GATE EE 2011

To solve the equation $x^{3+}x^{2+}x+1=0$, we can start by testing for simple integer roots. Let's try $x=-1$. Substituting this value gives $(-1)^{3+}(-1)^{2+}(-1)+1 = -1+1-1+1=0$. Since the equation holds true, $x=-1$ is one of the roots.

This means that $(x+1)$ is a factor of the polynomial. We can find the other factor by dividing the original polynomial by $(x+1)$, which gives a quotient of $x^{2+}1$.

So, the equation can be rewritten in factored form as $(x+1)(x^{2+}1)=0$. To find the roots, we set each factor to zero.
From $x+1=0$, we get $x=-1$.
From $x^{2+}1=0$, we get $x^2=-1$, which yields the imaginary roots $x = \pm\sqrt{-1}$. Using $j$ for the imaginary unit, these roots are $x = +j$ and $x = -j$.

Therefore, the three roots of the equation are $-1, +j, -j$.

To find the function $y$, we must reverse the differentiation by integrating the right-hand side of the equation with respect to $x$.
$y = \int e^{-3x} \, dx$
We use the standard integration formula for exponential functions, $\int e^{ax} \, dx = \frac{1}{a}e^{ax}$, where our constant $a$ is $-3$.

Applying this rule and including the constant of integration, $K$, yields:
$y = \frac{e^{-3x}}{-3} + K$
Simplifying the expression gives us the final solution.
$y = -\frac{1}{3}e^{-3x} + K$

Let's analyze the circuit based on the source voltage $v(t) = 1.0 \sin t$, which gives us an angular frequency of $\omega = 1$ rad/s.

First, we calculate the impedance of the series LC branch. The inductive reactance is $X_L = \omega L = (1)(1) = 1 \, \Omega$, and the capacitive reactance is $X_C = \frac{1}{\omega C} = \frac{1}{(1)(1)} = 1 \, \Omega$.

Since $X_L = X_C$, the total impedance of this branch is $Z_{LC} = j(X_L - X_C) = 0$. This means the LC branch acts as a short circuit at this frequency.

This short circuit effectively bypasses the middle 1 Ω resistor. The total current $i(t)$ from the source therefore flows through the loop formed by the voltage source and the bottom 1 Ω resistor.

Using Ohm's law, the current is $i(t) = \frac{v(t)}{R} = \frac{1.0 \sin t}{1} = \sin t$. The peak value of this current is $I_m=1$ A. The RMS value is found by dividing the peak value by $\sqrt{2}$, which gives $I_{rms} = \frac{1}{\sqrt{2}}$ A.

We can determine the components of the Fourier series by analyzing the signal's symmetries.

First, the signal $f(t)$ is symmetric about the vertical axis, meaning $f(t) = f(-t)$. This is the definition of an even function, for which all sine coefficients ($b_n$) are zero.

Second, the signal has a positive average value because its waveform is entirely above the time axis. This non-zero average corresponds to a non-zero DC coefficient, $a_0$.

Finally, if we subtract the DC component $a_0$, the remaining AC signal is a square wave that exhibits half-wave symmetry ($x(t+T/2) = -x(t)$). This type of symmetry means that only odd harmonics exist in the series.

Combining these properties, the Fourier series for $f(t)$ must contain a DC term ($a_0$) and cosine terms ($a_n$) for odd values of $n$ only.

A 3-point starter connects its "no-volt" holding coil in series with the motor's shunt field winding. This presents a problem when you try to increase the motor's speed using field weakening, which involves reducing the field current ($I_f$). A significantly reduced field current might not be strong enough to keep the holding coil energized, causing the starter to trip and shut down the motor.

A 4-point starter solves this by creating a separate, fourth connection point for the holding coil. This places the coil in a circuit parallel to the armature and field, making its operation independent of the field current. Therefore, the motor can run stably even when the field current is varied over a wide range for speed control.

The synchronous motor is tied to an infinite bus, which fixes the terminal voltage $V_t$ and demands a nearly constant net air-gap flux. As you reduce the field excitation current $I_f$, the rotor's magnetic field weakens. To maintain the constant total flux, the armature must draw a larger magnetizing current $I_a$ from the bus. This continues even as you reverse the field current, causing $I_a$ to steadily increase.

