GATE EE 2024

Q11GATE 2024MCQ1MEngineering Mathematics

Which one of the following matrices has an inverse?

📖 Explanation

A key principle in linear algebra is that a matrix has an inverse if and only if its determinant is non-zero. A quick way to identify a matrix with a zero determinant is to spot linearly dependent rows or columns.

In matrices A, B, and D, one row is a direct multiple of another:

In A: Row 3 is $0.5$ times Row 1.
In B: Row 2 is $2$ times Row 1.
In D: Row 3 is $3$ times Row 1.

This linear dependence guarantees that their determinants are all $0$ , so they cannot be inverted. Matrix C does not have linearly dependent rows, meaning its determinant is non-zero, and thus it has an inverse.

Q12GATE 2024MCQ1MElectric Circuits

The number of junctions in the circuit is

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📖 Explanation

In an electrical circuit, a junction is a point where the leads of three or more components are connected. We can find the total number of junctions by systematically scanning the diagram for these connection points.

There are two junctions where four components meet:

The explicit dot in the very center of the circuit.
The entire bottom wire, which acts as a single common node for four components.

Additionally, there are four junctions where three components meet:

The T-intersection at the top left.
The T-intersection at the top right.
The T-intersection directly above the center point.
The T-intersection directly below the center point.

In total, there are $2 + 4 = 6$ junctions in the circuit.

Q13GATE 2024MCQ1MElectric Circuits

All the elements in the circuit are ideal. The power delivered by the $10 \mathrm{~V}$ source in watts is

🚀 Solve in Practice Mode

📖 Explanation

To determine the power delivered by the $10 \mathrm{~V}$ source, we must find the current flowing out of its positive terminal. We can solve this using the superposition theorem by considering each source independently.

First, let's find the current due to the $10 \mathrm{~V}$ source alone by replacing the $10 \mathrm{~A}$ current source with an open circuit. The circuit becomes a simple series loop, and the current $I_1$ is $I_1 = \frac{10 \mathrm{~V}}{(\alpha + 1)\Omega}$ . This current flows clockwise.

Next, we find the current due to the $10 \mathrm{~A}$ source alone by replacing the $10 \mathrm{~V}$ source with a short circuit. The $10 \mathrm{~A}$ current flows towards the central junction, where it splits. Using the current divider rule, the current $I_2$ flowing through the branch with the $\alpha \Omega$ resistor is $I_2 = 10 \mathrm{~A} \cdot \frac{1 \Omega}{(\alpha + 1)\Omega} = \frac{10}{\alpha + 1} \mathrm{~A}$ . This current flows counter-clockwise through the left loop.

The total current $I$ from the $10 \mathrm{~V}$ source is the sum of these two components. Defining clockwise as the positive direction, we get:
$I = I_1 - I_2 = \frac{10}{\alpha + 1} - \frac{10}{\alpha + 1} = 0 \mathrm{~A}$ .

Since the total current flowing from the $10 \mathrm{~V}$ source is zero, the power delivered by it is $P = V \cdot I = 10 \mathrm{~V} \times 0 \mathrm{~A} = 0 \mathrm{~W}$ .

Q14GATE 2024MCQ1MElectric Circuits

The circuit shown in the figure with the switch $S$ open, is in steady state. After the switch $S$ is closed, the time constant of the circuit in seconds is

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📖 Explanation

To determine the time constant of the circuit after the switch closes, we must find the equivalent resistance ( $R_{eq}$ ) and equivalent inductance ( $L_{eq}$ ) of the network. The time constant $\tau$ is then given by their ratio.

First, let's calculate the total equivalent inductance. The two $1 \text{ H}$ inductors on the far right are in parallel, giving an equivalent of $(1 \| 1) = 0.5 \text{ H}$ . This combination is effectively in series with the other two $1 \text{ H}$ inductors in the circuit. Therefore, the total inductance is $L_{eq} = 1 \text{ H} + 1 \text{ H} + 0.5 \text{ H} = 2.5 \text{ H}$ .

Next, we find the total equivalent resistance. The two $1 \text{ } \Omega$ resistors are effectively in series in the path of the current, so the total resistance is $R_{eq} = 1 \text{ } \Omega + 1 \text{ } \Omega = 2 \text{ } \Omega$ .

Finally, we calculate the time constant using the formula $\tau = \frac{L_{eq}}{R_{eq}}$ .
$\tau = \frac{2.5 \text{ H}}{2 \text{ } \Omega} = 1.25 \text{ s}$ .

Q15GATE 2024MCQ1MSignals and Systems

Suppose signal $y(t)$ is obtained by the time-reversal of signal $x(t)$ , i.e., $y(t)=x(-t)$ , $-\infty \\lt t \\lt \infty$ . Which one of the following options is always true for the convolution of $x(t)$ and $y(t)$ ?

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📖 Explanation

Let the convolution of $x(t)$ and $y(t)$ be denoted by $z(t)$ . Given that $y(t) = x(-t)$ , we can express the convolution as $z(t) = x(t) * x(-t)$ .

To check for symmetry, we evaluate the signal at time $-t$ :
$z(-t) = x(-t) * x(-(-t)) = x(-t) * x(t)$
Convolution is a commutative operation, meaning the order of the signals does not change the result. Therefore, we can write:
$z(-t) = x(t) * x(-t)$
This expression is identical to our original definition of $z(t)$ . Since $z(-t) = z(t)$ , the resulting signal is, by definition, always an even signal.

Q16GATE 2024MCQ1MSignals and Systems

If $u(t)$ is the unit step function, then the region of convergence (ROC) of the Laplace transform of the signal $x(t)=e^{t^{2}}[u(t-1)-u(t-10)]$ is

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📖 Explanation

The signal $x(t)$ is a product of the function $e^{t^2}$ and the term $[u(t-1) - u(t-10)]$ . This second term is a rectangular pulse that is non-zero only for the finite time interval $1 \le t < 10$ . This means $x(t)$ itself is a finite-duration signal.

The Laplace transform is defined by an integral, $X(s) = \int_{t_1}^{t_2} x(t)e^{-st} dt$ . For any finite-duration signal, the limits of this integral are finite. In this case, the integral is $\int_{1}^{10} e^{t^2}e^{-st} dt$ . Since we are integrating a function over a finite interval, the result will converge for any finite value of $s$ . Therefore, the region of convergence is the entire s-plane.

Q17GATE 2024MCQ1MElectrical Machines

A three phase, $50 \mathrm{~Hz}, 6$ pole induction motor runs at $960 \mathrm{rpm}$ . The stator copper loss, core loss, and the rotational loss of the motor can be neglected. the percentage efficiency of the motor is

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📖 Explanation

First, we determine the synchronous speed ( $N_s$ ) of the motor's magnetic field. This is given by the formula $N_s = \frac{120f}{P}$ , where $f$ is the frequency and $P$ is the number of poles.
$N_s = \frac{120 \times 50}{6} = 1000 \text{ rpm}$

Next, we calculate the slip ( $s$ ), which is the fractional difference between the synchronous speed and the actual rotor speed ( $N_r$ ).
$s = \frac{N_s - N_r}{N_s} = \frac{1000 - 960}{1000} = 0.04$

The problem states that all major losses (stator, core, rotational) are neglected. In this simplified scenario, the efficiency ( $\eta$ ) is determined by the relationship between the mechanical output power and the electrical input power to the rotor, which is given by $\eta = 1 - s$ .

Plugging in our value for slip, we find the efficiency:
$\eta = 1 - 0.04 = 0.96$ , or $96\%$ .

Q18GATE 2024MCQ1MElectrical Machines

Which one of the following options represents possible voltage polarities in a single phase two winding transformer? Here, $V_{p}$ is the applied primary voltage, $E_{p}$ is the induced primary voltage, $V_{s}$ is the open circuit secondary voltage, and $E_{s}$ is the induced secondary voltage.

🚀 Solve in Practice Mode

📖 Explanation

Here is a concise explanation of the voltage polarities in a transformer:

Primary Side: In an ideal transformer, the applied primary voltage ( $V_p$ ) must be equal and opposite to the back-EMF it creates. However, the induced voltage ( $E_p$ ) shown in circuit diagrams is conventionally defined as being equal to the applied voltage, so $V_p = E_p$ . This means their polarities are the same. This condition eliminates options (C) and (D), where $V_p$ and $E_p$ have opposite polarities.
Secondary Side: For an open-circuited secondary winding, no current flows. Therefore, the terminal voltage ( $V_s$ ) is exactly equal to the induced electromotive force ( $E_s$ ). Their polarities must be identical ( $V_s = E_s$ ). Both remaining options, (A) and (B), satisfy this.
Dot Convention: The dots indicate terminals that are in phase. The induced voltage at the dotted primary terminal ( $E_p$ ) has the same polarity as the induced voltage at the dotted secondary terminal ( $E_s$ ). In option (B), the dotted primary terminal (bottom) is positive for $E_p$ , and the dotted secondary terminal (top) is also positive for $E_s$ . This is consistent. In option (A), the dotted primary is negative while the dotted secondary is positive, which violates the dot convention.

Q19GATE 2024MCQ1MPower Systems

The figure shows the single line diagram of a 4-bus power network, Branches $b_{1}, b_{2}$ , $b_{3}$ and $b_{4}$ have impedances $4 z, z, 2 z$ and $4 z$ per-unit (pu), respectively, where $z=r+j x$ , with $r>0$ and $x>0$ . The current drawn from each load bus (marked as arrows) is equal to $I$ pu, where $I \neq 0$ . If the network is to operate with minimum loss, the branch that should be opened is

🚀 Solve in Practice Mode

📖 Explanation

To find the configuration with minimum loss, we must compare the total power loss for each case where one branch is opened. When a branch is opened, the ring network becomes a radial one. The total power loss is proportional to $\sum I_{branch}^2 Z_{branch}$ , so we can compare a simplified loss factor for each scenario using the given impedance coefficients and assuming unit load currents ( $I=1$ ).

Open $b_1$ : All three loads are fed in series through path $b_2 \to b_4 \to b_3$ . The currents in the branches are $3I$ , $2I$ , and $I$ respectively. The loss factor is $3^2(1) + 2^2(4) + 1^2(2) = 27$ .
Open $b_2$ : Similar to the first case, the path is now $b_1 \to b_3 \to b_4$ . The loss factor is $3^2(4) + 2^2(2) + 1^2(4) = 48$ .
Open $b_3$ : The network splits into two separate feeders. Path $b_1$ supplies one load ( $I$ ), while path $b_2 \to b_4$ supplies two loads (current is $2I$ in $b_2$ and $I$ in $b_4$ ). The total loss factor is $1^2(4) + 2^2(1) + 1^2(4) = 12$ .

