Conductor Harmonic Motion - IIT JEE 2003

Problem:

Along horizontal wire AB, which is free to move in a vertical plane and carries a steady current of 20 A, is in equilibrium at a height of 0.01 m over another parallel long wire CD which is fixed in a horizontal plane and carries a steady current of 30 A. Show that when AB is slightly depressed, it executes simple harmonic motion. Find the period of oscillations.

Find value of Resistance - IIT JEE 2005

Problem:

In the figure both cells A and B are of equal emf. Find R for which potential difference across battery A will be zero, long time after the switch is closed. Internal resistance of batteries A and B are $r_{1}$ and $r_{2 }$respectively ($r_{1}>r_{2 }$).

Kinetic Theory of Gases - IIT JEE 1993

The following problem appeared in the Physics paper of IIT JEE 1993. It still considered to be one of the most formidable problems ever asked in the history of the exam.

Problem:

A neutron of kinetic energy 65 eV collides in-elastically with a singly ionized helium atom at rest. It is scattered at an angle of 90º with respect of its original direction.

(i) Find the allowed values of the energy of the neutron and that of the atom after the collision.

(ii) If the atom get de-excited subsequently by emitting radiation, find the frequencies of the emitted radiation. [Given: mass of He atom =4×(mass of neutron),Ionization energy of He atom = 13.6 eV]

Infinite Grid of Resistors

The following problem is well known among electrical engineering students. Its solution using superposition is also well known. Is there some other solution method without using the superposition argument??

Problem 

There is an infinite grid with square cells. The resistance of each wire between neighboring joints is equal to $R$. Find the resistance between two adjacent points on the grid. (Source: I.E Irodov)

Image : http://www.mathpages.com/home/kmath668/kmath668.htm

Mutual Inductance

One book every physics or engineering student must have in his collection is Aptitude Problems in Physics which is a collection of Moscow Physics Olympiad problems edited by S.S Krotov. The problems in the book are absolute gems. I will post a few questions from the Electricity and Magnetism section.

Problem:

Two long cylindrical coils with uniform winding of the same length and nearly the same radius have inductance $L_{1}$ and $L_{2}$. The coils are co-axially inserted into each other and connected to a current source. The directions of the current in the coils is such that the fluxes add . Determine the inductance L of such a composite coil.

Solution:

From the definition of inductance we have $\displaystyle L=\frac{N\phi}{I}$

$\displaystyle\phi =\frac{MMF}{Reluctance}=\displaystyle\frac{NIA}{\mu_{o}l}\implies L_{1}=kN_{1}^2$  and $L_{2}=kN_{2}^2$

When the coils are coaxially combined with flux additive composite inductance

$L_{c}=k(N_{1}+N_{2})^2$

$L_{c}=\displaystyle k\left(\sqrt{\frac{L_{1}}{k}}+\sqrt{\frac{L_{2}}{k}}\right)^2=L_{1}+L_{2}+2\sqrt{L_{1}L_{2}}$

Power Electronics GATE 2002

In the Fig. below , the ideal switch S is switched on and off with a switching frequency f = 10 kHz. The switching time period is $T=t_{ON}+t_{OFF}$ μs. The circuit is operated in steady state at the boundary of continuous and discontinuous conduction, so that the inductor current $i$ is as shown in Figure. What are the values of the on-time $t_{ON}$ of the switch and peak current $i_{p}$.

Solution:
When the switch is ON the Diode is reverse biased and the voltage across the inductor is $100$ V. Hence, the current through the inductor is given by.