During this process, the load angle $\delta$ increases. Eventually, the torque from the reversed field overpowers the machine's reluctance torque, causing the rotor to abruptly slip by one pole pitch (180 electrical degrees) to a new stable position. In this new alignment, the rotor and armature fields suddenly aid each other, creating an excessive magnetic flux. To counteract this over-flux and re-establish equilibrium at the bus voltage, the armature current must then decrease steeply.

To effectively transmit a large quantity of power, like 500 MW, across a long distance of 300 km, two key principles guide the design.

First, a high voltage level is essential. Using a higher voltage reduces the current for the same amount of power ($P = V \times I$), which in turn drastically cuts down on power lost as heat ($I^2R$). For this power level and distance, a 400 kV line is a standard and efficient choice.

Second, reliability is paramount, especially for a large power station. A double-circuit line provides redundancy. If one circuit is down for maintenance or due to a fault, the other can still deliver power, preventing a complete blackout at the load center.

To find the phase margin (PM), we first identify the gain crossover frequency where the magnitude $|G(j\omega)| = 1$. From the table, when the magnitude is 1.0, the corresponding phase angle is $-150^{\circ}$. The phase margin is the additional phase shift needed to reach $-180^{\circ}$, calculated as $PM = 180^{\circ} + (-150^{\circ}) = 30^{\circ}$.

To find the gain margin (GM), we identify the phase crossover frequency where the phase angle $\angle G(j\omega) = -180^{\circ}$. The table shows that at this frequency, the magnitude is $0.5$. The gain margin, expressed in decibels, is the reciprocal of this magnitude: $GM = 20\log_{10}\left(\frac{1}{0.5}\right) = 20\log_{10}(2) \approx 6 \text{ dB}$.

The steady-state error is found using the Final Value Theorem, $e_{ss} = \lim_{s \to 0} sE(s)$, where $E(s) = \frac{R(s)}{1+G(s)}$ for a unity feedback system. First, let's use the given information for a unit step input, $R(s) = 1/s$. We are told $e_{ss}=0.1$, so $0.1 = \lim_{s \to 0} s \frac{1/s}{1+G(s)} = \frac{1}{\lim_{s \to 0} (1+G(s))}$. This gives us a key characteristic of the system's low-frequency gain.

Now, consider the pulse input $r(t)$, which has the Laplace transform $R(s) = \frac{10}{s}(1-e^{-s})$. We apply the Final Value Theorem again for this new input.
$e_{ss} = \lim_{s \to 0} s E(s) = \lim_{s \to 0} s \frac{\frac{10}{s}(1-e^{-s})}{1+G(s)} = \lim_{s \to 0} \frac{10(1-e^{-s})}{1+G(s)}$

We can evaluate the limits of the numerator and denominator separately. The numerator approaches $10(1-e^0) = 0$. The denominator's limit, as we found earlier, is a finite constant (specifically, $1/0.1 = 10$). Therefore, the steady-state error is $0$.

In an ideal wattmeter, the current coil should only measure the current going to the load. However, the pressure (or voltage) coil also draws a small current, $I_P$, due to its own impedance. This current can add to the load current in the main current coil, causing the wattmeter to read high by including the power consumed by the pressure coil itself.

The compensating coil solves this problem. It is connected in series with the pressure coil, so it carries the exact same current, $I_P$. It is wound around the main current coil but in the opposite direction. This creates a counter-magnetic field that precisely cancels the magnetic effect of the unwanted current $I_P$. As a result, the net magnetic field influencing the meter's deflection is due only to the true load current, correcting the error. This means the coil compensates for the effect of the voltage coil circuit's impedance.

Let's analyze the behavior of each filter separately. The low-pass filter allows frequencies below its cut-off, $f_{L} = 30$ Hz, to pass through. The high-pass filter allows frequencies above its cut-off, $f_{H} = 20$ Hz, to pass.