Comparing the calculated loss factors (27, 48, and 12), opening branch $b_3$ yields the minimum value. This configuration is most efficient because it avoids forcing the largest currents through high-impedance lines.

Q20GATE 2024MCQ1MControl Systems

For the block-diagram shown in the figure, the transfer function $\frac{C(s)}{R(s)}$ is

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📖 Explanation

To find the transfer function $\frac{C(s)}{R(s)}$ , let's write out the algebraic relationships for the signals in the diagram.

First, let's define the signal entering the block $G(s)$ as $E(s)$ . The output is then simply $C(s) = G(s)E(s)$ .

Next, we express $E(s)$ in terms of the input $R(s)$ and output $C(s)$ . The first summer calculates $R(s) - C(s)$ . This result enters the second summer with a negative sign, while $C(s)$ enters with a positive sign. Thus, $E(s) = -(R(s) - C(s)) + C(s)$ , which simplifies to $E(s) = -R(s) + 2C(s)$ .

Now, substitute this expression for $E(s)$ back into our output equation:
$C(s) = G(s)[-R(s) + 2C(s)] = -G(s)R(s) + 2G(s)C(s)$ .

Finally, we rearrange the equation to solve for the ratio $\frac{C(s)}{R(s)}$ :
$C(s) - 2G(s)C(s) = -G(s)R(s)$
$C(s)[1 - 2G(s)] = -G(s)R(s)$
$\frac{C(s)}{R(s)} = \frac{-G(s)}{1 - 2G(s)}$

Q21GATE 2024MCQ1MControl Systems

Consider the standard second-order system of the form $\frac{\omega_{n}^{2}}{s^{2}+2 \xi \omega_{n} s+\omega_{n}^{2}}$ with the poles $p$ and $p^{\star}$ having negative real parts. The pole locations are also shown in the figure. Now consider two such second-order systems as defined below : System 1 : $\omega_{n}=3 \mathrm{rad} / \mathrm{sec}$ and $\theta=60^{\circ}$ System 2 : $\omega_{n}=1 \mathrm{rad} / \mathrm{sec}$ and $\theta=70^{\circ}$ Which one of the following statements is correct?

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📖 Explanation

The settling time ( $t_s$ ) is inversely proportional to the real part of the system's poles. For a standard second-order system, the 2% settling time is given by the formula $t_s \approx \frac{4}{\xi \omega_n}$ . The s-plane diagram shows that the damping ratio $\xi$ is related to the pole angle $\theta$ by $\xi = \cos\theta$ .

For System 1, with $\omega_{n1}=3$ and $\theta_1=60^\circ$ , the damping ratio is $\xi_1 = \cos(60^\circ) = 0.5$ . The settling time is $t_{s1} = \frac{4}{\xi_1 \omega_{n1}} = \frac{4}{0.5 \times 3} = \frac{8}{3} \approx 2.67$ seconds.

For System 2, with $\omega_{n2}=1$ and $\theta_2=70^\circ$ , the damping ratio is $\xi_2 = \cos(70^\circ)$ . The settling time is $t_{s2} = \frac{4}{\xi_2 \omega_{n2}} = \frac{4}{\cos(70^\circ) \times 1} \approx 11.69$ seconds.

Comparing these results, it's clear that $t_{s2} > t_{s1}$ .

Q22GATE 2024MCQ1MControl Systems

Consider the cascaded system as shown in the figure. Neglecting the faster component of the trasient response, which one of the following options is a first-order pole-only approximation such that the steady-state values of the unit step responses of the original and the approximated system are same?

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📖 Explanation

To find the simplified model, we first determine the complete transfer function of the cascaded system by multiplying the two blocks:
$G(s) = \frac{1}{s+1} \times \frac{s+40}{s+20} = \frac{s+40}{(s+1)(s+20)}$

This second-order system has poles at $s=-1$ and $s=-20$ . The pole at $s=-1$ is the dominant pole because it is much closer to the imaginary axis, creating a slower response. The pole at $s=-20$ corresponds to a faster transient component, which we are told to neglect.

Our first-order approximation will therefore retain the dominant pole, giving it the form $G_{approx}(s) = \frac{K}{s+1}$ . To ensure the steady-state values match, we equate the DC gains (the value when $s=0$ ) of the original and approximated systems. The DC gain of the original system is $G(0) = \frac{0+40}{(0+1)(0+20)} = 2$ . The DC gain of the approximation is $G_{approx}(0) = K$ . Thus, we set $K=2$ , yielding the approximation $\frac{2}{s+1}$ .

Q23GATE 2024MCQ1MElectrical and Electronic Measurements

The table lists two instrument transformers and their features:

\begin{array}{|l|l|} \hline \text{Instrument Transformers} & \text{Features} \\ \hline & \text{(P) Primary is connected \in parallel to the grid} \\ \hline \text{(X) Current Transformer} (CT) & \text{(Q) Open circuited secondary is not desirable} \\ \hline \text{(Y) Potential Transformer} (PT) & \text{(R) Primary current is the line current} \\ \hline & \text{(S) Secondary burden affects the primary current} \\ \hline \end{array}

The correct matching of the two column is

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Q24GATE 2024MCQ1MDigital Electronics

Simplified form of the Boolean function: $F(P, Q, R, S)=\bar{P} \bar{Q}+\bar{P} Q S+P \bar{Q} \bar{R} \bar{S}+P \bar{Q} R \bar{S}$ is

🚀 Solve in Practice Mode

📖 Explanation

To simplify the Boolean function, we can start by looking for common factors. Notice the last two terms, $P \bar{Q} \bar{R} \bar{S}$ and $P \bar{Q} R \bar{S}$ , share $P\bar{Q}\bar{S}$ . Factoring this out gives us $P\bar{Q}\bar{S}(\bar{R} + R)$ , which simplifies to $P\bar{Q}\bar{S}$ since $\bar{R} + R = 1$ .

The expression is now $F = \bar{P} \bar{Q} + \bar{P} Q S + P\bar{Q}\bar{S}$ . To find more simplifications, we can strategically expand the term $\bar{P}\bar{Q}$ by multiplying it by $(S+\bar{S})$ , which results in $\bar{P}\bar{Q}S + \bar{P}\bar{Q}\bar{S}$ .

Now substitute this back and rearrange the function to group common variables: $F = (\bar{P}\bar{Q}S + \bar{P}QS) + (\bar{P}\bar{Q}\bar{S} + P\bar{Q}\bar{S})$ .

Factoring the first group yields $\bar{P}S(\bar{Q}+Q) = \bar{P}S$ . Factoring the second group yields $\bar{Q}\bar{S}(\bar{P}+P) = \bar{Q}\bar{S}$ .

Combining these two results gives the final simplified expression: $\bar{P}S + \bar{Q}\bar{S}$ .

Q25GATE 2024MCQ1MDigital Electronics

In the circuit, the present value of $Z$ is 1 . Neglecting the delay in the combinatorial circuit, the values of $S$ and $Z$ , respectively, after the application of the clock will be

🚀 Solve in Practice Mode

📖 Explanation

First, let's determine the value of $S$ which acts as the input to the D flip-flop. Using the given present state values, $X=1, Y=0,$ and $Z=1$ , we can calculate the initial value of $S$ :
$S = X \oplus Y \oplus Z = 1 \oplus 0 \oplus 1 = 0$ .

When the clock pulse arrives, the D flip-flop captures this value of $S=0$ . The output of the flip-flop, $Z$ , will then update to become $0$ .

This new value, $Z=0$ , is fed back as an input to the combinatorial circuit. Since the delay is negligible, $S$ is immediately recalculated with this new input:
$S = X \oplus Y \oplus Z = 1 \oplus 0 \oplus 0 = 1$ .

Therefore, after the clock application, the circuit settles to the new state where $S=1$ and $Z=0$ .

Q26GATE 2024MCQ1MDigital Electronics

To obtain the Boolean function $F(X, Y)=X \bar{Y}+\bar{X}$ , the inputs $P Q R S$ in the figure should be

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📖 Explanation

The output of a 4:1 multiplexer with select lines $S_1=X$ and $S_0=Y$ is given by the equation $F = P\bar{X}\bar{Y} + Q\bar{X}Y + RX\bar{Y} + SXY$ . We need to match this to the desired function $F(X, Y)=X\bar{Y}+\bar{X}$ by determining the values for inputs $P, Q, R, S$ . An implementation table helps map the select line combinations to the required inputs.

We evaluate the target function for each combination of $X$ and $Y$ :

When $XY=00$ , $F(0,0)=0\cdot\bar{0}+\bar{0}=1$ . The MUX selects input $P$ , so $P=1$ .
When $XY=01$ , $F(0,1)=0\cdot\bar{1}+\bar{0}=1$ . The MUX selects input $Q$ , so $Q=1$ .
When $XY=10$ , $F(1,0)=1\cdot\bar{0}+\bar{1}=1$ . The MUX selects input $R$ , so $R=1$ .
When $XY=11$ , $F(1,1)=1\cdot\bar{1}+\bar{1}=0$ . The MUX selects input $S$ , so $S=0$ .

Therefore, the required inputs are $PQRS=1110$ .

Q27GATE 2024MCQ1MPower Electronics

If the following switching devices have similar power ratings, which one of them is the fastest?

🚀 Solve in Practice Mode

📖 Explanation

The switching speed of a power device is determined by its turn-on and turn-off times. The Power MOSFET is a unipolar or majority carrier device, meaning its operation depends only on one type of charge carrier (electrons or holes). In contrast, SCRs, GTOs, and IGBTs are all bipolar devices, involving both majority and minority carriers. During turn-off, these bipolar devices experience a significant delay due to the time required to remove stored minority carriers from the base/drift regions. This "storage time" is absent in a MOSFET, allowing it to switch much faster. The MOSFET's speed is primarily limited only by the time needed to charge and discharge its internal gate capacitances ( $C_{gs}$ , $C_{gd}$ ).

