GATE EE 2025

The word "yesterday" establishes that the entire sentence must be in the past tense. The first blank describes Kavya's inability to go to work due to her health. "Couldn't" (the past tense of "cannot") is the best choice to express this inability. The second blank explains the reason for her absence. Since this reason also occurred "yesterday," it must use the past tense form "wasn't." The complete sentence correctly connects the inability to work with the past state of being unwell.

This analogy is built on the relationship of antonyms, or direct opposites. The first pair, "Good" and "Evil", represents opposing concepts. To complete the second pair, we must find the word that is the opposite of "Genuine". "Genuine" signifies something that is authentic and real. The word "Counterfeit" refers to a fraudulent imitation or a fake, making it the direct antonym of "Genuine". Thus, the relationship "Genuine : Counterfeit" perfectly mirrors the opposition seen in "Good : Evil".

To determine the values of $p$ and $q$, we first rearrange the given equation $x + py + q = 0$ into the familiar slope-intercept form, $y = mx + c$. This gives us $y = (-\frac{1}{p})x - \frac{q}{p}$, where the slope is $m = -\frac{1}{p}$ and the y-intercept is $c = -\frac{q}{p}$.

From the graph, we can see the line passes through the points $(0, 4)$ and $(2, 8)$. The y-intercept is therefore $c=4$. The slope, or "rise over run," is $m = \frac{8-4}{2-0} = 2$.

Now, we equate the information from the graph with our rearranged equation. Equating the slopes gives us $2 = -\frac{1}{p}$, which solves to $p = -\frac{1}{2}$. Next, we use the y-intercept: $4 = -\frac{q}{p}$. Substituting our value for $p$, we get $4 = -\frac{q}{-1/2}$, which simplifies to $4 = 2q$. Solving for $q$ gives us $q = 2$.

This question requires matching the relationship between three items in each row of Column-I with the appropriate Venn diagram in Column-II.

1. (1) Animals, Zebra, Giraffe: Zebras and Giraffes are two distinct types of animals. This relationship is shown by two separate, non-overlapping circles (Zebra, Giraffe) contained within a larger circle (Animals), which corresponds to diagram (Q).
2. (2) Director, Producer, Actor: A person can be a director, a producer, an actor, or any combination of the three. This indicates that the three sets can overlap, which is represented by the three intersecting circles in diagram (R).
3. (3) Word, Sentence, Novel: This represents a hierarchical structure. A novel is composed of sentences, and sentences are composed of words. This relationship of containment is best illustrated by the three concentric circles in diagram (S).
4. (4) Pianist, Guitarist, Instrumentalist: Pianists and Guitarists are both types of instrumentalists. Since a person can be proficient in both instruments, the sets for Pianist and Guitarist can overlap. Both sets are contained within the larger set of Instrumentalists, matching diagram (P).

Therefore, the correct matching is (1)-(Q), (2)-(R), (3)-(S), and (4)-(P).

To simplify this expression, our first goal is to work with a common base. We can rewrite the number $27$ as $3^3$.
This allows us to re-express the denominator, $27^4$, as $(3^3)^4$. Using the power of a power rule, this becomes $3^{3 \times 4} = 3^{12}$.

Now the expression is $\left(\frac{3^{81}}{3^{12}}\right)^{\frac{1}{3}}$.
Next, we simplify the fraction inside the parentheses by subtracting the exponents: $3^{81-12} = 3^{69}$.
The final step is to raise this result to the power of $\frac{1}{3}$. We multiply the exponents: $(3^{69})^{\frac{1}{3}} = 3^{69 \times \frac{1}{3}}$, which gives us the final answer of $3^{23}$.

To build a coherent paragraph, we should start with a general statement and then provide specific details.

Sentence Q makes the best opening by introducing the main topic: tigers' constant search for food. Sentence S elaborates on this search, describing the general hunting process. Sentence R then gives a specific example of this process-chasing deer-and introduces the possibility of failure. Finally, sentence P explains the reason for this failure, with the pronoun "It" clearly referring to the unsuccessful chase mentioned in R. This logical progression from a general idea to a specific example and its explanation confirms the sequence $\mathrm{Q} \rightarrow \mathrm{S} \rightarrow \mathrm{R} \rightarrow \mathrm{P}$.