When these filters are cascaded, a signal must pass through both to be present at the output. This means the signal's frequency, $f$, must satisfy both conditions simultaneously: it must be higher than 20 Hz AND lower than 30 Hz.

The resulting passband is the range of frequencies $20 \text{ Hz} < f < 30 \text{ Hz}$. A filter that passes a specific band of frequencies while rejecting others is, by definition, a band-pass filter.

The circuit consists of two stages. The first stage is a differential amplifier with a gain of -1, as all its resistors are equal. Its output is simply the negative of the input voltage, or $-V_i$.

This signal, $-V_i$, is then fed into the second stage, which is an inverting Schmitt trigger. The trigger thresholds are determined by the voltage divider in the feedback loop and the op-amp's saturation voltages.

The threshold voltage is set at the inverting input as $V_{th} = \frac{R}{R+R}V_o = \frac{1}{2}V_o$. Since the op-amp's output $V_o$ saturates at $\pm 12\,\text{V}$, the switching thresholds for its input ($-V_i$) are $\pm \frac{12\,\text{V}}{2} = \pm 6\,\text{V}$.

Therefore, the entire circuit switches states when $-V_i = \pm 6\,\text{V}$, which means the primary input $V_i$ must cross thresholds of $\mp 6\,\text{V}$. This creates an inverting hysteresis loop, where the output switches from high to low at $V_i = -6\,\text{V}$ and from low to high at $V_i = +6\,\text{V}$.

The fundamental requirement for switches in a Current Source Inverter (CSI) is the ability to block reverse voltage. A standard power MOSFET contains an intrinsic anti-parallel body diode, which will conduct if a reverse voltage is applied across the switch. This inherent feature prevents the MOSFET from blocking voltage in the reverse direction. To provide this necessary reverse-blocking capability, a diode must be connected in series with the MOSFET. This configuration ensures the combined switch can withstand reverse voltage when it is off.

Let's analyze the properties of $Z$ and see how taking its reciprocal, $y = 1/Z$, transforms it.

First, consider the magnitude. The point $Z$ is inside the unit circle, which means its distance from the origin, or magnitude $|Z|$, is less than 1. The magnitude of $y$ is $|y| = |1/Z| = 1/|Z|$. Since $|Z| < 1$, its reciprocal $1/|Z|$ must be greater than 1. This tells us that the point $y$ lies outside the unit circle.

Next, consider the quadrant. Let $Z = a+bi$. Since $Z$ is in the first quadrant, both its real part $a$ and imaginary part $b$ are positive. We find $y$ by multiplying the numerator and denominator by the conjugate of $Z$:
$y = \frac{1}{a+bi} = \frac{a-bi}{a^{2+}b^2} = \frac{a}{a^{2+}b^2} - i\frac{b}{a^{2+}b^2}$
The real part of $y$ is positive (since $a>0$), and its imaginary part is negative (since $b>0$). A point with a positive real part and a negative imaginary part is in the fourth quadrant.

Combining these two results, the point $y$ must be outside the unit circle and in the fourth quadrant.

To determine the current's phasor representation, we first need to express both voltage and current in a consistent format, typically the cosine form. The voltage, $V(t) = 100\sqrt{2} \cos(100\pi t)$, is already in this form, so we can set its phase angle as our $0^\circ$ reference.

Next, we convert the current, $i(t) = 10\sqrt{2} \sin(100\pi t + \pi/4)$, to a cosine function using the identity $\sin(\theta) = \cos(\theta - \pi/2)$. This gives us $i(t) = 10\sqrt{2} \cos(100\pi t + \pi/4 - \pi/2)$, which simplifies to $i(t) = 10\sqrt{2} \cos(100\pi t - \pi/4)$.

Phasors are represented by their RMS magnitude and phase angle. The RMS magnitude of the current is the peak amplitude divided by $\sqrt{2}$, so $I_{rms} = \frac{10\sqrt{2}}{\sqrt{2}} = 10$ A. From our cosine expression for current, the phase angle is $-\pi/4$. Therefore, the current phasor is $10 \angle(-\pi/4)$ A.

