Linear Induction Motor

Problem:

The concept of a linear induction motor is used to develop a magnetically levitated high speed train. The engine car carries a 3 phase 10 pole winding placed in a linear configuration. The dimensions of the winding configuration of the car is of 10 m length ( in the direction of movement of the train) and 4 m wide excluding end winding. This winding is facing the track which is a linear dispossessed squirrel cage winding. The air gap is sandwiched between the above two. Determine the supply frequency to run the train near 400 Km/ hr. [IIT DELHI]

Square Loop in a Magnetic Field

Problem:

A rectangular frame ABCD made of auniform metal wire has a straight connection between E & F made of the same wire as shown in the figure. AEFD is a square of side 1 m & EB = FC = 0.5 m. The entire circuit is placed in a steadily increasing uniform magnetic field directed into the place ofthe paper & normal to it. The rate of change of the magnetic field is 1 T/s, the resistance per unit length ofthe wire is 1 $\Omega$/m. Find the current in segments AE, BE & EF. [IIT JEE'93]

Electron entering a capacitor

Problem:

An electron enters the region between the plates of a parallel plate capacitor at a point equidistant from either plate. The capacitor plates are $2* 10^{-2}$ m apart & $10^{-1}$ m long. A potential difference of 300 volt is kept across the plates. Assuming that the initial velocity of the electron is parallel to the capacitor plates, calculate the largest value of the velocity of the electron so that they do not fly out of the capacitor at the other end. [IIT JEE]

Capacitor Circuit

Problem:

Two capacitors A and B with capacities 3 pF and 2 pF are charged to a potential difference of 100 V and 180 V respectively. The plates of the capacitors are connected as shown in figure with one wire from each capacitor free. The upper plate of a is positive and that of B is negative, an uncharged 2 pF capacitor C with lead wires falls on the free ends to complete the circuit. Calculate the final charges on the three capacitors and the amount of electrostatic energy stored in the system before and after the completion of the circuit. [IIT JEE'97]

Capacitance

Problem:

The capacitance of a parallel plate capacitor with plate area $A$ & separation $d$ is $C$. The space between the plates is filled with two wedges of di-electric constant $K_{1}$ & $K_{2}$ respectively. Find the capacitance of the resulting capacitor. [IIT JEE'96]

Electricity & Magnetism S S Krotov

Problem:

A concealed circuit consisting of resistors has four terminals as shown in fig. If a voltage is applied between clamps 1 and 2 when 3 and 4 are disconnected power liberated is $P_{1}=$ 40 W. When clamps 3 and 4 are connected $P_{2}=$ 80 W. If the same source is connected to the clamps 3 and 4, the power liberated when 1 and 2 are disconnected $P_{3}=$ 20 W. Determine the power $P_{4}$ with 1 and 2 shorted with the same source applied between 3 and 4.

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 $\theta_{r}=\delta+\displaystyle\omega_{s}t$