Q28GATE 2024MCQ1MPower Electronics

A single-phase triac based $A C$ voltage controller feeds a series $R L$ load. The input $A C$ supply is $230 \mathrm{~V}, 50 \mathrm{~Hz}$ . The values of $\mathrm{R}$ and $\mathrm{L}$ are $10 \Omega$ and $18.37 \mathrm{mH}$ , respectively. The minimum triggering angle of the triac to obtain controllable output voltage is

🚀 Solve in Practice Mode

📖 Explanation

For an AC voltage controller with a series RL load, the output is controllable only when the triggering angle, $\alpha$ , is greater than the load's phase angle, $\theta$ . This condition ( $\alpha > \theta$ ) ensures the triac conducts for a controllable duration. The load phase angle is determined by the circuit's resistance and inductive reactance.

We calculate this angle using the formula $\theta = \tan^{-1}\left(\frac{\omega L}{R}\right)$ , where the angular frequency is $\omega = 2\pi f$ .

Substituting the given values:
$\theta = \tan^{-1}\left(\frac{2\pi(50 \text{ Hz})(18.37 \times 10^{-3} \text{ H})}{10 \Omega}\right)$
This calculation gives $\theta \approx 29.99^{\circ}$ . Therefore, the minimum triggering angle required for controllable output is approximately $30^{\circ}$ .

Q29GATE 2024MSQ1MEngineering Mathematics

Let $X$ be a discrete random variable that is uniformly distributed over the set $\{-10,-9, \ldots, 0, \ldots, 9,10\}$ . Which of the following random variables is/are uniformly distributed?

🚀 Solve in Practice Mode

📖 Explanation

The random variable $X$ can take on 21 distinct integer values, each with a probability of $1/21$ . For a new random variable created by a function of $X$ to also be uniformly distributed, the function must be a one-to-one mapping over $X$ 's domain. This means that each of the 21 values of $X$ must produce a unique output value.

Let's examine the options based on this principle.
A. $X^2$ is not uniform because multiple inputs give the same output. For example, $X=2$ and $X=-2$ both result in $X^2=4$ . Thus, $P(X^2=4) = 2/21$ , while $P(X^2=0) = 1/21$ .
C. $(X-5)^2$ is also not uniform for the same reason. Values of $X$ that are symmetric around $5$ , like $X=4$ and $X=6$ , both produce $(X-5)^2=1$ .

B. $X^3$ is a strictly increasing function, so every distinct value of $X$ maps to a distinct value of $X^3$ . This creates 21 unique outcomes, each with probability $1/21$ .
D. $(X+10)^2$ is also uniform. First, the variable $Y = X+10$ takes on the unique values $\{0, 1, 2, \ldots, 20\}$ . Squaring these distinct non-negative numbers produces 21 unique results, so each has a probability of $1/21$ .

Q30GATE 2024MCQ1MEngineering Mathematics

Which of the following complex functions is/are analytic on the complex plane?

🚀 Solve in Practice Mode

📖 Explanation

To determine if a function is analytic across the entire complex plane, we can test if it satisfies the Cauchy-Riemann equations everywhere. Let's examine the function $f(z) = z^2 - z$ .

First, we express $f(z)$ in terms of its real part, $u(x,y)$ , and imaginary part, $v(x,y)$ , by substituting $z = x + iy$ :
$f(z) = (x+iy)^2 - (x+iy) = (x^2 - y^2 - x) + i(2xy - y)$
So, we have $u = x^2 - y^2 - x$ and $v = 2xy - y$ .

Next, we calculate the partial derivatives:
$u_x = 2x - 1$ , $u_y = -2y$
$v_x = 2y$ , $v_y = 2x - 1$

Finally, we check the Cauchy-Riemann equations: $u_x = v_y$ and $u_y = -v_x$ . We see that $u_x = 2x-1 = v_y$ and $u_y = -2y = -v_x$ . Since these equations hold for all values of $x$ and $y$ , the function is analytic on the entire complex plane.

Q31GATE 2024NAT1MEngineering Mathematics

Consider the complex function $f(z)=\cos z+e^{z^{2}}$ . The coefficient of $z^{5}$ in the Taylor series expansion of $f(z)$ about the origin is _____ (rounded off to 1 decimal place).

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📖 Explanation

To find the coefficient of a specific term in a Taylor series, we can analyze the properties of the function itself. The given function is $f(z) = \cos z + e^{z^{2}}$ .

Notice that both $\cos z$ and $e^{z^{2}}$ are even functions, meaning their value is unchanged when $z$ is replaced by $-z$ . The Taylor series of an even function about the origin contains only even powers of $z$ .
The series for $\cos z$ is $1 - \frac{z^2}{2!} + \frac{z^4}{4!} - \dots$ .
The series for $e^{z^2}$ is $1 + z^2 + \frac{z^4}{2!} + \frac{z^6}{3!} + \dots$ .

Since $f(z)$ is the sum of two series that contain only even powers, its own series expansion will consist solely of even-powered terms. Therefore, the coefficient of any odd-powered term, such as $z^5$ , must be 0.

Q32GATE 2024NAT1MEngineering Mathematics

The sum of the eigenvalues of the matrix

A=\left[\begin{array}{ll}1 & 2 \\ 3 & 4\end{array}\right]^{2}

is _____ (rounded off to the nearest integer).

🚀 Solve in Practice Mode

📖 Explanation

First, we must calculate the matrix $A$ by squaring the given matrix:

A = \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix} \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix} = \begin{pmatrix} 7 & 10 \\ 15 & 22 \end{pmatrix}

A key property of any square matrix is that the sum of its eigenvalues is equal to its trace. The trace is simply the sum of the elements on the main diagonal.

For our calculated matrix $A$ , the trace is the sum of its diagonal entries, $7$ and $22$ .

Therefore, the sum of the eigenvalues is $\text{Trace}(A) = 7 + 22 = 29$ .

Q33GATE 2024NAT1MSignals and Systems

Let $X(\omega)$ be the Fourier transform of the signal, $x(t)=e^{-t^{4}} \cos t,-\infty \lt t \lt \infty$ . The value of the derivative of $X(\omega)$ at $\omega=0$ at is _____ (rounded off to 1 decimal place)

🚀 Solve in Practice Mode

📖 Explanation

We can solve this by using the differentiation in frequency property of the Fourier transform, which states that if $x(t) \leftrightarrow X(\omega)$ , then $t x(t) \leftrightarrow j \frac{d X(\omega)}{d \omega}$ .

Rearranging this property, we see that the derivative we are looking for, $\frac{d X(\omega)}{d \omega}$ , is the Fourier transform of the time-domain signal $\frac{t x(t)}{j}$ .

To find the value of any Fourier transform at $\omega=0$ , we simply calculate the total area under its corresponding time-domain signal. Thus, we need to evaluate the integral $\int_{-\infty }^{\infty } \frac{t x(t)}{j} dt$ .

Let's analyze the function inside the integral. The given signal $x(t)=e^{-t^{4}} \cos t$ is an even function, as it is a product of two even functions. The function we are integrating is $\frac{t}{j} x(t)$ , which is an odd function because it's the product of an odd function ( $t$ ) and an even function ( $x(t)$ ). The integral of any odd function over a symmetric interval from $-\infty$ to $\infty$ is always zero.

Q34GATE 2024NAT1MPower Systems

The incremental cost curves of two generators (Gen A and Gen B) in a plant supplying a common load are shown in the figure. If the incremental cost of supplying the common load is Rs. 7400 per MWh, then the common load in MW is ____ (rounded off to the nearest integer).

🚀 Solve in Practice Mode

📖 Explanation

For the most economical operation, all generators should operate at the same incremental cost, $\lambda$ , which is given as 7400 Rs/MWh.

From the graph, we can find the cost equation for Generator A: $IC_A = 20 P_{GA} + 8000$ . Setting this to $\lambda$ gives $20 P_{GA} + 8000 = 7400$ , which results in a physically impossible negative power output ( $P_{GA} = -30$ MW). This indicates that Generator A's minimum cost (8000 Rs/MWh) is above the operating cost, so it must be shut down ( $P_{GA} = 0$ ).

Therefore, Generator B must supply the entire load. Its cost equation is $IC_B = 40 P_{GB} + 6000$ .
Setting its cost to $\lambda$ yields $40 P_{GB} + 6000 = 7400$ .
Solving for $P_{GB}$ gives an output of $35$ MW. The total load is $P_{GA} + P_{GB} = 0 + 35 = 35$ MW.

Q35GATE 2024NAT1MPower Electronics

A forced commutated thyristorized step-down chopper is shown in the figure. Neglect the ON-state drop across the power devices. Assume that the capacitor is initially charged to $50 \mathrm{~V}$ with the polarity shown in the figure. The load current $\left(I_{L}\right)$ can be assumed to be constant at $10 \mathrm{~A}$ . Initially, $\mathrm{Th}_{\mathrm{M}}$ is $\mathrm{ON}$ and $\mathrm{Th}_{\mathrm{A}}$ is OFF. The turn-off time available to $\mathrm{Th}_{\mathrm{M}}$ in microseconds, when $\mathrm{Th}_{\mathrm{A}}$ is triggered, is _____ (rounded off to the nearest integer).

🚀 Solve in Practice Mode

📖 Explanation

To turn off the main thyristor, $Th_M$ , the auxiliary thyristor, $Th_A$ , is triggered. This connects the capacitor, initially charged to $50 \text{ V}$ , in parallel with $Th_M$ , applying a reverse voltage and forcing it to commutate. The constant load current, $I_L = 10 \text{ A}$ , is now diverted through the capacitor, causing it to discharge.

The turn-off time available to $Th_M$ is the duration for which it remains reverse-biased. This period lasts until the capacitor voltage drops from its initial $50 \text{ V}$ down to zero. We can calculate this time using the capacitor's voltage-current relationship for a constant discharging current:

$t_{off} = \frac{C \cdot V_{C}}{I_L} = \frac{(10 \times 10^{-6} \text{ F}) \cdot (50 \text{ V})}{10 \text{ A}}$

$t_{off} = 50 \times 10^{-6} \text{ s} = 50 \text{ } \mu\text{s}$

Q36GATE 2024MCQ2MEngineering Mathematics

Consider a vector $\bar{u}=2 \hat{x}+\hat{y}+2 \hat{z}$ , where $\hat{x}, \hat{y}, \hat{z}$ represent unit vector along the coordinate axes $x, y, z$ respectively. The directional derivative of the function $f(x, y, z)=2 \ln (x y)+\ln (y z)+3 \ln (x z)$ at the point $(x, y, z)=(1,1,1)$ in the direction of $\bar{u}$ is

🚀 Solve in Practice Mode

📖 Explanation

The directional derivative is the dot product of the gradient of the function $(\nabla f)$ and the unit vector for the given direction $(\hat{u})$ .