To transfer the maximum amount of power to the fixed $3\,\Omega$ load, we must maximize the current flowing through it. According to the power formula $P = I^2 R_L$, since the load resistance $R_L$ is constant, power is directly proportional to the square of the current.

To maximize the current, the total resistance of the circuit must be at its minimum. The total resistance is the sum of the load resistance and the equivalent resistance of the source network. We can only change the source resistance, which is the parallel combination of $R$ and $6\,\Omega$. To minimize this parallel equivalent resistance, we must set the variable resistor $R$ to its lowest possible value, which is $0\,\Omega$.

A powerful method for solving convolution problems is to use the Laplace transform. The convolution theorem states that convolution in the time domain corresponds to multiplication in the s-domain.

First, let's find the Laplace transforms of the individual signals, denoted as $X(s)$ and $Y(s)$:
$X(s) = \mathcal{L}\{e^{-t}\} = \frac{1}{s+1}$
$Y(s) = \mathcal{L}\{e^{-2t}\} = \frac{1}{s+2}$

The transform of the convolution, $Z(s)$, is the product of these two transforms:
$Z(s) = X(s)Y(s) = \frac{1}{s+1} \cdot \frac{1}{s+2}$

To find the time-domain signal $z(t)$, we must find the inverse Laplace transform of $Z(s)$. We can simplify $Z(s)$ using partial fraction expansion:
$Z(s) = \frac{1}{s+1} - \frac{1}{s+2}$

Finally, taking the inverse Laplace transform of each term gives us the result:
$z(t) = \mathcal{L}^{-1}\left\{\frac{1}{s+1}\right\} - \mathcal{L}^{-1}\left\{\frac{1}{s+2}\right\} = e^{-t} - e^{-2t}$

The sinusoidal supply voltage creates a sinusoidal magnetic flux ($\phi$) in the transformer core, as dictated by Faraday's law of induction ($v \propto d\phi/dt$). The key here is that the transformer is air-cored. Air is a linear magnetic material, meaning it does not saturate and has no hysteresis effects.

This linearity ensures a direct, proportional relationship between the magnetizing current ($i_m$) and the flux ($\phi$) it produces. In a linear system, a sinusoidal output (the flux) must be caused by a sinusoidal input (the magnetizing current). The distorted waveforms shown in the other options would only arise from the non-linear effects of an iron core.

Unbalanced system conditions, such as faults, produce negative sequence currents ($I_2$). In an alternator, these currents generate a magnetic field that rotates in the opposite direction to the rotor. This high relative speed induces double-frequency currents ($2f$) in the rotor's surface and windings. These currents lead to rapid and intense localized heating ($P \propto I_2^2$), which can quickly damage the rotor structure and insulation. Therefore, alternators are highly sensitive to negative sequence currents and require specific protection against them.

The power transfer capability of a transmission line is inversely proportional to its total series reactance, $X$. A long line is naturally inductive, which means it has a high inductive reactance, $X_L$, that limits power flow.

By connecting a capacitor in series, we introduce a capacitive reactance, $X_C$. This new reactance directly opposes the line's inductance, effectively reducing the total net reactance of the line to $X_{net} = X_L - X_C$.

Since power transfer is proportional to $1/X$, this reduction in the line's overall reactance directly increases the amount of power it can transmit.

To classify this system, we must examine the locations of its poles and zeros in the s-plane.

First, we determine stability by finding the poles from the denominator. The poles are at $s=-2$ and $s=-3$. Since both poles are in the left-half of the s-plane, the system is stable.

Next, we check the phase type by finding the zeros from the numerator. The single zero is at $s=1$. A system is classified as non-minimum phase if it has any zeros in the right-half of the s-plane. Because of the zero at $s=1$, this system is non-minimum phase.

This circuit is a Maxwell bridge, used for measuring an unknown inductive impedance $Z_x$. The quality factor of the inductor being measured is given by the equation $Q = \omega C_1 R_1$. This relationship highlights the bridge's primary limitation: measuring a high-Q inductor would require a very large and expensive variable resistor $R_1$. Therefore, the circuit is most practical and accurate for measuring inductors with a low Q-factor, typically in the range of 1 to 10. A low-Q inductor is defined as one where its internal series resistance ($R_x$) is significantly large compared to its inductive reactance ($\omega L_x$).

