GATE EE 2023 question paper with solutions

GATE EE 2023 answer key

General Aptitude Q.1 to Q.10

1. Q.1MCQ · 1 markOfficial answer: D

   General Aptitude › Grammar & vocabulary

   When the reporting verb is in the past tense ("told"), indirect speech moves the reported verb one tense back and keeps its aspect. Present continuous "am thinking" becomes past continuous "was thinking". The pronoun ("I" to "he") and the time expression ("this weekend" to "that weekend") shift along with it.

   WATCH OUT FOR

   Choosing (A) "thought": it is past tense, but it drops the continuous aspect, so it reports "I think", not "I am thinking". Choosing (B) "is thinking" is also wrong because it skips the backshift even though "this weekend" has already become "that weekend". (C) "am thinking" keeps the first-person verb after the pronoun has changed to "he".

   Full solution →
2. Q.2MCQ · 1 markOfficial answer: B

   General Aptitude › Reasoning

   Work out the relation in the complete pair first (here, Enforce : Relax are antonyms). The blank must then stand in the same relation to Permit, so you need Permit's antonym.

   WATCH OUT FOR

   Picking a synonym of Permit (Allow or License) without checking the relation in the complete pair. Another trap is choosing Reinforce because it sounds like Enforce. Sound-alike words are a common distractor in GATE analogy questions.

   Full solution →
3. Q.3MCQ · 1 markOfficial answer: A

   General Aptitude › Statistics & probability

   When events are independent, the probability that all of them happen ("and") is the product of their separate probabilities. For two rolls of a fair die the sample space is $6\times 6=36$ equally likely ordered pairs, and a fixed ordered result such as $(1,4)$ is exactly one of them.

   WATCH OUT FOR

   Adding the two probabilities instead of multiplying them ($\frac{1}{6}+\frac{1}{6}=\frac{1}{3}$, option D) confuses "and" with "or". Picking $\frac{1}{6}$ (option B) looks at only one roll. Another slip is counting $(4,1)$ as favourable as well and getting $\frac{2}{36}$. The order is fixed here: '1' first, then '4'.

   Full solution →
4. Q.4MCQ · 1 markOfficial answer: B

   General Aptitude › Reading

   Test a "certainty" claim with the extreme case. Even if all 30% of attempters were among the 65% who were advised, they would make up only $\frac{30}{65}\approx 46\%$ of that group. So more than half of the advised users must not have attempted, however the two groups overlap.

   WATCH OUT FOR

   One mistake is to treat "attempted to stop" as "successfully stopped" and pick (C), which just repeats the 30% figure. Another is to assume advice leads to attempts and pick (A) without checking that 30 attempters cannot be a majority of 65 advised users.

   Full solution →
5. Q.5MCQ · 1 markOfficial answer: C

   General Aptitude › Spatial

   Label every corner and crossing point, then sort the lines into parallel families: here there are four (horizontal, slant, and the two diagonal directions, with the inner parallelogram's sides parallel to the diagonals by the midpoint theorem). A triangle needs three sides from three different families. Count the triangles by size, then check the total by working through each choice of three families.

   WATCH OUT FOR

   Counting only the small triangles you can see at a glance and missing the large ones. If you leave out the four half-parallelograms ($ABC$, $ACD$, $ABD$, $BCD$) you get 16, which is option (B). If you also miss the four corner triangles ($AQL$, $DLP$, $CPR$, $BQR$) that a diagonal splits in two, you get 12, which is option (A). Some students also look for triangles inside the inner parallelogram. There are none, because every line inside it runs in one of only two directions.

   Full solution →
6. Q.6MCQ · 2 marksOfficial answer: D

   General Aptitude › Reasoning

   Registration is the only route to eligibility, and no HS student registered, so no HS student is eligible. That is all the passage tells us. It supports 'every eligible voter is from outside HS' and says nothing about how students of other departments did.

   WATCH OUT FOR

   Many candidates turn 'no HS student is eligible' around into 'every ineligible student is from HS' and choose (B). Others assume every other department registered because the passage calls the HS result surprising, and choose (A). The passage is silent about the other departments, so neither (i) nor (ii) is certain.

