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A regular tetrahedron is a pyramid with a triangular base. Six 10.0 - \(\Omega\) resistors are placed along its six edges, with junctions at its four vertices. A \(12.0-\mathrm{V}\) battery is connected to any two of the vertices. Find (a) the equivalent resistance of the tetrahedron between these vertices and (b) the current in the battery.

Short Answer

Expert verified
The equivalent resistance between the two vertices connected to the battery is 10.0 \(\Omega\), and the current in the battery is 1.2 A.

Step by step solution

01

- Analyze the Circuit

A regular tetrahedron has six edges. With a 10.0-\(\Omega\) resistor on each edge, there are essentially four nodes (the vertices of the tetrahedron) connected by resistors. When the battery is connected to two vertices, it forms two sets of parallel connections among those four resistors, with two resistors remaining in series with each set.
02

- Simplify the Network

Consider the tetrahedron as a combination of two separate parallel circuits (one for each pair of opposite edges) that are in series with each other. Each pair of parallel resistors can be replaced by their equivalent resistance. The formula for two resistors R1 and R2 in parallel is \(1/R_{eq} = 1/R1 + 1/R2\), which simplifies to \(R_{eq} = R1 \cdot R2 / (R1 + R2)\).
03

- Calculate the Equivalent Resistance for Each Parallel Pair

Using the formula for parallel resistance, calculate the equivalent resistance for a pair of 10.0-\(\Omega\) resistors: \(R_{eq} = 10.0 \ \Omega \cdot 10.0 \ \Omega / (10.0 \ \Omega + 10.0 \ \Omega) = 100 \ \Omega / 20 \ \Omega = 5.0 \ \Omega\).
04

- Combine the Equivalent Resistances in Series

Two equivalent resistances of 5.0 \(\Omega\) from the parallel pairs are now in series with each other. Resistors in series simply add together: \(R_{total} = 5.0 \ \Omega + 5.0 \ \Omega = 10.0 \ \Omega\).
05

- Determine Overall Equivalent Resistance

The overall equivalent resistance between the two vertices connected to the battery is 10.0 \(\Omega\).
06

- Calculate the Current in the Battery

Using Ohm’s law \(I = V/R\), calculate the current I that flows from the battery through the network. With V = 12.0 V and R = 10.0 \(\Omega\), the current is \(I = 12.0 \ \mathrm{V} / 10.0 \ \Omega = 1.2 \ \mathrm{A}\).

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Key Concepts

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

Equivalent Resistance
Understanding equivalent resistance is crucial when dealing with complex circuits. It refers to the total resistance a power source sees when current flows through an electrical network. Imagine simplifying a circuit down to a single resistor that has the same effect as the entire network combined. This is particilarly useful as circuits get more intricate, like with the tetrahedron network example.

In this specific case, we are looking at a network of resistors arranged on the edges of a tetrahedron. To simplify the network, certain resistors are combined into groups based on whether they're in series or in parallel. Two resistors in parallel have an equivalent resistance that can be calculated using the formula: \( R_{eq} = \frac{R1 \cdot R2}{R1 + R2} \). This allows us to systematically reduce complex networks into simpler versions, until we get a single equivalent resistance value.
Ohm's Law
At the heart of circuit analysis lies Ohm’s Law, a foundational principle expressing the relationship between voltage (V), current (I), and resistance (R) in an electrical circuit. This law states that \( I = \frac{V}{R} \) which means that the current flowing through a conductor between two points is directly proportional to the voltage across the two points, and inversely proportional to the resistance between them.

According to this principle, if you increase the voltage through a circuit while keeping the resistance constant, the current will also increase. Conversely, increasing resistance while maintaining the same voltage will lead to a decrease in current. In our tetrahedron problem, once you've found the equivalent resistance, you can easily find the current flowing from the battery using Ohm's law which will solidify your understanding of the interaction between these core electrical concepts.
Series and Parallel Circuits
Circuits can be organized in two fundamental ways: in series and in parallel. Resistor connections in series follow one-after-another, forming a single path for current flow, while resistors set in parallel have their ends connected to the same two points, forming multiple paths for electricity.

Resistors in series have their resistances simply added up \( (R_{total} = R1 + R2 + ... + Rn) \). In contrast, resistors in parallel require a different approach to find the total resistance, where \(\frac{1}{R_{eq}} = \frac{1}{R1} + \frac{1}{R2} + ... + \frac{1}{Rn}\).

It's key to correctly identify these configurations, as they affect the equivalent resistance of the circuit, something our tetrahedron exercise illustrates well. By knowing that two 5.0 \(\text{Ohm}\) resistors are in series, we simply add them together to get a total resistance of 10.0 \(\text{Ohm}\), significantly impacting the overall circuit behavior.
Electrical Resistors
Resistors are components that impede the flow of electric current, providing resistance in electric circuits. They’re designed to have a specific resistance value, which is measured in Ohms (\(\text{Ohm}\)). An understanding of resistors is important not just in theory but in practical circuit design and analysis.

In our tetrahedron circuit problem, resistors are placed along the edges, connecting at vertices to form a network. Each 10.0-\(\text{Ohm}\) resistor contributes to the total resistance seen by the battery. Without resistors, managing the current in circuits would be difficult, and creating circuits of various complexities, like our tetrahedron example, would be nearly impossible. Thus, resistors are essential both in controlling current and dividing voltage within a circuit.

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Most popular questions from this chapter

An automobile battery has an emf of 12.6 \(\mathrm{V}\) and an internal resistance of 0.0800\(\Omega\) . The headlights together present equivalent resistance 5.00\(\Omega\) (assumed constant). What is the potential difference across the headlight bulbs (a) when they are the only load on the battery and (b) when the starter motor is operated, taking an additional 35.0 A from the battery?

(a) What is the current in a \(5.60-\Omega\) resistor connected to a battery that has a \(0.200-\Omega\) internal resistance if the terminal voltage of the battery is 10.0 \(\mathrm{V} ?\) (b) What is the emf of the battery?

Two \(1.50-\mathrm{V}\) batteries - with their positive terminals in the same direction \(-\) are inserted in series into the barrel of a flashlight. One battery has an internal resistance of 0.255\(\Omega\) , the other an internal resistance of 0.153 . When the switch is closed, a current of 600 \(\mathrm{mA}\) occurs in the lamp. (a) What is the lamp's resistance? (b) What fraction of the chemical energy transformed appears as internal energy in the batteries?

A battery has an emf \(\mathcal{E}\) and internal resistance \(r\) A variable load resistor \(R\) is connected across the terminals of the battery. (a) Determine the value of \(R\) such that the potential difference across the terminals is a maximum. (b) Determine the value of \(R\) so that the current in the circuit is a maximum. (c) Determine the value of \(R\) so that the power delivered to the load resistor is a maximum. Choosing the load resistance for maximum power transfer is a case of what is called impedance matching in general. Impedance matching is important in shifting gears on a bicycle, in connecting a loudspeaker to an audio amplifier, in connecting a battery charger to a bank of solar photoelectric cells, and in many other applications.

The current in a loop circuit that has a resistance of \(R_{1}\) is 2.00 A. The current is reduced to 1.60 \(\mathrm{A}\) when an additional resistor \(R_{2}=3.00 \Omega\) is added in series with \(R_{1}\) . What is the value of \(R_{1} ?\)

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