$\boxed{i(t)=10^6t}$

Therefore peak current as seen from the graph $i_{p}=10^6T_{ON}$
At the instant of switching OFF the inductor back emf polarity is reversed and the diode starts conducting.
The volatge across the inductor now is $500$ V and the initial current being $i_{p}=10^6T_{ON}$ the current equation during OFF time is given by:

$\boxed{i(t)=10^6T_{ON}-5\cdot10^6t}$

From the graph @ $T_{OFF}$ current through inductor is zero

$\boxed{\implies i(T_{OFF})=10^6T_{ON}-5\cdot10^6T_{OFF}=0}$

Therefore $T_{ON}=5\cdot T_{OFF}$

$T=T_{ON}+T_{OFF}=\displaystyle\frac{1}{f}=100$ μs

$T_{ON}=63.33$ μs

$i_{p}=63.33$ Amp

Power Electronics GATE 2002

Problem: In the circuit shown in Figure, the source $I$ is a dc current source.The switch $S$ is operated with a time period $T$ and a duty ratio $D$. You may assume that the capacitance $C$ has a finite value which is large enough so that the voltage $V_{c}$ has negligible ripple, calculate the following under steady state conditions, in terms of $D$, $I$ and $R$.a) The voltage $V_{c}$, with the polarity shown in figure.b) The average output voltage $V_{o}$, with the polarity shown in figure.

The following solution is a little rigorous. A more intuitive and short solution is given at the end.

Solution 1 for (a):

Since the circuit is operating in steady state conditions, the amount of energy stored in its components has to be the same at the beginning and at the end of a commutation cycle. Capacitor voltage ripple is negligible.

 $\displaystyle\implies\boxed{\triangle {V_{c}}^{OFF}+ \triangle {V_{c}}^{ON}=0}$

When $S$ is OFF, the capacitor will be charged by the current source through the diode according to the equation $V_{c}(t)=V_{c}+\displaystyle\frac{I}{C}t$

 $\implies\boxed{\triangle{V_{c}}^{OFF}=\displaystyle\frac{I}{C}T_{OFF}}$

When $S$ is ON, the capacitor will be discharged through the resistor according to the equation $V_{c}(t)=V_{c}e^{\displaystyle\frac{-t}{RC}}$

 $\implies\boxed{\triangle{V_{c}}^{ON}=V_{c}e^{\displaystyle\frac{-T_{ON}}{RC}}-V_{c}}$

 But we have $\displaystyle\triangle {V_{c}}^{OFF}+ \triangle {V_{c}}^{ON}=0\implies\displaystyle\frac{I}{C}T_{OFF}+V_{c}e^{\displaystyle\frac{-T_{ON}}{RC}}-V_{c}=0$

$\implies\boxed{\displaystyle\frac{I}{C}T_{OFF}=V_{c}(1-e^{\displaystyle\frac{-T_{ON}}{RC}})}$

It is given that the value of $C$ is very large so we can neglect the higher order terms with degree $\geq2$ from the infinite series of  $e^{x}$. 

Therefore, $\boxed{\displaystyle\frac{I}{C}T_{OFF}=V_{c}\frac{T_{ON}}{RC}}$.
Substituting for $T_{ON}=DT$ and $T_{OFF}=(1-D)T$.

The capacitor voltage $\displaystyle\boxed{V_{c}=\displaystyle\frac{IR(1-D)}{D}}$
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 Solution 2 for (a):

The energy increase of $C$ during $OFF$ time $\approx V_{c}IT_{OFF}$

Energy decrease of $C$ during $ON$ time when it discharges through R $\approx\displaystyle\frac{V_{c}^2}{R}T_{ON}$

Both the above terms must be equal for steady state.

$\implies\boxed{V_{c}IT_{OFF}=\displaystyle\frac{V_{c}^2}{R}T_{ON}}$ leading to the same solution $\displaystyle\boxed{V_{c}=\displaystyle\frac{IR(1-D)}{D}}$
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Solution for (b):
Average value of $\boxed{V_{o}=\displaystyle\frac{-V_{c}T_{ON}+(0)T_{OFF}}{T}}$

Average $\boxed{V_{o}={-V_{c}T_{ON}}{T}\implies V_{o}=-V_{c}D} $

Therefore, average value of $\boxed{V_{o}=\displaystyle-IR(1-D)}$