In ALTernate (ALT) mode, the oscilloscope displays the complete trace of one channel during one full horizontal sweep, then switches and displays the other channel on the next sweep. This process requires the channel switching to be perfectly synchronized with the sweep cycle.

The horizontal sweep is controlled by the time base oscillator, which operates at a specific frequency, let's call it $f_{sweep}$. At the end of each sweep, the time base circuit generates a trigger pulse (the flyback or retrace blanking pulse) to initiate the next sweep. This pulse train has a frequency equal to $f_{sweep}$. This very signal is used to trigger the multiplexer's control logic (typically a flip-flop), causing it to switch to the alternate channel. Thus, the signal feeding the control logic has a frequency equal to the time base oscillator.

The logic circuit shows an XOR gate with its two inputs being the signal $X$ and its inverse, $\bar{X}$. The function of an XOR gate is to output a 1 if and only if its inputs are different.

Let's analyze the two possible cases for the input $X$:

1. If $X = 0$, the inputs to the XOR gate are $0$ and $\bar{0} = 1$. Since the inputs are different, the output $Y$ is 1.
2. If $X = 1$, the inputs to the XOR gate are $1$ and $\bar{1} = 0$. Again, the inputs are different, so the output $Y$ is 1.

In both cases, the output $Y$ is always 1, regardless of the value of $X$.

To turn off an SCR, simply forcing the anode current to zero isn't enough. The device contains stored charge carriers that must be removed before it can regain its ability to block forward voltage.

The circuit turn-off time is the duration for which the external commutation circuit applies a reverse voltage across the SCR. This reverse bias actively helps to sweep out the stored charges and forces the device to turn off completely. For a successful turn-off, this period must be greater than the SCR's intrinsic recovery time.

To solve this, we first need to find the Jacobian matrix, which is a key component of the Newton-Raphson method for systems of equations. Let's define our two functions based on the given equations:
$f_1(x_1, x_2) = 10x_2 \sin x_1 - 0.8$
$f_2(x_1, x_2) = 10x_2^2 - 10x_2 \cos x_1 - 0.6$

The Jacobian matrix, $J$, is a matrix of the partial derivatives of these functions.

Now, we evaluate this matrix at the given initial point $(x_1, x_2) = (0.0, 1.0)$.
Substituting these values and using $\cos(0) = 1$ and $\sin(0) = 0$, we get:

To find any local maxima or minima, we first look for critical points where the function's slope is zero. We do this by finding the first derivative of $f(x)=2x-x^{2}+3$ and setting it equal to zero.

The derivative is $f'(x) = 2 - 2x$. Setting this to zero gives $2 - 2x = 0$, which solves to $x=1$. This is our only critical point.

To classify this point, we use the second derivative test. The second derivative is $f''(x) = -2$. Since the value of the second derivative at our critical point, $f''(1) = -2$, is negative, the function is concave down at this point, indicating a local maximum. Because $x=1$ was the only critical point, the function has only a maximum.

The power dissipated by the lossy capacitor is entirely due to its equivalent parallel resistance, $R_p$. Using the rated voltage, this power loss is calculated as:
$P = \frac{V^2}{R_p} = \frac{(5 \times 10^3 \text{ V})^2}{1.25 \times 10^6 \, \Omega} = 20 \text{ W}$

The loss tangent, $\tan \delta$, quantifies the capacitor's inefficiency. For this parallel model, it is the ratio of current through the resistor to the current through the capacitor, which simplifies to the formula $\tan \delta = \frac{1}{\omega C_p R_p}$. Using the angular frequency $\omega = 2\pi f$, we find:
$\tan \delta = \frac{1}{2\pi(50 \text{ Hz})(0.102 \times 10^{-6} \text{ F})(1.25 \times 10^6 \, \Omega)} \approx 0.025$

The Laplace transform of a time-delayed function $f(t-\tau)$ is given by the time-shifting property, which states $F_2(s) = e^{-s\tau}F_1(s)$.