   Full solution →
7. Q.7MCQ · 2 marksOfficial answer: B

   General Aptitude › Numerical

   Match a graph to a formula by testing its visible features one at a time, each of which rules out options: the sign on each side of the origin, symmetry (odd or even), and the limits as $x\to \pm \infty$. A factor $|x|$ or $x^2$ can never be negative. A factor $2^{-x}$ with no modulus blows up as $x\to -\infty$. Only $x{2}^{-|x|}$ is odd, has the sign of $x$, and decays on both sides.

   WATCH OUT FOR

   Picking (D), $x{2}^{-x}$, because it matches the graph exactly for $x>0$ (there it is the same function as (B)). For negative $x$, $2^{-x}$ grows without bound, so $x{2}^{-x}\to -\infty$ instead of returning to zero. Another slip is to see the modulus in $2^{-|x|}$ and choose (A) without noticing that $x^2\ge 0$ can never produce the negative lobe.

   Full solution →
8. Q.8MCQ · 2 marksOfficial answer: D

   General Aptitude › Reading

   In a NOT-supported question, check each option against a specific sentence in the passage. The odd one out usually contradicts the passage outright, and an absolute word such as 'never' is a common signal. Here the passage itself shows modern cell biology encouraging a metaphor (the cell as a molecular machine), so a claim that it never uses figurative language is refuted by the text.

   WATCH OUT FOR

   Some students reject (B) because the passage never uses the word 'metaphor' about cell biology. But 'encourages us to imagine the cell as a molecular machine' is exactly the promotion of a machine metaphor, so (B) follows. Others misread the question as asking which option IS supported.

   Full solution →
9. Q.9MCQ · 2 marksOfficial answer: A

   General Aptitude › Numerical

   Only unit digits matter. Take $7^{999}$ out of $7^{1000}$ and pair it with $3^{999}$ to get $21^{999}\times 7$. A power of a number ending in 1 always ends in 1, so the answer is the unit digit of $7$. You can check it with the period-4 cycles of 3 ($3,9,7,1$) and 7 ($7,9,3,1$).

   WATCH OUT FOR

   Getting the cycle position wrong when the remainder is 0. $1000\mathrm{mod}4=0$ means the 4th term of the cycle, which is $1$. Picking the 1st term ($7$) gives $7\times 7=49$, unit digit $9$, which is option (D). Another slip is reading $3^{999}$ off at remainder 3 as $9$ (the 2nd term) and getting $9\times 1=9$.

   Full solution →
10. Q.10MCQ · 2 marksOfficial answer: B

    General Aptitude › Geometry

    Only one region is left white: the lens where the two semicircles overlap. So each shaded piece is a semicircle minus the whole lens, where the lens is two 90° segments of radius 3. That gives 9π/2 − (9π/2 − 9) = 9 per piece, so the π terms cancel. Equivalently, each shaded piece has the same area as a triangle with base 6 and height 3.

    WATCH OUT FOR

    Adding the two semicircles, 2 × 9π/2 = 9π, and picking (D) without removing the white overlap. A second trap is subtracting the lens only once, which gives 9π/2 + 9 ≈ 23.1. The lens sits inside both semicircles, so it must come out of each one, two times in all. A third trap is taking the lens to be a quarter-circle, when it is two circular segments.

    Full solution →

Electrical Engineering Q.11 to Q.65

1. Q.11MCQ · 1 markOfficial answer: D

   Engineering Mathematics › Linear algebra

   For a plane written as ${\mathbf{w}}^T\mathbf{x}=c$, i.e. $a{x}_1+b{x}_2+c^{\prime }{x}_3=c$, the coefficient vector $\mathbf{w}$ is orthogonal to the difference of any two points on the plane, so $\mathbf{w}$ (or any nonzero multiple of it) is the normal. The right-hand constant only moves the plane parallel to itself.

   WATCH OUT FOR

   Picking a vector whose dot product with $\mathbf{w}$ is zero, like (A) or (B), because "orthogonal" gets mixed up with "normal". A vector with ${\mathbf{v}}^T\mathbf{w}=0$ lies along the plane. The normal has to be parallel to $\mathbf{w}$. Another mistake is thinking the constant $1$ on the right-hand side changes the normal direction.