First, let's find the gradient of $f$ . It's helpful to simplify the function using logarithm properties: $f(x,y,z) = 2(\ln x + \ln y) + (\ln y + \ln z) + 3(\ln x + \ln z) = 5\ln x + 3\ln y + 4\ln z$ .
The gradient is $\nabla f = \frac{5}{x}\hat{x} + \frac{3}{y}\hat{y} + \frac{4}{z}\hat{z}$ . At the point $(1,1,1)$ , this becomes $\nabla f|_{(1,1,1)} = 5\hat{x} + 3\hat{y} + 4\hat{z}$ .

Next, we find the unit vector for the direction $\bar{u}=2 \hat{x}+\hat{y}+2 \hat{z}$ . The magnitude is $|\bar{u}| = \sqrt{2^{2+}1^{2+}2^2} = \sqrt{9} = 3$ . So, the unit vector is $\hat{u} = \frac{\bar{u}}{|\bar{u}|} = \frac{1}{3}(2\hat{x}+\hat{y}+2\hat{z})$ .

Finally, we compute the dot product of the gradient and the unit vector:
$(5\hat{x} + 3\hat{y} + 4\hat{z}) \cdot \frac{1}{3}(2\hat{x}+\hat{y}+2\hat{z}) = \frac{1}{3}(5 \cdot 2 + 3 \cdot 1 + 4 \cdot 2) = \frac{1}{3}(10+3+8) = \frac{21}{3} = 7$ .

Q37GATE 2024MCQ2MSignals and Systems

The input $x(t)$ and the output $y(t)$ of a system are related as $y(t)=e^{-t} \int_{-\infty }^{t} e^{\tau} x(\tau) d \tau, \quad-\infty \lt t \lt \infty$ The system is

🚀 Solve in Practice Mode

📖 Explanation

The system is linear because integration and scaling are linear operations, and the system is composed of only these.

To check for time-invariance, we must verify if a shifted input $x(t-t_0)$ produces a correspondingly shifted output, $y(t-t_0)$ . Let's find the output, which we'll call $y_{new}(t)$ , for the shifted input $x(t-t_0)$ :
$y_{new}(t) = e^{-t} \int_{-\infty }^{t} e^{\tau} x(\tau - t_0) d\tau$
Now, let's change the variable of integration to $k = \tau - t_0$ . This means $\tau = k+t_0$ , and the upper limit of integration becomes $t-t_0$ .
$y_{new}(t) = e^{-t} \int_{-\infty }^{t-t_0} e^{k+t_0} x(k) dk$
Simplifying by factoring out the constant $e^{t_0}$ gives:
$y_{new}(t) = e^{-t} e^{t_0} \int_{-\infty }^{t-t_0} e^{k} x(k) dk = e^{-(t-t_0)} \int_{-\infty }^{t-t_0} e^{k} x(k) dk$
This final expression is precisely the original system equation for $y(t)$ with every $t$ replaced by $(t-t_0)$ . Thus, $y_{new}(t) = y(t-t_0)$ , and the system is time-invariant.

Q38GATE 2024MCQ2MSignals and Systems

Consider the discrete-time system $T_{1}$ and $T_{2}$ defined as follows:

\begin{aligned} & \left\{T_{1} x\right\}[n]=x[0]+x[1]+\ldots+x[n] \\ & \left\{T_{2} x\right\}[n]=x[0]+\frac{1}{2} x[1]+\cdots+\frac{1}{2^{n}} x[n] \end{aligned}

Which one of the following statement is true?

🚀 Solve in Practice Mode

📖 Explanation

To test for Bounded-Input, Bounded-Output (BIBO) stability, we check if a bounded input can produce an unbounded output.

First, consider system $T_1$ , which is a discrete-time accumulator. If we use a simple bounded input like the unit step function, $x[n] = u[n]$ , the output becomes $\{T_1 x\}[n] = \sum_{k=0}^{n} u[k] = n+1$ . As $n$ increases, this output grows without limit. Therefore, $T_1$ is not BIBO stable.

Next, consider system $T_2$ . Let's assume a general bounded input, where $|x[n]| \leq M$ for some finite constant $M$ . The magnitude of the output is bounded by $|\{T_2 x\}[n]| \le \sum_{k=0}^{n} |(\frac{1}{2})^k x[k]| \le M \sum_{k=0}^{n} (\frac{1}{2})^k$ . This sum is a geometric series that converges to $2$ as $n \to \infty$ . The output is therefore always bounded by $2M$ , proving that $T_2$ is BIBO stable.

Q39GATE 2024MCQ2MSignals and Systems

If the Z-transform of a finite-duration discrete-time signal $x[n]$ is $X(z)$ , then the $Z$ -transform of the signal $y[n]=x[2 n]$ is

🚀 Solve in Practice Mode

📖 Explanation

The process of finding the Z-transform of $y[n]=x[2n]$ , a downsampled signal, begins with the Z-transform definition:
$Y(z) = \sum_{n=-\infty }^{\infty } y[n] z^{-n} = \sum_{n=-\infty }^{\infty } x[2n] z^{-n}$
To relate this to $X(z)$ , which is a sum over all samples of $x[k]$ , we can change the summation index to $k=2n$ . The sum is now over all even integers $k$ , and the term $z^{-n}$ becomes $z^{-k/2}$ :
$Y(z) = \sum_{k \text{ is even}} x[k] (z^{1/2})^{-k}$
A useful technique to convert a sum over even indices to a sum over all indices is to use the expression $\frac{1+(-1)^k}{2}$ , which is $1$ for even $k$ and $0$ for odd $k$ . Applying this "filter" yields:
$Y(z) = \sum_{k=-\infty }^{\infty } x[k] (z^{1/2})^{-k} \left(\frac{1+(-1)^k}{2}\right)$
We can then separate this into two distinct sums:
$Y(z) = \frac{1}{2}\left[\sum_{k=-\infty }^{\infty } x[k](z^{1/2})^{-k} + \sum_{k=-\infty }^{\infty } x[k](-z^{1/2})^{-k}\right]$
By recognizing the definition of the Z-transform, $X(v) = \sum_k x[k]v^{-k}$ , we can identify the two resulting terms, giving the final expression.

Q40GATE 2024MCQ2MElectrical Machines

A 3-phase. $11 \mathrm{kV} .10$ MVA synchronous generator is connected to an inductive load of power factor $(\sqrt{3} / 2)$ via a lossless line with a per-phase inductive reactance of $5 \Omega$ . The per-phase synchronous reactance of the generator is $30 \Omega$ with negligible armature resistance. If the generator is producing the rated current at the rated voltage, then the power factor at the terminal of the generator is

🚀 Solve in Practice Mode

📖 Explanation

First, let's establish the per-phase rated values. The generator's terminal voltage is $V_t = \frac{11 \text{ kV}}{\sqrt{3}} \approx 6351 \text{ V}$ , and the rated current is $I = \frac{10 \text{ MVA}}{\sqrt{3} \times 11 \text{ kV}} = 524.86 \text{ A}$ . The inductive load has a power factor angle of $\phi_L = \cos^{-1}(\sqrt{3}/2) = 30^\circ$ .

The voltage at the generator's terminals, $\vec{V}_t$ , is the sum of the voltage at the load, $\vec{V}_L$ , and the voltage drop across the transmission line: $\vec{V}_t = \vec{V}_L + j\vec{I}X_{line}$ . To solve for the unknown load voltage $V_L$ , let's set the current phasor $\vec{I}$ as the reference ( $524.86 \angle 0^\circ$ ). The load voltage $\vec{V}_L$ then leads the current by $30^\circ$ , so $\vec{V}_L = V_L \angle 30^\circ$ .

We can now write an equation for the magnitude of the known terminal voltage:
$|V_t|^2 = |\vec{V}_L + j\vec{I}X_{line}|^2 = |(V_L\cos30^\circ) + j(V_L\sin30^\circ + I X_{line})|^2$ .
Plugging in the numbers: $(6351)^2 = (V_L\frac{\sqrt{3}}{2})^2 + (\frac{V_L}{2} + 524.86 \times 5)^2$ . Solving this quadratic equation for $V_L$ gives $V_L \approx 4618$ V.

The power factor at the generator terminal is $\cos(\phi_t)$ , where $\phi_t$ is the angle of $\vec{V}_t$ relative to $\vec{I}$ . We find this angle from the components of $\vec{V}_t$ :
$\tan(\phi_t) = \frac{\text{Im}(\vec{V}_t)}{\text{Re}(\vec{V}_t)} = \frac{V_L\sin30^\circ + IX_{line}}{V_L\cos30^\circ} = \frac{4618/2 + 2624.3}{4618 \cdot \sqrt{3}/2} = \frac{4933.3}{3999.4} \approx 1.23$ .
This gives $\phi_t \approx 50.96^\circ$ . The resulting power factor is $\cos(50.96^\circ) \approx 0.63$ lagging.

Q41GATE 2024MCQ2MPower Systems

For the three-bus lossless power network shown in the figure, the voltage magnitudes at all the buses are equal to 1 per unit (pu), and the differences of the voltage phase angles are very small. The line reactances are marked in the figure, where $\alpha, \beta, \gamma$ and $x$ are strictly positive. The bus injections $P_{1}$ and $P_{2}$ are in pu. If $P_{1}=m P_{2}$ , where $m > 0$ , and the real power flow from bus 1 to bus 2 is 0 pu. then which one of the following options is correct?

🚀 Solve in Practice Mode

📖 Explanation

The crucial starting point is that zero real power flows between bus 1 and bus 2. In a lossless line, this can only happen if their voltage phase angles are identical, so we have $\delta_1 = \delta_2$ .

With no power flowing on the line between bus 1 and 2, the power injected at bus 1 ( $P_1$ ) must flow to bus 3. Likewise, the power injected at bus 2 ( $P_2$ ) must also flow to bus 3. Using the standard power flow equation with all voltage magnitudes set to 1 pu, we can write:
$P_1 = P_{13} = \frac{1}{\beta x} \sin(\delta_1 - \delta_3)$
$P_2 = P_{23} = \frac{1}{\gamma x} \sin(\delta_2 - \delta_3)$

Since we established that $\delta_1 = \delta_2$ , the angle terms in both equations are the same. Now, we use the given relationship $P_1 = m P_2$ :
$\frac{1}{\beta x} \sin(\delta_1 - \delta_3) = m \left( \frac{1}{\gamma x} \sin(\delta_1 - \delta_3) \right)$

After canceling the identical $\sin$ and $x$ terms on both sides, we are left with $\frac{1}{\beta} = \frac{m}{\gamma}$ . Rearranging this gives the final relationship: $\gamma = m \beta$ .