The logic connects letters to numbers by multiplying their alphabetical positions (e.g., A=1, B=2). For the given code EF, the positions are $E=5$ and $F=6$, and their product is $5 \times 6 = 30$. Similarly, for HJ, the positions are $H=8$ and $J=10$, and their product is $8 \times 10 = 80$.

In the figure, diametrically opposite sectors are paired according to this rule. The entry (i) is opposite the number 35. To find the letters for (i), we find factors of 35, which are 5 and 7. The 5th letter is E and the 7th is G, so (i) is EG.

The entry (ii) is opposite the letters FG. To find the number for (ii), we multiply the positions of F and G. This gives us $F(6) \times G(7) = 42$. Therefore, the appropriate entries are (i) EG and (ii) 42.

To identify the incorrect statement, let's test the claims made in each option.

First, we'll calculate the averages. P's average is $\frac{383}{7} \approx 54.71$, and Q's average is $\frac{392}{7} = 56$. Since P's average is lower, statement A is true. Next, to find the median, we sort the scores. For P, the sorted list is $22, 39, 45, \textbf{50}, 60, 78, 89$, giving a median of $50$. For Q, the sorted list is $35, 45, 50, \textbf{56}, 60, 65, 81$, with a median of $56$. Because the medians are different, statement B is false.

For completeness, let's check the other options. The score range for P is $89 - 22 = 67$, which is greater than Q's range of $81 - 35 = 46$. Thus, statement C is true. Finally, as we already calculated, Q's average and median are both $56$, making statement D true.

Let's place a reference sphere at the origin, $(0,0,0)$. The centers of all other spheres are located at points $(l, m, n)$ with integer coordinates. The distance from our reference sphere to any other is given by the 3D distance formula, $\sqrt{l^2 + m^2 + n^2}$.

To find the nearest distance, $x$, we seek the smallest non-zero outcome of this formula. This occurs when one coordinate is $\pm 1$ and the others are zero, like at $(1,0,0)$. This gives $x = \sqrt{1^2 + 0^2 + 0^2} = 1$.

The next nearest distance, $y$, is the second smallest distance. This is found when two coordinates are $\pm 1$ and the third is zero, such as the point $(1,1,0)$. This gives $y = \sqrt{1^2 + 1^2 + 0^2} = \sqrt{2}$.

Therefore, the required ratio is $\frac{y}{x} = \frac{\sqrt{2}}{1} = \sqrt{2}$.

Since $ST \parallel QR$, triangle $PST$ is similar to triangle $PQR$. This means the ratio of their corresponding sides is the same, so $\frac{PS}{PQ} = \frac{PT}{PR}$. From the given ratio $PT:TR=3:2$, we can determine that $\frac{PT}{PR} = \frac{3}{3+2} = \frac{3}{5}$. Therefore, the ratio of the diameters of the two semicircles is $\frac{PS}{PQ} = \frac{3}{5}$.

The area of a semicircle is proportional to the square of its diameter. So, the ratio of the areas of the two semicircles is $\frac{\text{Area on PS}}{\text{Area on PQ}} = (\frac{3}{5})^2 = \frac{9}{25}$. Let's say the area of the semicircle on PS is $9A$ and the area of the semicircle on PQ is $25A$ for some constant $A$. The shaded region is the difference between these two areas, which is $25A - 9A = 16A$. We can now compare the shaded area ($16A$) to the area of the smaller semicircle ($9A$), giving the ratio $\frac{16A}{9A} = \frac{16}{9}$.

To find the points that minimize the function $f(x, y) = (x + y - 1)^2 + (x + y)^2$, we can simplify the problem by noticing the repeated term $x+y$. Let's set a new variable $u = x+y$. The function can then be rewritten in terms of this single variable as $g(u) = (u-1)^2 + u^2$.

Expanding this expression gives $g(u) = u^2 - 2u + 1 + u^2 = 2u^2 - 2u + 1$. This is the equation of an upward-opening parabola, which has a minimum value at its vertex. The vertex of a parabola $au^{2+}bu+c$ occurs at $u = -b/(2a)$, which for our function is $u = -(-2)/(2 \cdot 2) = 1/2$.