Let's substitute this into the expression for $G(s)$:
$G(s) = \frac{(e^{-s\tau}F_1(s))F_{1}^*(s)}{|F_{1}(s)|^{2}} = e^{-s\tau} \frac{F_1(s)F_{1}^*(s)}{|F_{1}(s)|^{2}}$
A fundamental property of complex numbers is that the product of a function and its conjugate equals the square of its magnitude, i.e., $F_1(s)F_{1}^*(s) = |F_{1}(s)|^{2}$.

This allows the expression to simplify dramatically, as the numerator and denominator cancel out:
$G(s) = e^{-s\tau} \frac{|F_{1}(s)|^{2}}{|F_{1}(s)|^{2}} = e^{-s\tau}$
The inverse Laplace transform of $e^{-s\tau}$ is a time-shifted impulse (Dirac delta function) located at $t=\tau$, which is $\delta(t-\tau)$.

For a random signal $X$, the variance $\sigma^2$ is related to the mean square value $E[X^2]$ and mean $\mu$ by \sigma^2 = E\[X^2]$ - \mu^2$. Since the signal has zero mean ($\mu=0$), the variance is equal to the mean square value,$\sigma^2 = E[X^2]$. The r.m.s. value is defined as$\sqrt{E$[X^2]$}$, which is therefore equal to the standard deviation$\sigma$.

For a signal uniformly distributed on $[-a, a]$, its probability density function is $f(x) = \frac{1}{2a}$.

We calculate the mean square value (which is also the variance):
$E\$[X^2]\$ = \int_{-a}^{a} x^2 f(x) \,dx = \int_{-a}^{a} x^2 \frac{1}{2a} \,dx$
$E\$[X^2]\$ = \frac{1}{2a} \left[ \frac{x^3}{3} \right]_{-a}^{a} = \frac{1}{2a} \left( \frac{a^3}{3} - \frac{(-a)^3}{3} \right) = \frac{a^2}{3}$
The r.m.s. value is the square root of the mean square value:
$\text{r.m.s.} = \sqrt{E\$[X^2]\$} = \sqrt{\frac{a^2}{3}} = \frac{a}{\sqrt{3}}$

First, we calculate the initial back EMF ($E_{a1}$) using the motor voltage equation: $E_{a1} = V_t - I_{a1}R_a = 220 - (10)(1) = 210$ V.

The problem requires both torque and speed to remain constant. Since torque is proportional to the product of flux and armature current ($T \propto \phi I_a$), a 10% reduction in flux ($\phi_2 = 0.9\phi_1$) means the armature current must increase to maintain the same torque: $I_{a2} = I_{a1} / 0.9 = 10 / 0.9 \approx 11.11$ A.

Similarly, back EMF is proportional to flux and speed ($E_a \propto \phi N$). With constant speed, the new back EMF is directly proportional to the new flux: $E_{a2} = E_{a1} (\phi_2/\phi_1) = 210 \times 0.9 = 189$ V.

Finally, we apply the motor voltage equation for the second case, including the extra resistance $R_{ext}$. We can now solve for $R_{ext}$ using the new values:
$E_{a2} = V_t - I_{a2}(R_a + R_{ext})$
$189 = 220 - 11.11(1 + R_{ext})$
This gives an extra resistance of $R_{ext} \approx 1.79 \, \Omega$.

To find the most efficient generation schedule, we must identify the option that results in the minimum transmission line losses. The power loss in a line is proportional to its resistance and the square of the power it carries ($P_{loss} \propto P^2 R$).

First, let's establish the relative resistances of the lines. Since resistance is proportional to length, we can let the resistance of the 25 km line from G1 be $R$. Consequently, the resistance of the 75 km lines from G2 and G3 is $3R$.

The total system loss, $P_L$, is the sum of losses from each generator's path to the load. This gives us a total loss proportional to:
$P_L \propto P_1^2(R) + P_2^2(3R) + P_3^2(3R)$, which we can simplify to $P_L \propto P_1^2 + 3P_2^2 + 3P_3^2$.

