voltage across each resistor in parallel is the same.
- **Total Resistance:** Unlike series circuits, the total resistance decreases when more resistors are added. It's calculated through the reciprocal formula:
\[ \frac{1}{R_{total}} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} + \ldots \]
This means the more resistors you add, the lower the total resistance becomes.

With the resistors 3 惟, 5 惟, and 6 惟, some combinations include:

- 3 惟 and 5 惟 in parallel, giving 1.875 惟.
- All three in parallel result in approximately 1.176 惟.

By manipulating these combinations, we can tweak circuits for specific resistance values, which informs us about flexibility in design options.
Electrical Resistance
Electrical resistance is a measure of how strongly a material opposes the flow of electric current. It's an essential concept in understanding how circuits behave. The fundamentals include:

- **Ohm's Law:** This law relates the voltage (V), current (I), and resistance (R) in a simple formula:
\[ V = IR \]
This illustrates how voltage and current change in relation to resistance.

- **Material Influence:** Different materials have varying levels of resistance. Conductors have low resistance, whereas insulators have high resistance.

- **Temperature Impact:** Resistance can change with temperature. Typically, as temperature increases, resistance also increases for most materials.

Understanding resistance helps in exploring and managing how electrical devices operate. When designing circuits with specific needs, like those combining our given resistors, knowing each component's resistance helps achieve the desired outcome effectively.

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Most popular questions from this chapter

Find the equivalent resistance between points- \(a\) and \(-b\) for the combination shown The \(3.0-\Omega\) and \(2.0-\Omega\) resistors are in series and are equivalent to a 5.0-\Omega resistor. The equivalent \(5.0 \Omega\) is in parallel with the \(6.0 \Omega\), and their equivalent, \(R_{1}\), is $$ \frac{1}{R_{1}}=\frac{1}{5.0 \Omega}+\frac{1}{6.0 \Omega}=0.20+0.167=0.367 \Omega^{-1} \quad \text { or } \quad R_{1}=2.73 \Omega $$ The circuit thus far reduced is shown The \(7.0 \Omega\) and \(2.73 \Omega\) are equivalent to \(9.73 \Omega\). Now the \(5.0 \Omega\), \(12.0 \Omega\), and \(9.73 \Omega\) are in parallel, and their equivalent, \(R_{2}\), is $$ \frac{1}{R_{2}}=\frac{1}{5.0 \Omega}+\frac{1}{12.0 \Omega}+\frac{1}{9.73 \Omega}=0.386 \Omega^{-1} \quad \text { or } \quad R_{2}=2.6 \Omega $$ This \(2.6 \Omega\) is in series with the \(9.0-\Omega\) resistor. Therefore, the equivalent resistance of the combination is \(9.0 \Omega+2.6 \Omega=11.6\) \(\Omega\)

A galvanometer has a resistance of \(400 \Omega\) and deflects full scale for a current of \(0.20 \mathrm{~mA}\) through it. How large a shunt resistor is required to change it to a 3.0-A ammeter? We label the galvanometer \(\mathrm{G}\) and the shunt resistance \(R_{s}\). At full-scale deflection, the currents are as shown: The voltage drop from \(a\) to \(b\) across \(\mathrm{G}\) is the same as that across \(R_{s} .\) Therefore, $$ (2.9998 \mathrm{~A}) R_{\mathrm{s}}=\left(2.0 \times 10^{-4} \mathrm{~A}\right)(400 \Omega) $$ from which \(R_{s}=0.027 \Omega\).

A \(36-\Omega\) galvanometer is shunted by a resistor of \(4.0 \Omega\). What part of the total current will pass through the instrument?

The Wheatstone bridge shown is being used to measure resistance \(X\). At balance, the current through the galvanometer \(\mathrm{G}\) is zero and resistances \(L, M\), and \(N\) are \(3.0 \Omega, 2.0\) \(\Omega\), and \(10 \Omega\), respectively. Find the value of \(X\).

A \(120-\mathrm{V}\) house circuit has the following light bulbs turned on: \(40.0\) \(\mathrm{W}, 60.0 \mathrm{~W}\), and \(175.0 \mathrm{~W}\). Find the equivalent resistance of these lights. House circuits are so constructed that each device is connected in parallel with the others. From \(\mathrm{P}=V I=V^{2} / R\), for the first bulb $$ R_{1}=\frac{V^{2}}{P_{1}}=\frac{(120 \mathrm{~V})^{2}}{40.0 \mathrm{~W}}=360 \Omega $$ Similarly, \(R_{2}=240 \Omega\) and \(R_{3}=192 \Omega\). Because devices in a house circuit are in parallel, $$ \frac{1}{R_{\mathrm{eq}}}=\frac{1}{360 \Omega}+\frac{1}{240 \Omega}+\frac{1}{192 \Omega} \quad \text { or } \quad R_{\mathrm{eq}}=82.3 \Omega $$ As a check, note that the total power drawn from the line is \(40.0\) \(\mathrm{W}+60.0 \mathrm{~W}+75.0 \mathrm{~W}=75.0 \mathrm{~W}\). Then, using \(\mathrm{P}=V^{2} / R\) $$ R_{\mathrm{eq}}=\frac{V^{2}}{\text { total power }}=\frac{(120 \mathrm{~V})^{2}}{175.0 \mathrm{~W}}=82.3 \Omega $$
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