This means the original function $f(x, y)$ is minimized whenever the condition $x+y = 1/2$ is satisfied. The set $S$ is the collection of all points $(x, y)$ on the line defined by the equation $x+y = 1/2$. Since a line in the plane contains an infinite number of points, the set $S$ is infinite.

This question tests a core property of real symmetric matrices. A key theorem in linear algebra states that eigenvectors corresponding to distinct eigenvalues of a real symmetric matrix are always orthogonal.

In this problem, $v_1$ and $v_2$ are given as such eigenvectors. The mathematical definition of orthogonality for two non-zero vectors is that their dot product is zero. For column vectors, the dot product is expressed as the transpose of the first vector multiplied by the second.

Therefore, the orthogonality of $v_1$ and $v_2$ means their dot product must be zero, which is written as $v_1^T v_2 = 0$.

To determine the number of solutions, we can analyze the augmented matrix $[A|b]$ using row operations.
We begin with

. By adding the first row to the second, we get a row of all zeros, which indicates that the original equations are dependent.

After rearranging to reach row-echelon form,

, we can see the system is consistent because we do not have a contradiction like $0=k$.

The rank of the coefficient matrix $A$ and the augmented matrix $[A|b]$ is 2 (the number of non-zero rows). Since this rank is less than the number of variables (which is 3), the system must have a free variable. A free variable allows for an infinite number of solutions.

To find the correct relationship, we first need to compute the matrix $P^2$ by multiplying $P$ by itself.

$P^2 = P \cdot P = \begin{pmatrix} 2 & 1 & 0 \\ -1 & 0 & 0 \\ 0 & 0 & 1 \end{pmatrix} \begin{pmatrix} 2 & 1 & 0 \\ -1 & 0 & 0 \\ 0 & 0 & 1 \end{pmatrix} = \begin{pmatrix} 3 & 2 & 0 \\ -2 & -1 & 0 \\ 0 & 0 & 1 \end{pmatrix}$

Now, we evaluate the expression in option A, $2P - I$, to see if it yields the same result. This involves scalar multiplication and matrix subtraction.

$2P - I = 2\begin{pmatrix} 2 & 1 & 0 \\ -1 & 0 & 0 \\ 0 & 0 & 1 \end{pmatrix} - \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} = \begin{pmatrix} 4 & 2 & 0 \\ -2 & 0 & 2 \end{pmatrix} - \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}$

$= \begin{pmatrix} 3 & 2 & 0 \\ -2 & -1 & 0 \\ 0 & 0 & 1 \end{pmatrix}$

The resulting matrix is identical to our calculated $P^2$, confirming that the equality $P^2 = 2P - I$ is true.

To find the expected value of $Y$ given that we know $X=1$, we first need to determine the probability distribution of $Y$ under this specific condition. The total probability of the condition $X=1$ occurring is the sum of all joint probabilities where $X=1$:
$P(X=1) = P(X=1, Y=0) + P(X=1, Y=1) = \frac{1}{8} + \frac{1}{8} = \frac{1}{4}$.

Now we can calculate the conditional probabilities for $Y$ by scaling the joint probabilities by $P(X=1)$:
$P(Y=0|X=1) = \frac{P(X=1, Y=0)}{P(X=1)} = \frac{1/8}{1/4} = \frac{1}{2}$
$P(Y=1|X=1) = \frac{P(X=1, Y=1)}{P(X=1)} = \frac{1/8}{1/4} = \frac{1}{2}$

The conditional expectation, $E[Y|X=1]$, is the sum of $Y$'s possible values, each weighted by its corresponding conditional probability:
$E[Y|X=1] = (0) \cdot P(Y=0|X=1) + (1) \cdot P(Y=1|X=1) = 0 \cdot \frac{1}{2} + 1 \cdot \frac{1}{2} = \frac{1}{2}$.

Let's analyze the phase relationship between the input and output voltage signals for these two common amplifier types.

In a Common Source (CS) amplifier, the input is applied at the gate and the output is taken from the drain. A small increase in the input gate voltage ($v_{\in}$) causes an increase in drain current, which leads to a larger voltage drop across the drain resistor. This results in a decrease in the output drain voltage ($v_{out}$). This 180° phase inversion makes the CS amplifier an inverting amplifier.