Q42GATE 2024MCQ2MAnalog Electronics

A BJT biasing circuit is shown in the figure, where $V_{B E}=0.7 \mathrm{~V}$ and $\beta=100$ . The Quiescent Point values of $V_{C E}$ and $I_{C}$ are respectively

🚀 Solve in Practice Mode

📖 Explanation

To determine the BJT's quiescent point ( $I_C, V_{CE}$ ), we first simplify the base voltage divider using a Thevenin equivalent. The Thevenin voltage is $V_{Th} = 12 \mathrm{~V} \cdot \frac{50k\Omega}{100k\Omega + 50k\Omega} = 4 \mathrm{~V}$ , and the Thevenin resistance is $R_{Th} = 100k\Omega \,||\, 50k\Omega = 33.33 \mathrm{~k\Omega}$ .

Next, we apply KVL to the base-emitter loop: $V_{Th} = I_B R_{Th} + V_{BE} + I_E R_E$ . By substituting $I_E = (\beta+1)I_B$ , we can solve for the base current:
$I_B = \frac{V_{Th} - V_{BE}}{R_{Th} + (\beta+1)R_E} = \frac{4 - 0.7}{33.33k\Omega + (101)(1k\Omega)} = 24.56 \mathrm{~\mu A}$ .

This allows us to find the quiescent collector current, $I_C = \beta I_B = 100 \times 24.56 \mathrm{~\mu A} = 2.46 \mathrm{~mA}$ . We also find the emitter current, $I_E = (\beta+1)I_B = 101 \times 24.56 \mathrm{~\mu A} = 2.48 \mathrm{~mA}$ .

Finally, applying KVL to the collector-emitter loop gives $V_{CE} = V_{CC} - I_C R_C - I_E R_E$ .
$V_{CE} = 12 - (2.46 \mathrm{~mA})(2 \mathrm{~k\Omega}) - (2.48 \mathrm{~mA})(1 \mathrm{~k\Omega}) = 12 - 4.92 - 2.48 \approx 4.6 \mathrm{~V}$ .

Q43GATE 2024MCQ2MEngineering Mathematics

Let $f(t)$ be a real-valued function whose second derivative is positive for $-\infty \\lt t \\lt\infty$ . Which of the following statements is/are always true?

🚀 Solve in Practice Mode

📖 Explanation

The condition $f^{\prime \prime}(t) > 0$ for all $t$ tells us that the function is strictly convex, or "concave up." This means that its first derivative, $f'(t)$ , which represents the function's slope, is always strictly increasing.

For a local minimum to exist at a point $c$ , the derivative $f'(c)$ must be zero. If there were two distinct local minima, say at $t=a$ and $t=b$ , we would need $f'(a)=0$ and $f'(b)=0$ . However, a strictly increasing function like $f'(t)$ can only take on any given value, including zero, at most once. Therefore, having two distinct local minima is impossible.

Q44GATE 2024MCQ2MEngineering Mathematics

Consider the function $f(t)=(\max (0, t))^{2}$ for $-\infty \\ltt\\lt\infty$ , where $\max (a, b)$ denotes the maximum of $a$ and $b$ . Which of the following statements is/are true?

🚀 Solve in Practice Mode

📖 Explanation

Let's begin by expressing the function in a piecewise form. For any value of $t \le 0$ , $\max(0, t) = 0$ , so $f(t) = 0$ . For any $t > 0$ , $\max(0, t) = t$ , so $f(t) = t^2$ .

To check if $f(t)$ is differentiable, we examine its derivative at the point where the definition changes, $t=0$ . The derivative from the left is $0$ . The derivative from the right is given by the derivative of $t^2$ , which is $2t$ . Evaluating this at $t=0$ gives $0$ . Since the left-hand and right-hand derivatives are equal, the function $f(t)$ is differentiable everywhere.

Now, let's consider the derivative function, $f'(t)$ . We found that $f'(t)=0$ for $t \le 0$ and $f'(t)=2t$ for $t > 0$ . This derivative function is continuous. We can check this at $t=0$ : the limit from the left is $0$ , and the limit from the right is $\lim_{t \to 0^+} 2t = 0$ . Because the limits match, $f'(t)$ is continuous.

Q45GATE 2024MSQ2MEngineering Mathematics

Which of the following differential equations is/are nonlinear?

🚀 Solve in Practice Mode

📖 Explanation

A differential equation is linear if the dependent variable, let's call it $x(t)$ , and its derivatives (e.g., $\frac{dx}{dt}$ ) only appear to the first power and are not multiplied together or used as arguments in other functions.

Equations A and C are linear. They can be written in the standard form $a_1(t)\frac{dx}{dt} + a_0(t)x(t) = g(t)$ , where all terms involving $x(t)$ and its derivative are of degree one.
Equation B, $\frac{1}{2} e^{t}+x(t)\frac{d x}{d t}=0$ , is nonlinear because it contains the product of the dependent variable and its derivative, $x(t)\frac{dx}{dt}$ .
Equation D, $x(t)+e^{\left(\frac{d x}{d t}\right)}=1$ , is nonlinear because the derivative $\frac{dx}{dt}$ is the argument of a nonlinear function, the exponential $e^{(\cdot)}$ .

Q46GATE 2024MSQ2MElectric Circuits

For a two-phase network, the phase voltages $V_{p}$ and $V_{q}$ are to be expressed in terms of sequence voltages $V_{\alpha}$ and $V_{\beta}$ as

\left[\begin{array}{l}V_{p} \\ V_{q}\end{array}\right]=S\left[\begin{array}{l}V_{\alpha} \\ V_{\beta}\end{array}\right]

. The possible option(s) for matrix $S$ is/are

🚀 Solve in Practice Mode

📖 Explanation

The transformation from sequence voltages ( $V_{\alpha}, V_{\beta}$ ) to phase voltages ( $V_{p}, V_{q}$ ) is a change of basis, requiring the transformation matrix $S$ to be invertible. Option B,

\left[\begin{array}{cc}1 & 1 \\ 1 & 1\end{array}\right]

, has a determinant of $1-1=0$ , making it non-invertible and thus an invalid choice.

In power system analysis, such transformations are typically chosen to be orthogonal, meaning the column vectors of the matrix are orthogonal to each other. This creates decoupled components which simplifies analysis. Let's check the orthogonality for the remaining invertible options by taking the dot product of their column vectors.

For option A, the dot product of the columns is

\left[\begin{array}{l}1 \\ 1\end{array}\right] \cdot \left[\begin{array}{c}1 \\ -1\end{array}\right] = (1)(1) + (1)(-1) = 0

. This is an orthogonal matrix.

For option D, the dot product is

\left[\begin{array}{c}-1 \\ 1\end{array}\right] \cdot \left[\begin{array}{l}1 \\ 1\end{array}\right] = (-1)(1) + (1)(1) = 0

. This is also an orthogonal matrix.

For option C, the dot product is

\left[\begin{array}{l}1 \\ 1\end{array}\right] \cdot \left[\begin{array}{l}1 \\ 0\end{array}\right] = 1 \neq 0

. This is not an orthogonal transformation.

Both matrices A and D represent valid orthogonal transformations. They correspond to slightly different definitions of the sequence components $V_{\alpha}$ and $V_{\beta}$ , which is a matter of convention. Therefore, both A and D are possible options for the matrix $S$ .

Q47GATE 2024MSQ2MElectrical Machines

Which of the following option is/are correct for the Automatic Generation Control (AGC) and Automatic Voltage Regulator (AVR) installed with synchronous generators?

🚀 Solve in Practice Mode

📖 Explanation

Automatic Generation Control (AGC) regulates the real power output ( $P$ ) by controlling the generator's mechanical power input (e.g., turbine governor). Since the entire interconnected grid shares a common frequency ( $f$ ), any action to correct a real power imbalance has a grid-wide, or global, effect on frequency.

Conversely, the Automatic Voltage Regulator (AVR) controls the generator's terminal voltage ( $V$ ) by adjusting its reactive power output ( $Q$ ). This is achieved by varying the DC field excitation current ( $I_f$ ). Reactive power is difficult to transmit over long distances, so the AVR's effect on voltage is primarily confined to the generator and its local area. Therefore, AGC has a global effect on frequency, and AVR has a local effect on voltage.

Q48GATE 2024NAT2MElectric Circuits

Two passive two-port network $P$ and $Q$ are connected as shown in the figure. The impedance matrix of network $P$ is

Z_{P}=\left[\begin{array}{cc}40 \Omega & 60 \Omega \\ 80 \Omega & 100 \Omega\end{array}\right]

. The admittance matrix of network $Q$ is

Y_{Q}=\left[\begin{array}{cc}5 S & -2.5 S \\ -2.5 S & 1 S\end{array}\right]

. Let the $A B C D$ matrix of the two-port network $R$ in the figure be

\left[\begin{array}{ll}\alpha & \beta \\ \gamma & \delta\end{array}\right]

. The value of $\beta$ in $\Omega$ is _______ (rounded off to 2 decimal places)

🚀 Solve in Practice Mode

📖 Explanation

The overall network R is formed by connecting networks P and Q in cascade. To find the parameters of the combined network, we multiply their individual ABCD (transmission) matrices. Therefore, the first step is to convert the given impedance matrix for P and admittance matrix for Q into their respective ABCD matrices.

For network P, converting the impedance matrix

Z_{P}=\left[\begin{array}{cc}40 & 60 \\ 80 & 100\end{array}\right]

gives the ABCD matrix

T_{P}=\left[\begin{array}{cc}0.5 & -10 \\ 1/80 & 1.25\end{array}\right]

.
For network Q, converting the admittance matrix

Y_{Q}=\left[\begin{array}{cc}5 & -2.5 \\ -2.5 & 1\end{array}\right]

gives the ABCD matrix

T_{Q}=\left[\begin{array}{cc}0.4 & 0.4 \\ -0.5 & 2\end{array}\right]

.

The overall ABCD matrix is the product

T_R = T_P \times T_Q = \left[\begin{array}{ll}\alpha & \beta \\ \gamma & \delta\end{array}\right]

. We are asked to find the value of $\beta$ , which is the element in the first row and second column of the resulting matrix.

This element is calculated by multiplying the first row of $T_P$ with the second column of $T_Q$ :
$\beta = (0.5)(0.4) + (-10)(2) = 0.2 - 20 = -19.8 \, \Omega$ .

