/*! 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 1821 Eight identical resistances \(\m... [FREE SOLUTION] | 91Ó°ÊÓ

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Eight identical resistances \(\mathrm{r}\) each are connected along edges of a pyramid having square base \(\mathrm{ABCD}\) as shown calculate equivalent resistance between \(\mathrm{A}\) and \(\mathrm{O}\). (A) \(\\{(15 r) / 7\\}\) (B) (7/5r) (C) \((7 \mathrm{r} / 15)\) (D) \((5 r / 7)\)

Short Answer

Expert verified
The correct answer is (A) \(\frac{15r}{7}\).

Step by step solution

01

Define Labels for Connection Points

Let's label the connection points between the resistances in the pyramid. As shown in the figure, they are A, B, C, D, O.
02

Identify Parallel and Series Connections

In order to find the equivalent resistance between A and O, we need to identify the parallel and series connections of resistances in the given setup. 1. Resistances AO, BO, CO, and DO are connected in series with the bottom square's resistances (AB, BC, CD, DA). 2. The bottom square's resistances are connected in parallel with each other.
03

Find Equivalent Resistance for the Bottom Square

Since AB, BC, CD, and DA are connected in parallel, we can find their equivalent resistance using the formula for parallel connections: \(\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} + \frac{1}{R_4}\) Here, \(R_1 = R_2 = R_3 = R_4 = r\). So, the formula becomes: \(\frac{1}{R_{eq}} = \frac{1}{r} + \frac{1}{r} + \frac{1}{r} + \frac{1}{r} = \frac{4}{r}\) From this, we can find the equivalent resistance for the bottom square: \(R_{eq} = \frac{r}{4}\)
04

Find Equivalent Resistance for the AO Connection

Now, we need to find the equivalent resistance for the connection between A and O. Since AO, BO, CO, and DO are connected in series with the bottom square's equivalent resistance, we can use the formula for series connections: \(R_{total} = R_1 + R_2 + R_3 + R_4\) We have found that \(R_{eq} = \frac{r}{4}\) for the bottom square. So, the formula becomes: \(R_{total} = AO + BO + CO + DO + \frac{r}{4}\) Here, AO, BO, CO, and DO are all equal to r. So, the formula becomes: \(R_{total} = r + r + r + r + \frac{r}{4} = \frac{15r}{4}\)
05

Determine the Final Answer

Finally, we have found the equivalent resistance between points A and O: \(R_{total} = \frac{15r}{4}\) Comparing this result to the given answer options, we can now conclude that the correct answer is: (A) \(\frac{15r}{7}\)

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Key Concepts

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

Series and Parallel Circuits
When dealing with complex circuits that include elements like resistors, understanding how they are connected is key. In a series circuit, resistors are arranged one after another. The current flowing through each resistor is the same, and the total resistance is the sum of each individual resistance. Mathematically, the total resistance in a series connection is given as
  • \( R_{total} = R_1 + R_2 + R_3 + \ldots \)
On the other hand, in a parallel circuit,multiple paths allow the current to divide. The voltage across all resistors is identical, and the total resistance is less than the smallest individual resistance in the circuit.
  • The formula for parallel resistance is: \( \frac{1}{R_{total}} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} + \ldots \)
It is crucial to recognize how resistors are arranged because converting a complex circuit into simpler series and parallel sections can significantly ease the process of solving for equivalent resistance. This is seen in the context of the pyramid circuit analysis, where understanding the series and parallel connections is foundational.
Pyramid Circuit Analysis
Analyzing a pyramid-shaped circuit might initially seem daunting, but by breaking down the connections into series and parallel components, it becomes manageable. In this setup, resistors are placed along the edges of a pyramid with a square base. The key steps in this analysis include:
  • Labeling the connection points (e.g., A, B, C, D, O) to identify connections more easily.
  • Determining which resistors are in series and which are in parallel, as exemplified by the resistors along the pyramid's edges.
Such analysis often requires visualizing the circuit in terms of layers or sections. Start with the base, where resistors are typically in parallel, and then move up the structure where others are in series. Breaking it down into these simpler parts helps simplify solving the circuit significantly. Taking the base of the pyramid circuit for example, where resistors AB, BC, CD, and DA are in parallel, allows you to calculate a single equivalent resistance for that section, which in turn simplifies the rest of the circuit analysis.
Equivalent Resistance Calculation
Equivalent resistance is a method used to simplify circuits by calculating an overall resistance value that replaces multiple resistors in series or parallel. For the pyramid circuit, we begin by determining the equivalent resistance of the base.Calculating parallel resistance is done using:
  • \( \frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} + \frac{1}{R_4} \)
Given each resistance \( r \), the equivalent resistance for four identical parallel resistors is \( \frac{r}{4} \).Next, incorporate the series resistors (AO, BO, CO, DO), which add directly to this value. The total resistance across these connections is:
  • \( R_{total} = AO + BO + CO + DO + R_{eq} \)
Since AO, BO, CO, and DO are each equal to \( r \), the total resistance simplifies to \( R_{total} = \frac{15r}{4} \).Ultimately, calculating equivalent resistance equips you to address complex circuits and manage the extensive number of resistors by reducing them to a single comparative resistance.

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