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Chapter 3: Problem 8

Infinite square lattice \(*\) * Consider a two-dimensional infinite square lattice of \(1 \Omega\) resistors. That is, every lattice point in the plane has four \(1 \Omega\) resistors connected to it. What is the equivalent resistance between two adjacent nodes? This problem is a startling example of the power of symmetry and superposition. Hint: If you can determine the voltage drop between two adjacent nodes when a current of, say, 1 A goes in one node and comes out the other, then you are done. Consider this setup as the superposition of two other setups.

Short Answer

Expert verified
The equivalent resistance between two adjacent nodes in the given infinite square lattice of 1 \( \Omega \) resistors is 1 \( \Omega \).

Step by step solution

01

Visualization of the Superposition Principle

First, the problem is divided into two simpler situations. Imagine two setups can be superimposed: In Setup A, a current of 1 A goes in one node and is equally divided into two currents of 0.5 A that come out from two symmetric nodes on the opposite side. Setup B is similar to A, with the inflow and outflow nodes shifted to an adjacent position.
02

Calculation of Voltage Difference for Each Setup

Due to symmetry, the voltage drop between inflow and outflow nodes would be the same for both setups A and B. So, the potential difference (V) can be calculated across the resistor using Ohm's Law (V=IR). Since we've considered 1 A current and each resistor is of 1 \( \Omega \), the voltage difference between two nodes would also be 1 V.
03

Superposition of Both Setups

The final step is to superimpose both setups. The currents in the common part of the network (between the nodes we are interested in) add up for both setups, becoming 1 A, the same as the initial problem. So, the superposition of the currents is identical to the initial setup and we can infer that the equivalent resistance is also 1 \( \Omega \).

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Key Concepts

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

Ohm's Law
Ohm's Law is fundamental in the study of electricity and magnetism. It explains the relationship between voltage, current, and resistance in an electrical circuit. Ohm's Law is expressed as \( V = IR \), where \( V \) represents voltage, \( I \) is current, and \( R \) stands for resistance. This equation helps to calculate one variable if the other two are known.
For example, in an electric circuit with a known resistance and voltage, Ohm's Law can determine the current flowing through the circuit.
This principle is crucial in analyzing circuits, such as in the exercise of the infinite square lattice of resistors. By applying Ohm's Law, the voltage drop across each resistor can be understood when a certain current is applied.
Understanding Ohm's Law allows students to solve complex circuit problems by simplifying the relationships between electrical quantities.
Superposition Principle
The superposition principle is a powerful method used in electric circuit analysis. It allows for simplifying complex circuits by considering each independent source separately.
In the context of resistor networks, superposition can be used to determine the combined effect of all sources by calculating the effect of each source independently and then summing the results.
In the infinite lattice exercise, superposition is used to simplify the analysis by breaking down the complex network into two simpler setups. Each of these setups involves applying the superposition principle, which helps to analyze the individual contributions of current in different paths.
By superimposing the effects, it becomes clear how the network behaves as a whole. This approach illustrates the elegance and efficiency of using the superposition principle to solve intricate circuit problems.
Electric Circuit Analysis
Electric circuit analysis involves reviewing and understanding circuits to determine voltage, current, and resistance across components. It involves using laws and principles such as Ohm's Law and Kirchhoff's Laws.
In the electric circuit analysis of an infinite lattice of resistors, techniques like symmetry and superposition are crucial.
Symmetry helps to identify equivalent parts of the circuit, simplifying calculations and ensuring that findings are consistent throughout the network. Leveraging symmetries often reduces the complexity of the system under study.
For the analysis of complex circuits, a systematic approach is key. Breaking the problem into smaller, manageable parts鈥� as done in the original problem鈥� helps solve for the equivalent resistance effectively.
This detailed analysis deepens understanding of how electricity flows through varied pathways in intricate networks.

