Power dissipated in resistance networks

JEE Advanced 2008 Paper 1, Question 25

The figure below shows three resistor configurations R1, R2 and R3 connected to 3 \mathrm{~V} battery. If the power dissipated by the configuration R1, R2 and R3 is P1, P2 and P3 respectively, then

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  1. P1 > P2 > P3
  2. P1 > P3 > P2
  3. P2 > P1 > P3
  4. P3 > P2 > P1


The power dissipated by a resistance network is

(1)   \begin{equation*}   P = \frac{V^2}{R} .  \end{equation*}

The potential applied to all three networks arre the same, so we just need to find the resistances. Of circuits R2 and R3, the latter has a greater resistance since

(2)   \begin{equation*}   % R_3 = 1 \, \Omega + \frac{2 \, \Omega \cdot 2 \, \Omega}{2 \, \Omega + 2 \, \Omega} = 2 \, \Omega ,   R_3 = 1 \, \Omega + \left( \frac{1}{2 \Omega} + \frac{1}{2 \Omega} \right)^{-1} = 2 \, \Omega , \end{equation*}


(3)   \begin{equation*}   % R_2 = \frac{1 \, \Omega \cdot 1 \, \Omega}{1 \, \Omega + 1 \, \Omega} = 0.5  \, \Omega .   R_2 = \left( \frac{1}{1 \Omega} + \frac{1}{2 \Omega} + \frac{1}{2 \Omega} \right)^{-1} = 0.5 \, \Omega. \end{equation*}

The first circuit can be redrawn in the following way to make the comparison easier.…

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Refracting surface with a coating

JEE Advanced 2014 Paper 1, Question 9

A transparent thin film of uniform thickness and refractive index n_{1}=1.4 is coated on the convex spherical surface of radius R at one end of a long solid glass cylinder of refractive index n_{2}=1.5, as shown in the figure. Rays of light parallel to the axis of the cylinder traversing through the film from air to glass get focused at distance f_{1} from the film, while rays of light traversing from glass to air get focused at distance f_{2} from the film. Then

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  1. \left|f_{1}\right|=3 R
  2. \left|f_{1}\right|=2.8 R
  3. \left|f_{2}\right|=2 R
  4. \left|f_{2}\right|=1.4 R

Related Problem: Lensing by oil on water


We can solve this problem using …

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Lensing by oil on water

JEE Advanced 2011 Paper 2, Question 36

Water (with refractive index =\frac{4}{3} ) in a tank is 18 \mathrm{~cm} deep. Oil of refractive index \frac{7}{4} lies on water making a convex surface of radius of curvature R=6 {\rm ~cm} as shown. Consider oil to act as a thin lens. An object ‘S’ is placed 24 \mathrm{~cm} above water surface. The location of its image is at ‘x’ cm above the bottom of the tank. What is ‘x’?

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To find the location of the final image we consider the oil and the water in the tank as a combination of two elements: a lens and a …

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Lenses II: Image formation

Continuing our discussion from the previous chapter, we can use the thin lens equation and our knowledge of the signs of f, to determine the position, orientation, and magnification of the image for different object distances p. This is similar to the analysis we did for mirrors here. The results are summarized in the table below.

Converging lens

The focal length of a converging lens is positive. That means light from infinity will be brought to focus behind the lens. We will begin our analysis there.

Starting from p = \infty

Consider an object kept in front of a converging lens. …

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Lenses I: The thin lens equation

A lens is a refracting element with two curved surfaces. Light changes direction as it refracts through each of these surfaces. If the thickness of the lens is small compared to the radius of curvature of each surface, we can think of light as bending just once at the central plane of the lens. This approximation is justified more rigorously in chapter 27 of the Feynman lectures. For our present discussion we start with equation (27.12) from that book, called the thin lens equation,

(1)   \begin{equation*}   \frac{1}{p} + \frac{1}{q} = \frac{1}{f} ,  \end{equation*}

where p is the object distance from the lens, q is the image distance, and …

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