   Full solution →
2. Q.12MCQ · 1 markOfficial answer: B

   Control Systems › Block diagrams & SFG

   Name the output of each summing junction and write one equation per junction using the signs printed on the diagram, then solve for $Y/R$. In Mason's formula, both forward paths end at the summing junction that sits on the only loop, so ${\Delta }_1={\Delta }_2=1$. Because the feedback is positive, $\Delta =1-\frac{1}{s}$, which puts the pole at $s=+1$.

   WATCH OUT FOR

   Reaching for the negative-feedback formula $\frac{G}{1+GH}$ out of habit, even though both signs at the first junction are $+$. That gives $\frac{3s+2}{s+1}$, which is the trap option (D). Swapping which gain goes to the integrator path as well gives $\frac{2s+3}{s+1}$, option (A). Another slip is treating the direct path of gain $3$ as not touching the loop and using ${\Delta }_2=1-\frac{1}{s}$. That gives $\frac{3s-1}{s-1}$, which is not one of the options.

   Full solution →
3. Q.13MCQ · 1 markOfficial answer: C

   Control Systems › Nyquist

   On the Nyquist contour, the arc at infinity maps to ${\lim }_{|s|\to \infty }G(s)H(s)$. If the numerator and denominator have the same degree, that limit is the ratio of their leading coefficients, so the whole arc shrinks to one finite point on the real axis. If the function is strictly proper, the point is the origin. Here the ratio is $3/1=3$.

   WATCH OUT FOR

   The usual mistake is to pick $0$ out of the habit that "the infinite arc maps to the origin". That is only true for strictly proper functions, and this one has equal degrees. The other trap is $-5$, which comes from putting $s=0$ into the function. That gives the start of the $j\omega$-axis image, not the image of the arc at infinity.

   Full solution →
4. Q.14MCQ · 1 markOfficial answer: D

   Control Systems › Compensators & PID

   The PI controller stabilises the loop (closed-loop poles at $s=-1$, twice), and its integrator forces the steady-state error to zero, so $y(\infty )=1$. At steady state the plant still needs an input: $u(\infty )=y(\infty )/G(0)=1/(-1)=-1$. Zero error does not mean zero controller output, because the integrator holds whatever value the plant needs.

   WATCH OUT FOR

   Three common traps. (1) Seeing the unstable plant pole or the integrator and choosing $\infty ,\infty$ without checking the closed-loop characteristic equation, which is $(s+1{)}^2$ and stable. (2) Reasoning that the controller output must be 0 because the error is 0. The integral term keeps a nonzero value, here $-1$, because $G(0)=-1$ is negative. (3) Getting the right values and then reversing the order: the question asks for the controller output first, which gives (D), not (C).

   Full solution →
5. Q.15MCQ · 1 markOfficial answer: A

   Electrical Machines › Induction machines

   Slip $s=\frac{N_s-N_r}{N_s}$ takes different ranges depending on the relative direction and speed of the rotor with respect to the (possibly reversed) rotating field: $0$ to $1$ in motoring, negative in generating (rotor driven faster than sync speed, same direction as the field), and above $1$ (up to $2$) in plugging, where reversing two supply leads reverses the field direction while the rotor keeps turning the old way.

   WATCH OUT FOR

   Confusing the sign convention: assuming plugging gives a negative slip (like generator action) instead of realizing that reversing two supply leads reverses the direction of the rotating field, so the rotor's own direction becomes "backward" relative to the new field, pushing slip past 1 (up to 2) rather than below 0.

   Full solution →
6. Q.16MCQ · 1 markOfficial answer: B

   Electrical Machines › Induction machines

   By the double revolving field theory, a single-phase induction motor's pulsating mmf is resolved into two counter-rotating fields (forward and backward), each behaving like a separate 3-phase-equivalent field. If the rotor slip with respect to the forward field is $s$, its slip with respect to the backward field is $(2-s)$, so the backward-field rotor current frequency is $f_b=(2-s)f$.

   WATCH OUT FOR

   Using slip $s$ (instead of $2-s$) with the line frequency for the backward-field current, which gives $f_b=sf=5\text{Hz}$ — this is actually the frequency due to the forward field's own slip-frequency rotor current, not the backward field's.