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Most popular questions from this chapter

Stationary rod and moving charge A charge \(q\) moves with speed \(v\) parallel to a long rod with linear charge density \(\lambda\), as shown in Fig. 5.30. The rod is at rest. If the charge \(q\) is a distance \(r\) from the rod, the force on it is simply \(F=q E=q \lambda / 2 \pi r \epsilon_{0}\) Now consider the setup in the frame that moves along with the charge \(q\). What is the force on the charge \(q\) in this new frame? Solve this by: (a) transforming the force from the old frame to the new frame, without caring about what causes the force in the new frame; (b) calculating the electric force in the new frame.

Combining the current densities We have \(5 \cdot 10^{16}\) doubly charged positive ions per \(\mathrm{m}^{3}\), all moving west with a speed of \(10^{5} \mathrm{~m} / \mathrm{s}\). In the same region there are \(10^{17}\) electrons per \(\mathrm{m}^{3}\) moving northeast with a speed of \(10^{6} \mathrm{~m} / \mathrm{s}\). (Don't ask how we managed it!) What are the magnitude and direction of \(\mathbf{J}\) ?

Two ways of calculating energy \(* *\) * A capacitor consists of two arbitrarily shaped conducting shells, with one inside the other. The inner conductor has charge \(Q\), the outer has charge \(-Q\). We know of two ways of calculating the energy \(U\) stored in this system. We can find the electric field \(E\) and then integrate \(\epsilon_{0} E^{2} / 2\) over the volume between the conductors. Or. if we know the potential difference \(\phi\), we can write \(U=Q \phi / 2\) (or equivalently \(U=C \phi^{2} / 2\) ). (a) Show that these two methods give the same energy in the case of two concentric shells. (b) By using the identity \(\nabla \cdot(\phi \nabla \phi)=(\nabla \phi)^{2}+\phi \nabla^{2} \phi\), show that the two methods give the same energy for conductors of any shape.

Average of six points Let \(\phi(x, y, z)\) be any function that can be expanded in a power series around a point \(\left(x_{0}, y_{0}, z_{0}\right)\). Write a Taylor series expansion for the value of \(\phi\) at each of the six points \(\left(x_{0}+\delta, y_{0}, z_{0}\right),\left(x_{0}-\right.\) \(\left.\delta, y_{0}, z_{0}\right),\left(x_{0}, y_{0}+\delta, z_{0}\right),\left(x_{0}, y_{0}-\delta, z_{0}\right),\left(x_{0}, y_{0}, z_{0}+\delta\right),\left(x_{0}, y_{0},\right.\), \(z_{0}-\delta\) ), which symmetrically surround the point \(\left(x_{0}, y_{0}, z_{0}\right)\) at a distance \(\delta\). Show that, if \(\phi\) satisfies Laplace's equation, the average of these six values is equal to \(\left(x_{0}, y_{0}, z_{0}\right)\) through terms of the third order in \(\delta\).

Electron in an oscilloscope * The deflection plates in a high-voltage cathode ray oscilloscope ure two rectangular plates, \(4 \mathrm{~cm}\) long and \(1.5 \mathrm{~cm}\) wide, and spaced, \(.8 \mathrm{~cm}\) apart. There is a difference in potential of \(6000 \mathrm{~V}\) between he plates. An electron that has been accelerated through a potenial difference of \(250 \mathrm{kV}\) enters this deflector from the left, moving, of find the position of the electron and its direction of motion when t leaves the deflecting field at the other end of the plates. We shall lates is uniform right up to the end. The rest energy of the electron nay be taken as \(500 \mathrm{keV}\). (a) First carry out the analysis in the lab frame by answering the following questions: \- What are the values of \(\gamma\) and \(\beta ?\) \- What is \(p_{x}\) in units of \(m c ?\) \- How long does the electron spend between the plates? (Neglect the change in horizontal velocity discussed in Exercise 5.25.) \- What is the transverse momentum component acquired, in units of \(m c\) ? \- What is the transverse velocity at exit? \- What is the vertical position at exit? \- What is the direction of flight at exit?(b) Now describe this whole process as it would appear in an inertial frame that moved with the electron at the moment it entered the deflecting region. What do the plates look like? What is the field between them? What happens to the electron in this coordinate system? Your main object in this exercise is to convince yourself that the two descriptions are completely consistent.
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