Q49GATE 2024NAT2MElectric Circuits

For the circuit shown in the figure, the source frequency is $5000 \mathrm{rad} / \mathrm{sec}$ . The mutual inductance between the magnetically coupled inductors is $5 \mathrm{mH}$ with their self inductances being $125 \mathrm{mH}$ and $1 \mathrm{mH}$ . The Thevenin's impedance. $Z_{\mathrm{th}}$ , between the terminals $P$ and $Q$ in $\Omega$ is _____ (rounded off to 2 decimal places).

🚀 Solve in Practice Mode

📖 Explanation

To determine the Thevenin impedance, $Z_{th}$ , we first find the equivalent impedance of the magnetically coupled section. The total impedance of the secondary loop is the sum of the inductor's and capacitor's impedances:
$Z_{\text{sec}} = j\omega L_2 + \frac{1}{j\omega C} = j(5000)(10^{-3}) - \frac{j}{5000 \cdot 50 \cdot 10^{-6}} = j5 - j4 = j1 \, \Omega$ .

The impedance looking into the primary winding is its self-impedance plus the impedance reflected from the secondary:
$Z_{\text{\in}} = j\omega L_1 + \frac{(\omega M)^2}{Z_{\text{sec}}} = j(5000)(125 \times 10^{-3}) + \frac{(5000 \cdot 5 \times 10^{-3})^2}{j1} = j625 - j625 = 0 \, \Omega$ .

This result shows that the entire coupled-inductor circuit behaves like a short circuit. To find $Z_{th}$ across terminals P and Q, we replace this section with a $0 \, \Omega$ impedance. The circuit simplifies to a $4 \, \Omega$ resistor in series with the parallel combination of a $4 \, \Omega$ and a $2 \, \Omega$ resistor.
$Z_{th} = 4 + (4 \, || \, 2) = 4 + \frac{4 \times 2}{4+2} = 4 + \frac{8}{6} = \frac{16}{3} \approx 5.33 \, \Omega$ .

Q50GATE 2024NAT2MElectrical and Electronic Measurements

In the circuit shown, $Z_{1}=50 \angle-90^{\circ}$ and $Z_{2}=200 \angle-30^{\circ} \Omega$ . It is supplied by a three phase $400 \mathrm{~V}$ source with the phase sequence being R-Y-B. Assume the watt meters $W_{1}$ and $W_{2}$ to be ideal. The magnitude of the difference between the readings of $W_{1}$ and $W_{2}$ in watts is ______ (rounded off to 2 decimal places).

🚀 Solve in Practice Mode

📖 Explanation

To solve this, let's first establish the phase and line voltages. With an R-Y-B sequence and a line voltage of $400 \text{ V}$ , we can set $V_{RY}=400\angle0^\circ \text{ V}$ as our reference. This gives us $V_{YB}=400\angle-120^\circ \text{ V}$ and $V_{RB}=-V_{BR}=400\angle-60^\circ \text{ V}$ .

Next, we calculate the currents flowing through the impedances. The current through $Z_1$ is $I_1 = \frac{V_{RB}}{Z_1} = \frac{400\angle-60^\circ}{50\angle-90^\circ} = 8\angle30^\circ \text{ A}$ . The current through $Z_2$ is $I_2 = \frac{V_{RY}}{Z_2} = \frac{400\angle0^\circ}{200\angle-30^\circ} = 2\angle30^\circ \text{ A}$ .

Wattmeter $W_1$ measures the power using the voltage $V_{RB}$ and the total line current $I_R = I_1 + I_2 = 10\angle30^\circ \text{ A}$ . The reading is $W_1 = |V_{RB}||I_R|\cos(\angle V_{RB} - \angle I_R) = (400)(10)\cos(-60^\circ - 30^\circ) = 4000\cos(-90^\circ) = 0 \text{ W}$ .

Wattmeter $W_2$ is connected to read the power using voltage $V_{YB}$ and current $I_2$ . Its reading is $W_2 = |V_{YB}||I_2|\cos(\angle V_{YB} - \angle I_2) = (400)(2)\cos(-120^\circ - 30^\circ) = 800\cos(-150^\circ) \approx -692.82 \text{ W}$ .

The magnitude of the difference between these readings is $|W_1 - W_2| = |0 - (-692.82)| = 692.82 \text{ W}$ .

Q51GATE 2024NAT2MElectromagnetic Theory

In the $(x, y, z)$ coordinate system, three point-charges $Q, Q$ and $\alpha Q$ are located in free space at $(-1,0,0),(1,0,0)$ and $(0,-1,0)$ respectively. The value of $\alpha$ for the electric field to be zero at $(0,0.5,0)$ is _____ (rounded off to 1 decimal places).

🚀 Solve in Practice Mode

📖 Explanation

For the net electric field at the point $P(0, 0.5, 0)$ to be zero, the vector sum of the electric fields from the three charges must be zero. By symmetry, any field components in the x-direction must cancel out, so we only need to consider the y-components.

First, let's find the combined field from the two identical charges $Q$ at $(\pm 1, 0, 0)$ . The distance from each of these charges to $P$ is $r = \sqrt{1^2 + 0.5^2} = \sqrt{1.25}$ . Their individual x-components cancel, and their y-components add. The total upward field from this pair is $E_{2Q,y} = 2 \times \frac{kQ}{r^2} \times (\text{y-component of unit vector}) = 2 \times \frac{kQ}{1.25} \times \frac{0.5}{\sqrt{1.25}}$ .

Next, the field from the charge $\alpha Q$ at $(0, -1, 0)$ is entirely along the y-axis. The distance to point $P$ is $d = 0.5 - (-1) = 1.5$ . This field is $E_{\alpha Q, y} = \frac{k(\alpha Q)}{d^2} = \frac{k\alpha Q}{1.5^2}$ .

Setting the sum of the y-components to zero gives:
$2 \frac{kQ(0.5)}{1.25 \sqrt{1.25}} + \frac{k\alpha Q}{1.5^2} = 0$

After canceling $kQ$ and evaluating the constants, we get:
$0.7156 + 0.4444\alpha = 0$
Solving for $\alpha$ gives $\alpha = -\frac{0.7156}{0.4444} \approx -1.61$ .

Q52GATE 2024NAT2MElectromagnetic Theory

The given equation represents a magnetic field strength $\bar{H}(r, \theta, \phi)$ in the spherical coordinate system, in free space. Here, $\hat{r}$ and $\hat{\theta}$ represent the unit vectors along $r$ and $\theta$ , respectively. The value of $P$ in the equation should be _____ (rounded off to the nearest integer). $\bar{H}(r, \theta, \phi)=\frac{1}{r^{3}}(\hat{r} P \cos \theta+\hat{\theta} \sin \theta)$

🚀 Solve in Practice Mode

📖 Explanation

A fundamental property of any magnetic field is that it must be divergenceless, as stated by Gauss's law for magnetism ( $\nabla \cdot \bar{B} = 0$ ). In free space, this means the magnetic field strength $\bar{H}$ must also satisfy $\nabla \cdot \bar{H} = 0$ .

To find the value of $P$ , we must enforce this condition by applying the divergence operator in spherical coordinates:
$\nabla \cdot \bar{H} = \frac{1}{r^2}\frac{\partial}{\partial r}(r^2 H_r) + \frac{1}{r\sin\theta}\frac{\partial}{\partial\theta}(H_\theta \sin\theta) = 0$
Substituting the field's components, $H_r = \frac{P \cos\theta}{r^3}$ and $H_\theta = \frac{\sin\theta}{r^3}$ , and performing the differentiation gives:
$-\frac{P \cos\theta}{r^4} + \frac{2 \cos\theta}{r^4} = 0$
For this equation to hold true for all values of $r$ and $\theta$ , it must be that $-P + 2 = 0$ . Therefore, we find that $P=2$ .

Q53GATE 2024NAT2MSignals and Systems

If the energy of a continuous-time signal $x(t)$ is $E$ and the energy of the signal $2 x(2 t-1)$ is $c E$ , then $c$ is ______ (rounded off to 1 decimal place).

🚀 Solve in Practice Mode

📖 Explanation

Let's determine the energy of the transformed signal, $2x(2t-1)$ , by analyzing the effect of each operation. The initial energy of $x(t)$ is $E$ .

First, a time shift of $-1$ to get $x(t-1)$ does not change the signal's energy. Next, time-scaling by a factor of 2 to get $x(2t-1)$ compresses the signal, which divides the energy by that same factor, resulting in an energy of $E/2$ . Finally, multiplying the signal's amplitude by 2 increases the energy by the square of this factor, so we multiply by $2^2 = 4$ .

Combining these effects, the final energy is $4 \times (E/2) = 2E$ . Since this is given as $cE$ , we find that $c=2.0$ .

Q54GATE 2024NAT2MElectrical Machines

A 3-phase star connected slip ring induction motor has the following parameters referred to the stator: $R_{s}=3 \Omega, X_{s}=2 \Omega, X_{r}^{\prime}=2 \Omega, R_{r}^{\prime}=2.5 \Omega,$ The per phase stator to rotor effective turns ratio is $3: 1$ . The rotor winding is also star connected. The magnetizing reactance and core loss of the motor can be neglected. To have maximum torque at starting, the value of the extra resistance in ohms (referred to the rotor side) to be connected in series with each phase of the rotor winding is _____ (rounded off to 2 decimal places).

🚀 Solve in Practice Mode

📖 Explanation

To achieve maximum torque at starting (where slip $s=1$ ), the total rotor resistance referred to the stator, $R'_{r,total}$ , must equal the magnitude of the motor's total standstill impedance. The total series reactance is $X_{eq} = X_s + X'_r = 2 + 2 = 4 \, \Omega$ .

Thus, the required total rotor resistance is $R'_{r,total} = \sqrt{R_s^2 + X_{eq}^2} = \sqrt{3^2 + 4^2} = 5 \, \Omega$ .

This total resistance is composed of the inherent rotor resistance and the added external resistance, both referred to the stator: $R'_{r,total} = R'_r + R'_{ext}$ . The required external resistance referred to the stator is therefore $R'_{ext} = 5 \, \Omega - 2.5 \, \Omega = 2.5 \, \Omega$ .

Finally, to find the actual resistance value on the rotor side, we divide by the square of the effective turns ratio ( $a=3$ ):
$R_{ext} = \frac{R'_{ext}}{a^2} = \frac{2.5}{3^2} = \frac{2.5}{9} \approx 0.278 \, \Omega$ .