   Full solution →
7. Q.17MCQ · 1 markOfficial answer: C

   Signals and Systems › LTI & convolution

   For a first-order system $\frac{dy}{dt}+ay=bx$ that starts at rest, $H(s)=\frac{b}{s+a}$, so $h(t)=b{e}^{-at}u(t)$. The coefficient of $y$ sets the decay rate, the coefficient of $x$ sets the gain, and starting at rest (causality) supplies the $u(t)$.

   WATCH OUT FOR

   Picking (D), $2e^{-3t}$, because the exponent and gain are right, and forgetting that a system starting at rest is causal, so its impulse response must carry $u(t)$. The other common slip is swapping the coefficients, which gives the $e^{-2t}$ terms in (A) and (B).

   Full solution →
8. Q.18MCQ · 1 markOfficial answer: A

   Signals and Systems › Fourier transform

   Time and bandwidth trade off. A rectangular spectrum of half-width $W_0$ gives $x(t)=\sin (W_{0}t)/(\pi t)$, whose peak is $W_0/\pi$, main-lobe width is $2\pi /W_0$ and area is $X(0)=1$. As the band widens without limit, the pulse gets taller and narrower while its area stays 1, which is an impulse. Equivalently, a flat spectrum $X(\omega )=1$ is the transform of $\delta (t)$.

   WATCH OUT FOR

   Forgetting the $1/(2\pi )$ factor, or taking $x(t)=\sin (W_{0}t)/t$ or $\sin (W_{0}t)/(\pi t)\cdot 2$. At $t=\pi /(2W_0)$ that leads students to a $W_0$-free value like $\pm 1/\pi$, so they pick (C) or (D). The correct value, $2W_0/{\pi }^2$, depends on $W_0$. A second trap is thinking $x(0)$ falls as $W_0$ grows because 'the sinc gets narrower'. The pulse does narrow, but its peak $W_0/\pi$ rises.

   Full solution →
9. Q.19MCQ · 1 markOfficial answer: B

   Signals and Systems › z-transform

   The DTFT exists exactly when the ROC of $X(z)$ contains the unit circle $|z|=1$. With poles at radii $\frac{1}{5}$, $\frac{2}{3}$ and $3$, only the annulus $\frac{2}{3}<|z|<3$ contains $|z|=1$, so the sequence must be two-sided: right-sided for the poles inside the unit circle, left-sided for the pole outside it.

   WATCH OUT FOR

   Choosing (D) on the reflex that "causal/right-sided means stable". A right-sided sequence has ROC $|z|>3$ (outside the outermost pole), and since the pole at $z=3$ lies outside the unit circle, that ROC misses $|z|=1$. Stability plus causality needs every pole inside the unit circle, and this $X(z)$ fails that.

   Full solution →
10. Q.20MCQ · 1 markOfficial answer: A
11. Q.21MCQ · 1 markOfficial answer: A
12. Q.22MCQ · 1 markOfficial answer: D

    Electrical Machines › DC machines

    A full-pitch rotor coil's net e.m.f. is the algebraic difference of the motional e.m.f.s at its two diametrically-opposite sides. That difference is finite only if the stator-produced radial flux density reverses sign 180° away (the ordinary 2-pole case); here, because A & B (0°,180°) carry the same-direction current and A′ & B′ (90°,270°) carry the same-direction current, the flux density instead repeats identically every 180°, so the difference — and hence the coil e.m.f. — is zero at every instant for both coils.

    WATCH OUT FOR

    Assuming any arrangement with two conductors "into the page" and two "out of the page" 90° apart must be a standard 2-pole field (where diametrically opposite points see opposite-sign flux), and therefore concluding the coil EMFs add up to a finite value — missing that here it is conductors 180° apart (A & B, A' & B') that share the SAME current direction, giving a field pattern that repeats every 180° instead of reversing sign.

    Full solution →
13. Q.23MCQ · 1 markOfficial answer: B
14. Q.24MCQ · 1 markOfficial answer: A

    Engineering Mathematics › Linear algebra

    With four options to choose from, check $\mathbf{Au}=\mathbf{v}$ by multiplying each option by $\mathbf{u}$. As a sanity check: $\mathbf{u}$ and $\mathbf{v}$ are both 5 long, and turning $\mathbf{u}$ clockwise by $\theta$ (with $\cos \theta =4/5$, $\sin \theta =3/5$) lands it on $\mathbf{v}$. That rotation is $R(-\theta )=\left[\begin{array}{cc}\cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{array}\right]$.