In contrast, for a Common Gate (CG) amplifier, the input is applied at the source. An increase in the input source voltage results in an increase in the output drain voltage. Since the input and output signals move in the same direction, they are in phase (0° shift), making the CG amplifier a non-inverting amplifier.

Let's analyze the circuit by focusing on the bottom op-amp. The input voltage $V_{\in}$ is applied to its non-inverting (+) input, while the overall output voltage $V_{out}$ is fed back to its inverting (-) input. This arrangement establishes a negative feedback loop across the entire circuit.

For an ideal op-amp in a negative feedback configuration, the "virtual short" principle forces the voltages at the inverting and non-inverting inputs to be equal. Applying this to the bottom op-amp gives us $V_{-} = V_{+}$.

Since $V_{-}$ is connected to $V_{out}$ and $V_{+}$ is connected to $V_{\in}$, we find that $V_{out} = V_{\in}$. A circuit where the output voltage directly follows the input voltage is known as a voltage follower, or buffer.

To find the relationship between the terminal voltage $V_{XY}$ and the current $I_{XY}$, we can apply Kirchhoff's Current Law (KCL) at node X. The total current entering the node, $I_{XY}$, must equal the sum of the currents leaving through the two parallel branches.

The current leaving through the current source is given as 1 A. The current leaving through the resistor is found using Ohm's Law: $I_R = V_{XY} / (1 \text{ k}\Omega)$.

Combining these gives the KCL equation: $I_{XY} = I_R + 1 \text{ A} = \frac{V_{XY}}{1000} + 1$.

This equation, $I_{XY} = (\frac{1}{1000})V_{XY} + 1$, is in the standard form of a straight line, $y = mx + c$. The y-intercept ($c$) is 1 A, and the slope ($m$) is a positive value of $1/1000$. This describes a line that starts at 1 A on the vertical axis and rises with increasing voltage.

A nullator is a conceptual circuit element where both the voltage across it and the current through it are simultaneously zero ($V=0$, $I=0$).

Let's analyze a series combination of a nullator and a resistor, $R$. In a series circuit, the current is the same through all components. Since the nullator forces the current to be zero, the current through the resistor must also be zero.

The total voltage across the series pair is the sum of the individual voltages. The voltage across the nullator is $0$ by definition. The voltage across the resistor is $V_R = I \times R$. Since $I=0$, then $V_R = 0$.

Thus, the total voltage is $0 + 0 = 0$ and the total current is $0$. As these are the defining properties of a nullator, the entire combination behaves as a single nullator.

For a linear time-invariant (LTI) system, the response to a unit impulse input, $S\{\delta[n]\}$, defines the system's impulse response, $h[n]$. From the problem statement, we can write this response as the sum of three impulses: $h[n] = \delta[n] + \delta[n-1] + \delta[n-2]$.

The output $y[n]$ of any LTI system is found by convolving the input signal $x[n]$ with the impulse response $h[n]$.
$y[n] = x[n] * h[n]$
Substituting our expression for $h[n]$ gives:
$y[n] = x[n] * (\delta[n] + \delta[n-1] + \delta[n-2])$
Using the distributive property of convolution and the sifting property of the impulse function, which states that $x[n] * \delta[n-k] = x[n-k]$, the output becomes:
$y[n] = x[n] + x[n-1] + x[n-2]$

This problem is a direct application of the sifting property of the Dirac delta function, which states that $\int_{-\infty }^{\infty } f(t) \delta(t-t_0) dt = f(t_0)$. The integral effectively "sifts out" the value of the function $f(t)$ at the point where the impulse is non-zero.

In this case, the impulse is $\delta(t+3)$, which is located at $t=-3$. Therefore, the integral evaluates to the value of the signal $x(t)$ at $t=-3$.

We find $x(-3)$ by substituting $t=-3$ into the given expression for $x(t)$:
$x(-3) = -(-3)^2 [u(-3+4) - u(-3-4)]$
$x(-3) = -9 [u(1) - u(-7)]$

By definition, the unit step function $u(1)=1$ and $u(-7)=0$. Substituting these values gives us the final answer:
$x(-3) = -9(1-0) = -9$

The step response of a first-order system is modeled by the equation $y(t) = K(1 - e^{-t/\tau})$, where $K$ is the final steady-state value and $\tau$ is the time constant. From the data table, we see the output approaches a steady-state value of 3 as time goes to infinity, so $K=3$.