Now, we can substitute the generation values from each option (which sum to the 120 MW load) into this expression to find the scenario with the minimum relative loss:

• A: $80^2 + 3(20^2) + 3(20^2) = 6400 + 1200 + 1200 = 8800$
• B: $60^2 + 3(30^2) + 3(30^2) = 3600 + 2700 + 2700 = 9000$
• C: $40^2 + 3(40^2) + 3(40^2) = 1600 + 4800 + 4800 = 11200$
• D: $30^2 + 3(45^2) + 3(45^2) = 900 + 6075 + 6075 = 13050$

Comparing these results, the generation schedule in option A yields the smallest proportional loss value, making it the most efficient way to supply the load.

As the gain $k$ approaches infinity, the roots of the characteristic equation move along the branches of the root locus. These branches terminate at the open-loop zeros or approach infinity along asymptotes.

The system has three poles ($P=3$) and one zero ($Z=1$). This means one branch of the locus will terminate at the finite zero, $s = -2/3$. The other two branches ($P-Z=2$) will approach infinity.

The asymptotes for these two branches intersect the real axis at a centroid given by $\sigma_A = \frac{\Sigma(\text{poles}) - \Sigma(\text{zeros})}{P-Z} = \frac{(0+0-2) - (-2/3)}{2} = -2/3$. The angles of the asymptotes are $\pm 90^\circ$, forming a vertical line at $Re(s) = -2/3$.

Therefore, as $k \to \infty$, one root is at $s=-2/3$, and the other two roots travel along the line $Re(s)=-2/3$. All three roots are in the left-half plane and share nearly the same real part.

The CALL SUB instruction, being 3 bytes long and starting at address $LP$, pushes the address of the next instruction, $LP+3$, onto the stack as the return address. The goal is to change this return point to $LP+DISP+3$. To achieve this, we must modify the address on top of the stack within the subroutine.

The correct sequence first retrieves the original return address from the stack into the H-L register pair with POP H. Then, DAD D adds the displacement value $DISP$ (which was pre-loaded into the D-E pair) to the H-L pair. The result in H-L is now the desired return address, $LP+3+DISP$. Finally, PUSH H places this new address back onto the stack, ensuring the RET instruction sends the program to the intended location.

To find the collector current ($I_C$), we should first determine the maximum possible current that the output circuit can support. This occurs when the transistor is in saturation, acting like a nearly closed switch. This maximum value is the saturation current, $I_{C(sat)}$.

We can find this current by applying Kirchhoff's Voltage Law to the collector-emitter loop. The total voltage difference from the collector's ground connection (0 V) to the emitter's supply (-12 V) is 12 V. This voltage is dropped across the collector resistor ($R_C$) and the transistor itself ($V_{CE}$).

In saturation, the voltage across the transistor is given as $V_{CE(sat)} = 0.2 V$. The remaining voltage must be across the resistor. Using Ohm's Law:
$I_C = I_{C(sat)} = \frac{V_{Total} - V_{CE(sat)}}{R_C} = \frac{(0 - (-12)) - 0.2}{2.2 \, k\Omega} = \frac{11.8 \, V}{2.2 \, k\Omega} \approx 5.36 \, \text{mA}$
While we are given $\beta$ and other base circuit values, they serve to confirm that the base current is strong enough to push the transistor into this saturation mode. Thus, the collector current is limited by the output loop to this saturation value.

The current through the main thyristor, $i_M$, is composed of two parts: the constant load current, $i_L$, and an oscillatory current that charges the commutation capacitor, $C$. The peak of this oscillatory current is $I_P = V_s \sqrt{C/L}$. Plugging in the values, we get $I_P = 200 \sqrt{0.1 \times 10^{-6} / (1 \times 10^{-3})} = 2$ A. The maximum current through thyristor M occurs when this peak adds to the maximum load current: $i_{M,max} = i_{L,max} + I_P = 10 \text{ A} + 2 \text{ A} = 12 \text{ A}$.

The auxiliary thyristor, A, is used to turn off the main thyristor. When A is triggered, it must immediately carry the entire load current that was previously flowing through M. Therefore, the maximum current through the auxiliary thyristor is simply the maximum load current: $i_{A,max} = i_{L,max} = 10 \text{ A}$.