    WATCH OUT FOR

    Picking (B) because it looks like the textbook rotation matrix $\left[\begin{array}{cc}\cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{array}\right]$. That matrix rotates anticlockwise, but $\mathbf{u}$ has to turn clockwise to reach $\mathbf{v}$. (B) actually sends $\mathbf{u}$ to $(7/5,24/5)$. A second trap is checking only that the y-component comes out 0: (D) passes that test but gives $(7/5,0)$, not $(5,0)$.

    Full solution →
15. Q.25MCQ · 1 markOfficial answer: C

    Engineering Mathematics › Probability & statistics

    When the mean is known (here zero), $\frac{1}{N}\sum x_n^2$ is exactly unbiased for ${\sigma }^2$, because $E(x_n^2)={\sigma }^2$ and expectation passes through a sum. Bessel's $N-1$ only compensates for subtracting the sample mean $\bar{x}$. Expectation does not pass through a square root, so an unbiased variance estimate does not give an unbiased standard-deviation estimate.

    WATCH OUT FOR

    Picking (A) out of habit, because "dividing by $N-1$ gives the unbiased variance". That rule is for $\sum (x_n-\bar{x}{)}^2$, where the sample mean is subtracted. Here nothing is subtracted, so dividing by $9999$ overestimates ${\sigma }_o^2$. A second trap is assuming that an unbiased ${\hat{\sigma }}^2$ makes $\hat{\sigma }$ unbiased, as in (B).

    Full solution →
16. Q.26MSQ · 1 markOfficial answer: A, D
17. Q.27MSQ · 1 markOfficial answer: B

    Signals and Systems › LTI & convolution

    Causality means $h[n]=0$ for $n<0$. The step response is the running sum $s[n]={\sum }_{k\le n}h[k]$, so $s[n]=0$ for $n<0$ exactly when $h[n]=0$ for $n<0$ (since $h[n]=s[n]-s[n-1]$). BIBO stability means $\sum |h[n]|<\infty$. A finite-duration $h[n]$ gives a finite sum. A bound on each sample does not, and causality has no bearing on it.

    WATCH OUT FOR

    Thinking causality implies stability (A), or that $|h[n]|<1$ for every sample makes the system stable (D). Stability needs the whole sum $\sum |h[n]|$ to be finite, and a bound on each sample does not give that. A second slip is disproving (D) with a one-sided sequence such as $\frac{1}{n+2}u[n]$. It is zero for $n<0$, so it breaks the condition $0<|h[n]|$ and is not a valid counterexample.

    The question and its official answer →
18. Q.28MSQ · 1 markOfficial answer: A, C
19. Q.29NAT · 1 markOfficial answer: 7.9 to 8.1

    Networks (Electric Circuits) › Transients

    At $t=0^{+}$, treat the inductor as a $6\text{ A}$ current source and the capacitor as a $12\text{ V}$ voltage source, both values taken from the $t<0$ DC steady state. Once $K$ opens, the $10\text{ A}$ source current has only two paths. The inductor fixes its branch at $6\text{ A}$, so the $R_2$–$C$ branch must take $4\text{ A}$. That sets the node voltage at $2\times 4+12=20\text{ V}$, and $V_L$ is whatever $R_1$ leaves over: $20-12=8\text{ V}$.

    WATCH OUT FOR

    Treating the node voltage $V_{AB}$ as continuous (keeping $12\text{ V}$ at $0^{+}$) gives $V_L=12-12=0$. Only $v_C$ and $i_L$ are continuous; $V_{AB}$ jumps to $20\text{ V}$ because the capacitor branch now takes $4\text{ A}$ through $R_2$. Other slips: forgetting the $R_2$ drop, sending current through the $R_2$–$C$ branch in the $t<0$ DC analysis (the capacitor is open), or letting $i_L$ jump straight to $10\text{ A}$.

