An infinite ladder of resistors

Problem 3.152 of Irodov

The figure below shows an inifite circuit formed by the repetition of the same link, consisting of resistance R_1 = 4.0 \ \Omega and R_2 = 3.0 \ \Omega. Find the resistance of this circuit between points A and B.

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Solution

Let’s denote the resistance between the points A and B by R_{AB}. Since the circuit is infinite, removing the first R_1 and R_2 resistors gives the same arrangement back again — the arrangement is self-similar. That means, the resistance between the points C and D is just R_{AB} without the left-most R_1 and R_2 resistors, and we may redraw the circuit as shown below.

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It is now straightforward to calculate the resistance,

(1)

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Electric field in a hollow region

IIT-JEE 2007 Paper 2, Question 6

(This problem is was repeated in JEE Advanced 2015 Paper 2, Question 11)

A spherical portion has been removed from a solid sphere having a charge distributed uniformly in its volume as shown in the figure. The electric field inside the emptied space is

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  1. zero everywhere
  2. non-zero and uniform
  3. non-uniform
  4. zero only at its center

Related Problems:
Electric field from a sphere with a cavity
Flux from a charged shell
Charge at one corner of a cube

Solution

As explained in a related problem we can think of the given charge distribution as a …

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Delay in a fiber optic cable

JEE Advanced 2019 Paper 1, Question 18

A planar structure of length L and width W is made of two different optical media of refractive indices n_{1}=1.5 and n_{2}=1.44 as shown in figure. If L \gg W, a ray entering from end {\rm AB} will emerge from end {\rm CD} only if the total internal reflection condition is met inside the structure. For L=9.6 m, if the incident angle \theta is varied, the maximum time taken by a ray to exit the plane \mathrm{CD} is t \times 10^{-9} s. Determine t.

[ Speed of light c=3 \times 10^{8} m/s. ]

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Solution

To ensure that the ray spends the maximum possible time inside the media, we should arrange it so …

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Charge at one corner of a cube

A charge q sits at one corner of a cube of side length a as shown in the figure below. What is the electric flux through the shaded side?

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Related problems:
Electric field from a sphere with a cavity
Electric field in a hollow region
Flux from a charged shell

Solution

Since the problem is asking us to find the electric flux it is natural to guess that we need to apply Gauss’s Law. However, we need a closed surface with appropriate symmetry to be able to use Gauss’s Law (see article). So we consider the situation shown below.

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We have placed …

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Electric field from a sphere with a cavity

A sphere of radius a is filled with positive charge with uniform density \rho. Then a smaller sphere of radius a/2 is carved out, as shown in the figure below, and left empty. What are the direction and magnitude of the electric field at A? At B?

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Related problems:
Electric field in a hollow region
Charge at one corner of a cube
Flux from a charged shell

Solution

This problem can be solved by using the principle of superposition. For instance, consider a point charge +q at some point P in space. It creates an electric field everywhere. However, if you place a …

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