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 $\boxed{V_{o}=\displaystyle-IR(1-D)}$

A Power Sytem Stability Problem From GATE 2002

The following problem appeared in GATE 2002 conducted by IISc Bangalore. GATE 2002 was the last GATE where subjective numericals were posed.The Electrical Engineering paper especially had some really nice problems. One such problem is given below. Anybody who can solve this problem without help, can rest assured that he/she has a really good understanding of power system stability concepts. Unfortunately such problems no longer appear since the test pattern was changed to objective type/ multiple choice since 2003.

Problem

A synchronous generator is to be connected to an infinite bus through a transmission line of reactance X = 0.2 pu as shown in the fig below. The generator data is as follows:

X'= 0.1 pu, E' = 1.0 pu, H = 5MJ/MVA, mechanical power $P_{m}$ = 0.0 pu, $\omega=2\pi 50$ rad/sec. All quantities are expressed on a common base.

The generator is initially running on an open circuit with the frequency of the open circuit voltage slightly higher than that of the infinite bus. If at the instant of closure $\displaystyle\delta=0$ and $\displaystyle\omega=\displaystyle\frac{d\delta}{dt}=\displaystyle\omega_{o}$. compute the maximum value of  $\displaystyle\omega_{o}$ so that the generator pulls into synchronism.

Solution:

The swing equation derived in most text books is for the transient operation of the generator due to load fluctuations or faults. However, in this particular numerical, at the instant of closure the load angle $\delta=0$. After switching in as the generator tries to synchronize with the grid, $\delta$ increases and during this period the rotor angle $\theta_{r}=\delta$*.

Therefore, rotor speed during this period will be $\omega_{r}=\displaystyle\frac{d\theta_{r}}{dt}=\displaystyle\frac{d\delta}{dt}$

Using the swing equation for per unit quantities and the above equations we get

$\boxed{\displaystyle\frac{2H}{\omega_{B}^2}\omega_{r}\frac{d\omega_{r}}{dt}=P_{m}-\frac{EV}{X}\sin{\delta}}$

$P_{m}=0$ p.u. is given.

$\boxed{\displaystyle\frac{2H}{\omega_{B}^2}\omega_{r}\frac{d\omega_{r}}{dt}=-\frac{EV}{X}\sin{\delta}}$

$\boxed{\displaystyle\frac{2H}{\omega_{B}^2}\omega_{r}\frac{d\omega_{r}}{d\delta}\frac{d\delta}{dt}=-\frac{EV}{X}\sin{\delta}}$

$\boxed{\displaystyle\frac{2H}{\omega_{B}^2}\omega_{r}^2\frac{d\omega_{r}}{d\delta}=-\frac{EV}{X}\sin{\delta}}$

The maximum value of $\omega$ will occur when $\displaystyle\frac{d\omega_{r}}{d\delta}=0$. Substituting in above we get $\sin{\delta}=0\implies\delta=\frac{\pi}{2}$

$\boxed{\displaystyle\int_{\omega_{o}}^{\omega_{B}}\frac{2H}{\omega_{B}^2}\omega_{r}^2{d\omega_{r}}=-\int_{0}^{\delta}\frac{EV}{X}\sin{\delta}{d\delta}}$

$\boxed{\displaystyle\int_{\omega_{o}}^{\omega_{B}}\frac{2H}{\omega_{B}^2}\omega_{r}^2{d\omega_{r}}=-\int_{0}^{\displaystyle\frac{\pi}{2}}\frac{EV}{X}\sin{\delta}{d\delta}}$

Solving the above integral we get the solution as

$\boxed{\displaystyle\omega_{o}=\omega_{B}\left(1+\frac{3EV}{2HX\omega_{B}}\right)^{\displaystyle\frac{1}{3}}}$

$\omega_{o}=2{\pi}$(50.053) rad/sec. Slip frequency limit is seen to be 0.053 Hz which is agreeable with practical situations.

*Note that this is different from the equation for rotor angle in the case when a change in load or fault condition causes transients in a synchronized generator $\theta_{r}=\delta+\displaystyle\omega_{s}t$

A nice Geometry Problem

The following problem is an example of how deceptive a seemingly simple geometry problem can be. Angle chasing is not going to lead anywhere.

Problem.

Let $\triangle ABC$ be isosceles with $AB=AC$ and $\angle BAC = 20^\circ$. Point $D$ is on side $AC$ such that $\angle DBC = 60^\circ$. Point $E$ is on side $AB$ such that $\angle ECB = 50^\circ$. Find, with proof, the measure of $\angle EDB$.

Source: Mathematical Gazette 1922