    Full solution →
20. Q.30NAT · 1 markOfficial answer: 1340 to 1345

    Electrical Machines › DC machines

    For a separately excited DC motor, $E_b=V-I_a{R}_a$ and $E_b\propto N\varphi$. With field current unaltered, $\varphi$ is constant. A constant-torque load means $T=k\varphi I_a$ stays constant, so with $\varphi$ fixed, $I_a$ stays fixed too — it does NOT change just because the supply voltage changes. So only $V$ changes, and $N_2=N_1\times \frac{E_{b2}}{E_{b1}}$.

    WATCH OUT FOR

    Assuming the armature current changes when the supply voltage drops, or forgetting that a "constant torque load with unaltered field current" pins $I_a$ at its rated value throughout, and instead recomputing $I_a$ from the new terminal voltage.

    Full solution →
21. Q.31NAT · 1 markOfficial answer: 4

    Signals and Systems › LTI & convolution

    Convolving with a unit rectangular pulse on $(-1,1)$ gives a sliding-window integral: $z(t)={\int }_{t-1}^{t+1}y(\lambda )d\lambda$. $z$ is largest when the window holds the most positive area and none of the negative area. $y$ is positive only on $(3,5)$, and that interval is exactly as wide as the window, so the best window is $[3,5]$. Its centre gives $t=4$. The same result follows from $dz/dt=y(t+1)-y(t-1)$ changing from $+2$ to negative at $t=4$.

    WATCH OUT FOR

    Answering $5$ because that is where $y(t)$ peaks. The maximum of a convolution is not at the peak of either signal: at $t=5$ the window $[4,6]$ loses the area on $(5,6)$, so $z(5)=1.5<2$. Another mistake is answering $3$, the centre of $y$'s support. There the positive and negative areas cancel and $z(3)=0$. A third error is to use the window $[t,t+2]$ or $[t-2,t]$ instead of $[t-1,t+1]$. That forgets the pulse is centred at the origin and moves the answer by one second.

    Full solution →
22. Q.32NAT · 1 markOfficial answer: 5.9 to 6.1

    Networks (Electric Circuits) › Laws & node/mesh

    An ideal voltage source fixes the voltage across its own terminals. Once you see that the junction under $a$ is wired straight to $b$, $V_1$ sits directly across $P$ and $b$. The $I_1$ branch (in series with the two $2\Omega$ resistors) is connected across $V_1$, so it cannot change any voltage in the rest of the circuit. What is left is an $8\text{V}$ source driving $0.5\Omega$ in series with $3\parallel 3=1.5\Omega$, and the voltage divider rule gives $6\text{V}$.

    WATCH OUT FOR

    Missing the plain wire from the junction under $a$ to $b$. A student who misses it treats the vertical $3\Omega$, the middle $2\Omega$ and $V_1$ as meeting at a separate node and solves a much harder, wrong circuit. A second trap is letting $I_1$ push current through the divider: it only changes the current through the ideal source $V_1$. Reading the battery backwards (long plate towards $b$) gives $-6\text{V}$. The long plate is the positive terminal, and here it faces the top-left node.

    Full solution →
23. Q.33NAT · 1 markOfficial answer: 83 to 84
24. Q.34NAT · 1 markOfficial answer: 0.48 to 0.52

    Engineering Mathematics › Numerical methods (IN)

    When dy/dt depends only on t, the initial-value problem is really a definite integral: y(1) = y(0) + ∫₀¹ f(t) dt. Here α = 0.01 is small, so the integrand barely moves from 1/2 across [0, 1] and y(1) ≈ 0.50. Simpson's rule, the trapezoidal rule and a single Euler step all agree on this to 2 decimal places.

    WATCH OUT FOR

    Reporting the slope at t = 1, f(1) ≈ 0.4926, as the value of y(1). y is the area accumulated under the slope curve from 0 to 1, not the slope at the end point. That answer only lands near the key here because f is almost constant. The other common slip is to assume the right-hand side depends on y and start a multi-step Runge–Kutta calculation. Nothing here needs that: it is a single quadrature.

    Full solution →
25. Q.35NAT · 1 markOfficial answer: 0.9 to 1.1

    Engineering Mathematics › Probability & statistics

    For a least-squares line $y=a+bx$ the slope is $b=\frac{n\sum xy-\sum x\sum y}{n\sum x^2-(\sum x{)}^2}$. When the $x$-values are symmetric about zero ($\sum x=0$), this reduces to $b=\sum xy/\sum x^2$, and the intercept is just the mean of $y$.