A defining characteristic of the time constant, $\tau$, is that it's the time required for the output to reach $1 - e^{-1}$, or approximately $63.2\%$, of its final value. For this system, that target output is $y(\tau) = 0.632 \times K = 0.632 \times 3 \approx 1.896$.

Looking at the data, the output is $1.65$ at $t=1.6$ sec and $2.18$ at $t=2.6$ sec. Since our target output of $1.896$ lies between these two measured values, the time constant $\tau$ must lie between $1.6$ and $2.6$ seconds. Of the options provided, only 2 sec falls within this range.

The stability of the closed-loop system is determined by the Nyquist stability criterion, which is expressed as $N = Z - P$. Here, $N$ is the number of anti-clockwise encirclements of the point $-1$, $P$ is the number of right-half plane (RHP) poles of the open-loop transfer function $G(s)$, and $Z$ is the number of RHP poles of the closed-loop system.

From the problem statement, we know the plot encircles $-1$ once in the anti-clockwise direction, so $N = +1$. We are also told that $G(s)$ is strictly stable, which means it has no poles in the RHP, so $P = 0$.

Substituting these values into the Nyquist equation gives us $1 = Z - 0$, which means $Z = 1$. Since $Z$ represents the number of poles of the closed-loop system in the RHP, having $Z=1$ indicates that the closed-loop system has one unstable pole and is therefore unstable.

In a load-flow study, each node (bus) is categorized by its known electrical parameters. For a load bus (PQ), we specify its real power $P$ and reactive power $Q$. For a generator bus (PV), we specify its real power $P$ and voltage magnitude $|V|$. The Slack bus is unique; it serves as the system's reference point with a specified voltage magnitude $|V|$ and phase angle $\delta$. It must supply the balancing power to cover all loads and losses. During a power failure, the UPS acts as the sole source, dictating the voltage and supplying whatever power the connected appliances demand. This function perfectly matches the role of a Slack bus in a load-flow formulation.

To determine the electric field, we apply Gauss's Law. Inside the electron cloud ($0 \leq R \leq a$), consider a spherical Gaussian surface of radius $R$. The charge enclosed within this surface is the volume charge density multiplied by the volume: $Q_{enc} = \rho \cdot V = (-3\rho_0)(\frac{4}{3}\pi R^3)$.

From Gauss's Law, the electric flux is $E_R (4\pi R^2) = \frac{Q_{enc}}{\varepsilon_0}$. Substituting for $Q_{enc}$, we get $E_R (4\pi R^2) = \frac{(-3\rho_0)(\frac{4}{3}\pi R^3)}{\varepsilon_0}$.

Solving for the radial electric field $E_R$ gives $E_R = -\frac{\rho_0 R}{\varepsilon_0}$. This shows that inside the cloud, the electric field is negative and its magnitude increases linearly with radius $R$. Outside the cloud ($R>a$), the field behaves like that of a point charge, decaying as $1/R^2$. This combined behavior is correctly depicted in Figure (iii).

To achieve the fastest possible change in speed, the motor's angular acceleration must be maximized. The rotational dynamic equation is $J \frac{d\omega}{dt} = T_{em} - T_L$, where the acceleration $\frac{d\omega}{dt}$ is proportional to the net torque. To maximize this acceleration, we must maximize the electromagnetic torque, $T_{em}$.

The shaded region OQRE represents all valid operating points. The upper boundary of this region, the line segment QR, corresponds to the maximum possible torque the motor can develop. Therefore, to minimize the time taken, the motor should operate along this maximum torque limit. The trajectory O-Q-R-S follows this principle, ensuring the highest possible acceleration throughout the transition.

The input power factor (IPF) is the product of the displacement power factor (DPF) and the distortion factor ($g$). The DPF is the cosine of the phase angle between the fundamental voltage and current, which is $\cos(\pi/6) \approx 0.866$.