Q55GATE 2024NAT2MElectrical Machines

A $5 \mathrm{~kW}$ . $220 \mathrm{~V}$ DC shunt motor has $0.5 \Omega$ armature resistance including brushes. The motor draws a no-load current of $3 \mathrm{~A}$ . The field current is constant at $1 \mathrm{~A}$ . Assuming that the core and rotational losses are constant and independent of the load, the current (in amperes) drawn by the motor while delivering the rated load, for the best possible efficiency, is _____ (rounded off to 2 decimal places).

🚀 Solve in Practice Mode

📖 Explanation

First, we determine the motor's constant rotational losses from the no-load test. The no-load armature current is the total current minus the field current, so $I_{a,nl} = 3 \text{ A} - 1 \text{ A} = 2 \text{ A}$ . The corresponding back EMF is $E_{b,nl} = 220 \text{ V} - (2 \text{ A} \times 0.5 \Omega) = 219 \text{ V}$ . The power developed at no-load equals the rotational losses: $P_{rot} = E_{b,nl} \times I_{a,nl} = 219 \text{ V} \times 2 \text{ A} = 438 \text{ W}$ .

Next, under rated load, the motor's shaft output is $5000 \text{ W}$ . The total internal power developed by the armature, $P_{dev}$ , must cover both this output and the constant rotational losses. Therefore, $P_{dev} = 5000 \text{ W} + 438 \text{ W} = 5438 \text{ W}$ .

This developed power can also be expressed in terms of the load armature current, $I_a$ , as $P_{dev} = (V - I_a R_a)I_a$ . Substituting the known values gives the quadratic equation $(220 - 0.5 I_a)I_a = 5438$ , which simplifies to $0.5 I_a^2 - 220 I_a + 5438 = 0$ .

Solving for $I_a$ yields the armature current as approximately $26.28 \text{ A}$ . The total current drawn by the motor is the sum of the armature and field currents: $I_L = I_a + I_f = 26.28 \text{ A} + 1 \text{ A} = 27.28 \text{ A}$ .

Q56GATE 2024NAT2MPower Systems

The single line diagram of a lossless system is shown in the figure. The system is operating in steady-state at a stable equilibrium point with the power output of the generator being $P_{\max } \sin \delta$ . where $\delta$ is the load angle and the mechanical power input is $0.5 P_{\max }$ . A fault occurs on line 2 such that the power output of the generator is less than 0.5. $P_{\max }$ during the fault. After the fault is cleared by opening line 2. The power output of the generator is $\left\{P_{\max } / \sqrt{2}\right\} \sin \delta$ . If the critical fault clearing angle is $\pi / 2$ radians, the accelerating area on the power angle carve is ______ times $P_{\max }$ (rounded off to 2 decimal places)

🚀 Solve in Practice Mode

📖 Explanation

The question asks for the accelerating area ( $A_1$ ) on the power-angle curve. For the system to be stable at the critical clearing angle ( $\delta_{cr}$ ), the equal area criterion must be satisfied. This means the accelerating area ( $A_1$ ) gained during the fault must be exactly equal to the maximum decelerating area ( $A_2$ ) available after the fault is cleared.

We can calculate this decelerating area, $A_2$ , by integrating the net power from the clearing angle $\delta_{cr} = \pi/2$ to the maximum angle of swing $\delta_{\max}$ . To find $\delta_{\max}$ , we equate the post-fault electrical power to the mechanical power: $\frac{P_{\max}}{\sqrt{2}}\sin\delta_{\max} = 0.5P_{\max}$ . This yields $\sin\delta_{\max} = \frac{1}{\sqrt{2}}$ , so $\delta_{\max} = \pi - \frac{\pi}{4} = \frac{3\pi}{4}$ or $135^\circ$ .

The accelerating area is therefore given by the integral for the decelerating area:
$A_1 = A_2 = \int_{\pi/2}^{3\pi/4} \left( \frac{P_{\max}}{\sqrt{2}}\sin\delta - 0.5P_{\max} \right) d\delta$
Evaluating this integral results in:
$A_1 = P_{\max} \left[ -\frac{\cos\delta}{\sqrt{2}} - 0.5\delta \right]_{\pi/2}^{3\pi/4} = P_{\max}\left(0.5 - \frac{\pi}{8}\right) \approx 0.107 P_{\max}$

Q57GATE 2024NAT2MControl Systems

Consider the closed-loop system shown in the figure with $G(s)=\frac{K\left(s^{2}-2 s+2\right)}{\left(s^{2}+2 s+5\right)}$ . The root locus for the closed-loop system is to be drawn for $0 \leq K \\lt \infty$ . The angle of departure (between $0^{\circ}$ and $360^{\circ}$ ) of the root locus branch drawn from the pole $(-1+j 2)$ , in degree, is ______ (rounded off to the nearest integer).

🚀 Solve in Practice Mode

📖 Explanation

To find the angle of departure for a root locus branch from a complex pole, we apply the angle condition. The formula is $\phi_d = 180^\circ + (\text{\sum of angles from zeros}) - (\text{\sum of angles from other poles})$ . The system's poles are at $p_{1,2} = -1 \pm j2$ and its zeros are at $z_{1,2} = 1 \pm j1$ .

We are finding the departure angle from the pole $p_1 = -1+j2$ . First, we calculate the angles from the two zeros to this pole:
$\angle(p_1-z_1) + \angle(p_1-z_2) = \angle(-2+j1) + \angle(-2+j3) = 153.43^\circ + 123.69^\circ = 277.12^\circ$ .

Next, we find the angle from the other pole, $p_2 = -1-j2$ :
$\angle(p_1-p_2) = \angle(j4) = 90^\circ$ .

Substituting these values into the departure angle formula gives:
$\phi_d = 180^\circ + 277.12^\circ - 90^\circ = 367.12^\circ$ .

This angle is equivalent to $367.12^\circ - 360^\circ = 7.12^\circ$ . Rounding to the nearest integer, the angle of departure is $7^\circ$ .

Q58GATE 2024NAT2MControl Systems

Consider the stable closed-loop system shown in the figure. The asymptotic Bode magnitude plot of $G(s)$ has a constant slope of $-20 \mathrm{~dB} /$ decade at least till $100 \mathrm{rad} / \mathrm{sec}$ with the gain crossover frequency being $10 \mathrm{rad} / \mathrm{sec}$ . Hie asymptotic Bode phase plot remains constant at $-90^{\circ}$ at least till $\omega=10 \mathrm{rad} / \mathrm{sec}$ . The steady-state error of the closed-loop system for a unit ramp input is _____ (rounded off to 2 decimal places).

🚀 Solve in Practice Mode

📖 Explanation

The Bode plot characteristics provided-a magnitude slope of $-20 \mathrm{~dB} /$ decade and a constant phase of $-90^{\circ}$ -are the signature of a system with a single integrator at the origin. Thus, the open-loop transfer function has the form $G(s) = \frac{K}{s}$ . This is a Type 1 system.

The gain crossover frequency is the frequency at which the open-loop gain is unity, or $0 \mathrm{~dB}$ . We are given this frequency is $10 \mathrm{rad/sec}$ .
For our transfer function, the magnitude is $|G(j\omega)| = \frac{K}{\omega}$ . Setting the magnitude to 1 at $\omega=10$ gives $\frac{K}{10} = 1$ , which means the gain is $K=10$ .

For a Type 1 system subjected to a unit ramp input, the steady-state error is $e_{ss} = \frac{1}{K_v}$ , where $K_v$ is the velocity error constant.
The velocity error constant is calculated as $K_v = \lim_{s \to 0} sG(s) = \lim_{s \to 0} s \left(\frac{10}{s}\right) = 10$ .
Therefore, the steady-state error is $e_{ss} = \frac{1}{10} = 0.1$ .

Q59GATE 2024NAT2MControl Systems

Consider the stable closed-loop system shown in the figure. The magnitude and phase values of the frequency response of $G(s)$ are given in the table. The value of the gain $K_{1}( > 0)$ for a $50^{\circ}$ phase margin is (rounded off to 2 decimal places).

\begin{array}{|c|c|c|} \hline \omega \text{ \in rad/sec} & \text{Magnitude \in dB} & \text{Phase \in degrees} \\ \hline 0.5 & -7 & -40 \\ \hline 1.0 & -10 & -80 \\ \hline 2.0 & -18 & -130 \\ \hline 10.0 & -40 & -200 \\ \hline \end{array}

🚀 Solve in Practice Mode

📖 Explanation

To achieve a specific phase margin, we first identify the frequency at which the system's phase meets the required condition, and then adjust the gain $K_I$ to make this frequency the gain crossover frequency.

The open-loop transfer function is $L(s) = \frac{K_I}{s}G(s)$ . The phase margin (PM) is defined as $PM = 180^{\circ} + \angle L(j\omega_{gc})$ . For a $50^{\circ}$ PM, the total phase at the gain crossover frequency $\omega_{gc}$ must be $-130^{\circ}$ . The controller $\frac{K_I}{s}$ contributes a phase of $-90^{\circ}$ , so the phase from $G(s)$ must be $\angle G(j\omega_{gc}) = -130^{\circ} - (-90^{\circ}) = -40^{\circ}$ .

Looking at the table, a phase of $-40^{\circ}$ for $G(s)$ occurs at $\omega = 0.5$ rad/sec. This frequency is our gain crossover frequency, $\omega_{gc}$ . At this frequency, the table shows the magnitude of $G(s)$ is $-7$ dB. For the total open-loop magnitude $|L(j\omega_{gc})|$ to be $0$ dB, the magnitude of the controller term $\frac{K_I}{s}$ must be $+7$ dB.

We can now solve for $K_I$ by setting the magnitude of the controller at $\omega_{gc}=0.5$ :
$20 \log_{10} \left( \frac{K_I}{0.5} \right) = 7$ dB
This gives $\frac{K_I}{0.5} = 10^{7/20} \approx 2.239$ .
Therefore, $K_I \approx 0.5 \times 2.239 \approx 1.12$ .

Q60GATE 2024NAT2MAnalog Electronics

In the given circuit, the diodes are ideal. The current $I$ through the diode $D_{1}$ in milliamperes is ______ (rounded off to two decimal places).

🚀 Solve in Practice Mode

📖 Explanation

To find the current $I$ through diode $D_1$ , we first assume that both ideal diodes, $D_1$ and $D_2$ , are conducting (ON). We can then determine the voltage $V_X$ at the central node where the components meet.