    WATCH OUT FOR

    Dropping the minus sign in $(-1)(0.8)$, which gives $\sum xy=3.6$ and a wrong slope of $1.8$. A second mistake is to regress $x$ on $y$, which minimises horizontal distances instead of vertical ones and generally gives a different line. The question asks for $y$ fitted against $x$.

    Full solution →
26. Q.36MCQ · 2 marksOfficial answer: D

    Control Systems › Bode & margins

    A pure time delay $e^{-sT}$ has unit magnitude at every frequency and phase $-\omega T$. So a flat magnitude plot together with a phase that falls in a straight line against $\omega$ (linear axis) means gain times delay. Read the gain from the dB level ($K=10^{\text{dB}/20}$) and the delay from the slope of the phase line, working in radians.

    WATCH OUT FOR

    Taking the delay as $T=60$ because the phase is $-60^{\circ }$ at $\omega =1\text{rad/s}$: $\omega T$ is in radians, so $T=\pi /3\approx 1.047\text{s}$. A second slip is reading the dB value as the gain itself, or converting with $10^{8/10}$ (a power ratio) instead of $10^{8/20}$. $10^{8/10}\approx 6.3$ matches no option, but the ${\log }_{10}\omega$ axis on the magnitude plot can also push students to treat the phase plot as logarithmic, where a delay does not plot as a straight line.

    Full solution →
27. Q.37MCQ · 2 marksOfficial answer: D

    Analog Electronics › Op-amps

    Each feedback branch conducts in only one direction. For $V_{in}>0$, the $1\text{k}\Omega$–$D_1$ branch makes the circuit a unity-gain inverting amplifier, so $V_O=-V_{in}$ and it reaches $-10\text{V}$. For $V_{in}<0$, the $D_2$–Zener branch clamps the output at $V_{-}+V_Z=+5\text{V}$. Neither case reaches a rail, so the $\pm 15\text{V}$ supplies are distractors.

    WATCH OUT FOR

    Thinking the op-amp saturates at $+15\text{V}$ in the negative half-cycle because $D_1$ is off, which leads to option (A). This overlooks the $D_2$–Zener path, which closes the loop at $V_O=+5\text{V}$. Two other slips give the wrong numbers: misreading the Zener's orientation, so the clamp lands on the negative half (option C), or letting the positive half-cycle run to the $-15\text{V}$ rail (option B) instead of following the unity-gain relation $V_O=-V_{in}$.

    Full solution →
28. Q.38MCQ · 2 marksOfficial answer: B

    Control Systems › Compensators & PID

    For a lead network $\frac{1+s/a}{1+s/(\beta a)}$, the maximum phase lead occurs at the geometric mean of the corners, ${\omega }_m=a\sqrt{\beta }$. The gain there is $\sqrt{\beta }$, which is $10{\log }_{10}\beta$ dB, or half the high-frequency gain of $20{\log }_{10}\beta$ dB. The asymptotic and exact values agree at this frequency: $|K(j{\omega }_m)|=\sqrt{1+\beta }/\sqrt{1+1/\beta }=\sqrt{\beta }$. One condition fixes $\beta$, and the other then fixes $a$.

    WATCH OUT FOR

    Using $20{\log }_{10}\beta$ (the high-frequency gain, reached only above $\omega =\beta a$) as the gain at ${\omega }_m$. Setting $20{\log }_{10}\beta =6$ gives $\beta \approx 2$, and $a\sqrt{\beta }=4$ then puts the answer near (D). Option (A) catches students who check only $a\sqrt{\beta }=4$ and ignore the gain condition. Another slip is placing ${\omega }_m$ at the arithmetic mean $(a+\beta a)/2$ of the corners instead of the geometric mean.

    Full solution →
29. Q.39MCQ · 2 marksOfficial answer: C

    Networks (Electric Circuits) › Three-phase (EE)

    With its current coil in one line (R) and its pressure coil across the other two lines (Y to B), one wattmeter reads $V_L{I}_L\sin \varphi$, not $V_L{I}_L\cos \varphi$. For sequence R-Y-B, $V_{YB}$ lags $V_{RN}$ by exactly $90^{\circ }$, so the reading measures reactive power: $Q_{\text{total}}=\sqrt{3}W$. The sign of the reading tells you whether the load lags or leads.