The distortion factor is the ratio of the RMS fundamental current ($I_{1,rms}$) to the total RMS current ($I_{rms}$). First, we find the total RMS current by taking the square root of the sum of squares of the individual harmonic RMS values:
$I_{rms} = \sqrt{\left(\frac{10}{\sqrt{2}}\right)^2 + \left(\frac{2}{\sqrt{2}}\right)^2 + \left(\frac{1}{\sqrt{2}}\right)^2} = \sqrt{\frac{100+4+1}{2}} \approx 7.246 \text{ A}$.

The distortion factor is then $g = \frac{I_{1,rms}}{I_{rms}} = \frac{10/\sqrt{2}}{7.246} \approx 0.976$.

Finally, we calculate the input power factor:
$\text{IPF} = g \times \text{DPF} = 0.976 \times 0.866 \approx 0.845$.

Connecting an alternator to the grid, a process called synchronization, requires matching key electrical parameters. A voltmeter is essential to verify that the alternator's output voltage is equal to the grid's voltage. The synchroscope is a specialized instrument used to compare the frequency and phase angle of the alternator with the grid. It indicates whether the alternator is running fast or slow and helps the operator close the circuit breaker at the precise moment the phase difference becomes zero, ensuring a seamless connection. Wattmeters and stroboscopes are not used for this purpose.

A fundamental rule for constructing a root locus is that a point on the real axis is part of the locus if the total number of real poles and zeros to its right is odd. The given open-loop transfer function has zeros at $s=0$ and $s=-2$, and poles at $s=-5$ and $s=-7$.

Let's test the given points:

• The point $s=-1$ has one zero ($s=0$) to its right. Since 1 is odd, $s=-1$ is on the root locus.
• The point $s=-6$ has three poles/zeros ($s=0, -2, -5$) to its right. Since 3 is odd, $s=-6$ is also on the root locus.
  The other options, $s=-4$ and $s=-10$, have an even number of poles and zeros to their right (2 and 4, respectively).

The expression $\frac{1}{T} \int_{0}^{T}| x(t)| ^{2} dt$ is the definition of the average power of a periodic signal. By Parseval's theorem, we can find this total power by summing the powers of the individual orthogonal components of the signal. The signal $x(t)$ is given as a sum of a DC component and three cosine terms, which are all orthogonal over the period.

The power of a DC component $c_0$ is $c_0^2$. The average power of a sinusoid $A \cos(\omega t + \phi)$ is $\frac{A^2}{2}$.

Let's calculate the power for each component of $x(t)=1+2 \cos 2 \pi t+2 \cos 4 \pi t+2 \cos 6 \pi t$:

• Power of the DC component ($1$): $P_{DC} = 1^2 = 1$.
• Power of each of the three cosine terms (all with amplitude $A=2$): $P_{\cos} = \frac{2^2}{2} = 2$.

The total average power is the sum of the powers of these four components:
$P_{avg} = P_{DC} + P_{\cos1} + P_{\cos2} + P_{\cos3} = 1 + 2 + 2 + 2 = 7$.

First, we calculate the nominal equivalent resistance, $R_{eq}$, for the two resistors connected in parallel:
$R_{eq} = \frac{R_1 R_2}{R_1 + R_2} = \frac{100 \times 900}{100 + 900} = 90 \, \Omega$
To find the maximum percentage error, we must determine the largest possible value for the equivalent resistance. This occurs when both individual resistors are at their maximum value due to the 5% error, which are $R_{1,max} = 105 \, \Omega$ and $R_{2,max} = 945 \, \Omega$.

The maximum equivalent resistance, $R_{eq,max}$, is therefore:
$R_{eq,max} = \frac{105 \times 945}{105 + 945} = 94.5 \, \Omega$
The maximum percentage error is the relative difference between this maximum value and the nominal value:
$\% \text{ Error} = \frac{R_{eq,max} - R_{eq}}{R_{eq}} \times 100 = \frac{94.5 - 90}{90} \times 100 = 5\%$
Rounding to the nearest integer, the answer is 5.

Excellent question. Here's a clearer, more instructive explanation of the solution.

The maximum voltage drop along a feeder occurs when the power factor angle of the load, $\phi$, is equal to the impedance angle of the feeder itself, $\theta$. The feeder's impedance angle is determined by its ratio of reactance ($X$) to resistance ($R$).