By applying Kirchhoff's Current Law (KCL) at node $V_X$ , we can write an equation for the sum of currents leaving the node. The underlying analysis model yields the equation $\frac{V_X - 10 \text{ V}}{1 \text{ k}\Omega} + \frac{V_X - 3 \text{ V}}{1 \text{ k}\Omega} + \frac{V_X}{1 \text{ k}\Omega} = 0$ . Solving this gives $3V_X = 13$ , which means $V_X \approx 4.33 \text{ V}$ .

Now, we use this node voltage to find the current $I$ through $D_1$ . Based on a current balance equation at the node, we have $I + I_1 = I_2$ , where $I_1 = \frac{4.33 \text{ V} - 3 \text{ V}}{1 \text{ k}\Omega} \approx 1.33 \text{ mA}$ and $I_2$ is determined to be $3 \text{ mA}$ .

Therefore, the current $I$ can be calculated as $I = 3 \text{ mA} - 1.33 \text{ mA} = 1.67 \text{ mA}$ .

Q61GATE 2024NAT2MAnalog Electronics

A difference amplifier is shown in the figure. Assume the op-amp to be ideal. The CMRR (in $\mathrm{dB}$ ) of the difference amplifier is _____(rounded off to 2 decimal places).

🚀 Solve in Practice Mode

📖 Explanation

To determine the Common Mode Rejection Ratio (CMRR), we first find the output voltage $v_o$ as a function of the two inputs, $v_{in1}$ and $v_{in2}$ . Using superposition, we can write the output as $v_o = A_{v1}v_{in1} + A_{v2}v_{in2}$ .

The gain for the inverting input $v_{in1}$ is $A_{v1} = -\frac{101 \text{ k}\Omega}{10.5 \text{ k}\Omega} \approx -9.619$ .
The gain for the non-inverting input $v_{in2}$ is the product of the voltage divider and the non-inverting gain: $A_{v2} = \left(\frac{101.5}{9.5+101.5}\right) \times \left(1+\frac{101}{10.5}\right) \approx 9.71$ .

From these gains, we find the differential gain $A_d = \frac{A_{v2} - A_{v1}}{2} = \frac{9.71 - (-9.619)}{2} \approx 9.6645$ .
The common-mode gain is $A_c = A_{v2} + A_{v1} = 9.71 + (-9.619) = 0.091$ .

The CMRR is the magnitude of the ratio of the differential gain to the common-mode gain: $\text{CMRR} = \left|\frac{A_d}{A_c}\right| = \frac{9.6645}{0.091} \approx 106.20$ .
Finally, converting this value to decibels gives $\text{CMRR}_{\text{dB}} = 20 \log_{10}(106.20) \approx 40.52 \text{ dB}$ .

Q62GATE 2024NAT2MPower Electronics

A single-phase half-controlled bridge converter supplies an inductive load with ripple free load current. The triggering angle of the converter is $60^{\circ}$ . The ratio of the rms value of the fundamental component of the input current to the rms value of the total input current of the bridge is ______ (rounded off to 3 decimal places).

🚀 Solve in Practice Mode

📖 Explanation

The question asks for the input current distortion factor, which is the ratio of the RMS value of the fundamental component of the input current ( $I_{s1,rms}$ ) to the total RMS input current ( $I_{s,rms}$ ).

For a half-controlled converter with a ripple-free load current ( $I_0$ ) and firing angle $\alpha$ , this ratio is given by:
$\frac{I_{s1,rms}}{I_{s,rms}} = \frac{\frac{2\sqrt{2}}{\pi} I_0 \cos(\frac{\alpha}{2})}{I_0 \sqrt{\frac{\pi-\alpha}{\pi}}} = \frac{2\sqrt{2} \cos(\frac{\alpha}{2})}{\sqrt{\pi(\pi-\alpha)}}$
We are given $\alpha = 60^{\circ}$ , which must be in radians ( $\pi/3$ ) for the formula. Substituting this value:
$\text{Ratio} = \frac{2\sqrt{2} \cos(30^{\circ})}{\sqrt{\pi(\pi-\frac{\pi}{3})}} = \frac{2\sqrt{2} (\frac{\sqrt{3}}{2})}{\sqrt{\frac{2\pi^2}{3}}} = \frac{\sqrt{6}}{\pi\sqrt{2/3}}$
Simplifying this expression gives:
$\text{Ratio} = \frac{\sqrt{6} \cdot \sqrt{3}}{\pi \sqrt{2}} = \frac{\sqrt{18}}{\pi \sqrt{2}} = \frac{3\sqrt{2}}{\pi \sqrt{2}} = \frac{3}{\pi} \approx 0.955$

Q63GATE 2024NAT2MPower Electronics

A single-phase full bridge voltage source inverter (VSI) feeds a purely inductive load. The inverter output voltage is a square wave in $180^{\circ}$ conduction mode. The fundamental frequency of the output voltage is $50 \mathrm{~Hz}$ . If the DC input voltage of the inverter is $100 \mathrm{~V}$ and the value of the load inductance is $20 \mathrm{mH}$ , the peak-to-peak load current in amperes is ______ (rounded off to the nearest integer).

🚀 Solve in Practice Mode

📖 Explanation

For a purely inductive load, the voltage-current relationship is $v(t) = L \frac{di}{dt}$ . The inverter produces a square wave voltage. During the first half-cycle, from $t=0$ to $t=T/2$ , the output voltage is constant and equal to the DC supply voltage, $V_S$ .

This constant voltage causes the current to change linearly. The total change in current over this half-period corresponds to the peak-to-peak current, $I_{pk-pk}$ . We can find this by rearranging and integrating the inductor equation:
$\Delta i = I_{pk-pk} = \frac{1}{L} \int_{0}^{T/2} V_S \,dt = \frac{V_S}{L} \left(\frac{T}{2}\right)$

Since the frequency is $f = 1/T$ , we can write the peak-to-peak current as:
$I_{pk-pk} = \frac{V_S}{2fL}$

Substituting the given values:
$I_{pk-pk} = \frac{100 \text{ V}}{2 \times 50 \text{ Hz} \times 20 \times 10^{-3} \text{ H}} = \frac{100}{2} = 50 \text{ A}$

Q64GATE 2024NAT2MPower Electronics

In the DC-DC converter shown in the figure, the current through the inductor is continuous, The switching frequency is $500 \mathrm{~Hz}$ . The voltage $\left(V_{0}\right)$ across the load is assumed to be constant and ripple free. The peak inductor current in amperes is ______(rounded off to the nearest integer).

🚀 Solve in Practice Mode

📖 Explanation

This circuit is a boost converter. First, we determine the duty cycle ( $\alpha$ ) using the voltage gain relationship $V_o = V_i / (1-\alpha)$ . Substituting the given values, we get $40 = 20 / (1-\alpha)$ , which yields a duty cycle of $\alpha=0.5$ .

Next, we calculate the average inductor current, $I_L$ . The average output current is $I_o = V_o/R = 40/10 = 4 \text{ A}$ . For a boost converter, the average inductor current is $I_L = I_o / (1-\alpha) = 4 / (1-0.5) = 8 \text{ A}$ .

The peak-to-peak ripple of the inductor current is found using $\Delta I_L = \frac{V_i \alpha}{fL}$ . This gives $\Delta I_L = \frac{20 \times 0.5}{500 \times 2 \times 10^{-3}} = 10 \text{ A}$ .

Finally, the peak inductor current is the average current plus half of the ripple current: $I_{L, \text{peak}} = I_L + \frac{\Delta I_L}{2} = 8 + \frac{10}{2} = 13 \text{ A}$ .

Q65GATE 2024NAT2MPower Electronics

A single-phase full-controlled thyristor converter bridge is used for regenerative braking of a separately excited DC motor with the following specifications:

\begin{array}{|l|l|} \hline \text{Rated armature voltage} & 210 \mathrm{~V} \\ \hline \text{Rated armature current} & 10 \mathrm{~A} \\ \hline \text{Rated speed} & 1200 \mathrm{rpm} \\ \hline \text{Armature resistance} & 1 \Omega \\ \hline \text{Input to the converter bridge} & 240 \mathrm{~V} at 50 \mathrm{~Hz} \\ \hline \begin{array}{l} \text{The armature of the DC motor is fed from the} \\ \text{full-controlled bridge and the field current is kept} \\ \text{constant.} \\ \end{array} & \\ \hline \end{array}

Assume that the motor is running at $600 \mathrm{rpm}$ and the armature terminals of the motor are suitably reversed for regenerative braking. If the armature current of the motor is to be maintained at the rated value, the triggering angle of the converter bridge in degrees should be _____ (rounded off to 2 decimal places).

🚀 Solve in Practice Mode

📖 Explanation

Excellent question! Let's break down the solution with a clearer, step-by-step approach.

First, we need to find the motor's back-EMF constant, $K_m$ . We can calculate this from the motor's rated operating point. The rated terminal voltage must overcome both the back-EMF ( $E_b$ ) and the resistive voltage drop:
$E_{b, \text{rated}} = V_t - I_a R_a = 210 \text{ V} - (10 \text{ A} \times 1 \Omega) = 200 \text{ V}$ .
Since $E_b = K_m \omega$ , we find $K_m = \frac{200 \text{ V}}{1200 \text{ rpm} \times (2\pi/60) \text{ rad/s}} \approx 1.5915 \text{ V-s/rad}$ .

Next, we determine the conditions for regenerative braking at $600 \text{ rpm}$ . The motor, now acting as a generator, produces a back-EMF of $E_b' = K_m \times (600 \times 2\pi/60) = 100 \text{ V}$ . With the armature terminals reversed, the converter must produce a negative voltage to absorb power. The required converter voltage is $V_0 = -E_b' + I_a R_a = -100 \text{ V} + (10 \text{ A} \times 1 \Omega) = -90 \text{ V}$ .

Finally, we find the triggering angle $\alpha$ that produces this voltage. For a single-phase full-controlled converter, the average DC voltage is $V_0 = \frac{2V_m}{\pi}\cos\alpha$ . Given an AC input of $240 \text{ V}$ (RMS), the peak voltage is $V_m = 240\sqrt{2} \text{ V}$ . Setting the expressions for $V_0$ equal:
$-90 = \frac{2(240\sqrt{2})}{\pi}\cos\alpha$
Solving this equation for $\alpha$ gives $\alpha = \arccos\left(\frac{-90\pi}{480\sqrt{2}}\right) \approx 114.62^\circ$ .

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