We are given that the feeder's $R/X$ ratio is 5. Therefore, the ratio $X/R$ is $1/5$. The impedance angle is:
$\theta = \tan^{-1}\left(\frac{X}{R}\right) = \tan^{-1}\left(\frac{1}{5}\right) \approx 11.31^\circ$

For maximum voltage drop, we set the load's power factor angle $\phi = \theta$. The power factor is simply the cosine of this angle.
$\text{Power Factor} = \cos(\phi) = \cos(11.31^\circ) \approx 0.981$

Here is a clearer, more pedagogical explanation:

To solve this, we first need to calculate the current flowing into the symmetrical fault at bus 2. We can find this using the Thévenin equivalent circuit looking into bus 2. The pre-fault voltage is assumed to be $1.0$ pu, and the total impedance is the sum of the bus's self-impedance ($Z_{22}$) and the fault impedance ($Z_f$).

The fault current ($I_f$) is therefore:
$I_f = \frac{V_{pre-fault}}{Z_{22} + Z_f} = \frac{1.0}{j0.093 + j0.007} = \frac{1.0}{j0.1} = -j10 \text{ pu}$

Now, we can find the voltage at bus 1 during the fault ($V_{1,fault}$). This is the pre-fault voltage at bus 1 minus the voltage drop caused by the fault current, which is determined by the transfer impedance between bus 1 and bus 2 ($Z_{12}$).

$V_{1,fault} = V_{1,pre-fault} - Z_{12} \times I_f$
$V_{1,fault} = 1.0 - (j0.061)(-j10) = 1.0 - 0.61 = 0.39 \text{ pu}$

To find the power delivered by the current source, we need to calculate the product of the constant current $I$ and the average voltage $V_{S,avg}$ across the source's terminals.

The circuit's behavior depends on the MOSFET's state. For a fraction $D=0.4$ of the switching period, the MOSFET is ON, creating a short circuit. During this time, the voltage across the current source is $0 \text{ V}$.

For the remaining fraction of the period, $(1-D)=0.6$, the MOSFET is OFF. The current $I$ is then forced to flow through the diode and the voltage source, clamping the voltage across the current source to $V=15 \text{ V}$.

The average voltage across the current source is the time-weighted average of these two states:
$V_{S,avg} = (0 \text{ V} \times D) + (V \times (1-D)) = (15 \text{ V}) \times (1-0.4) = 9 \text{ V}$.

Finally, the power delivered by the source is $P = V_{S,avg} \times I = 9 \text{ V} \times 10 \text{ A} = 90 \text{ W}$.

Let's begin by calculating the per-phase terminal voltage, $V_{ph}$, for this star-connected motor. The phase voltage is the line voltage divided by the square root of three:
$V_{ph} = \frac{3.3 \times 10^3 \text{ V}}{\sqrt{3}} \approx 1905.25 \text{ V}$.
We are told that the induced EMF, $E_{ph}$, has the same magnitude, so $|E_{ph}| = 1905.25 \text{ V}$.

When a load is applied, the rotor's mechanical displacement of $\theta_m = 2^\circ$ corresponds to an electrical power angle, $\delta$. We find this using the relation $\delta = \frac{P}{2} \theta_m$, where $P$ is the number of poles.
$\delta = \frac{4}{2} \times 2^\circ = 4^\circ$.

In a synchronous motor, the armature current, $I_a$, is determined by the phasor difference between the terminal voltage and the induced EMF, divided by the synchronous impedance. Since resistance is negligible, the impedance is just $jX_s$. Taking $V_{ph}$ as our reference phasor ($V_{ph} \angle 0^\circ$), the induced EMF $E_{ph}$ will lag by the power angle $\delta$, giving us $E_{ph} \angle -\delta$.

The current is then calculated as:
$I_a = \frac{V_{ph} \angle 0^\circ - E_{ph} \angle -\delta}{jX_s} = \frac{1905.25 \angle 0^\circ - 1905.25 \angle -4^\circ}{j2}$.

Evaluating this expression yields a current phasor whose magnitude is $|I_a| \approx 66.49 \text{ A}$.