    WATCH OUT FOR

    There are three common mistakes. (1) Using the phase voltage $400/\sqrt{3}\approx 231\text{V}$ instead of the line voltage across the pressure coil. That gives $\sin \varphi =-0.866$ and the trap option (B), 0.5 leading. (2) Reading the pressure coil as $V_{BY}$ instead of $V_{YB}$ (ignoring the + mark on C), or taking the sequence as R-B-Y. Either one flips the sign and gives 0.866 lagging. (3) Treating the reading as $V_L{I}_L\cos \varphi$ as if it were an active-power meter.

    Full solution →
30. Q.40MCQ · 2 marksOfficial answer: C

    Digital Electronics › ADC & DAC

    In the figure the first code change comes at half an LSB (9.8 mV), and each code is centred on a whole number of LSBs. So this ADC rounds to the nearest code: output = round(Vin / LSB), where 1 LSB = V_FS/(2^n − 1) = 5/255 V ≈ 19.6 mV, as the graph shows.

    WATCH OUT FOR

    Truncating 101.6 to 101 gives 65H, which is option (B). Truncation is what an ADC with no offset would do, but here the figure puts the first code change at 9.8 mV (half an LSB), so the ADC rounds and 101.6 becomes 102 (66H). A second slip is using 1 LSB = 5/256 V = 19.53 mV. That does not match the 19.6 mV spacing printed on the graph, and it only reaches 102 because the result, 102.0006, happens to sit on a code boundary.

    Full solution →
31. Q.41MCQ · 2 marksOfficial answer: D
32. Q.42MSQ · 2 marksOfficial answer: A, B, D

    Engineering Mathematics › Calculus

    $|x_1{|}^p+|x_2{|}^p\le 1$ is the unit ball of the $p$-norm. At $p=1$ it is a diamond (area 2), at $p=2$ the unit circle (area $\pi$), and as $p\to \infty$ it fills out the square $[-1,1{]}^2$ (area 4). As $p\to 0$ it collapses onto the two axes (area 0). Every one of these regions lies inside the square $[-1,1{]}^2$, and in general $A(p)=4\Gamma (1+1/p{)}^2/\Gamma (1+2/p)$.

    WATCH OUT FOR

    The usual slip is taking the $p\to 0$ limit to be a unit square or some other finite shape and ticking (C). For any point off the axes, $|x{|}^p\to 1$ as $p\to 0$, so the sum tends to 2 and the point drops out: the region shrinks to zero area. At $p=1$, students also often forget the factor of 4 from symmetry and give $\frac{1}{2}$, or take the diamond's side as 1 instead of $\sqrt{2}$ and get an area of 1.

    Full solution →
33. Q.43MSQ · 2 marksOfficial answer: Marks to all

    Electromagnetics › Magnetostatics

    When uniform E and B fields are at right angles, a charge released from rest feels no force along B, so it stays in the plane perpendicular to B. In that plane it traces a cycloid: its velocity oscillates at the cyclotron frequency $\omega =qB/m$ on top of a steady drift of speed $E/B$ along $\mathbf{E}\times \mathbf{B}$.

    WATCH OUT FOR

    Picking (C) because "a charge in a magnetic field moves in a circle". That is true only with no E field and velocity perpendicular to B. With E present, a charge released from rest follows a cycloid, not a circle. Picking (A) is the opposite mistake: forgetting that the magnetic force turns the motion along E into motion along y, so z keeps coming back to zero. Also, the sentence and the figure give different field directions. Decide which one you are using before you work out the plane of motion and the drift direction.

    Full solution →
34. Q.44MSQ · 2 marksOfficial answer: B, C

    Analog Electronics › Diode circuits

    An ideal current source pushes its current whatever the voltage across it. So KCL at a node fed by a current source forces at least one diode leaving that node to conduct. Once you know which diode must carry the current, use the ideal-diode rule (a conducting diode has $V_{AK}=0$, and no diode can have $V_{AK}>0$) to decide which of the rival diodes is the one.

    WATCH OUT FOR

    Taking the 10 V source as the voltage of node P. It is in series with the 4 A source, so it fixes neither $V_P$ nor the current. The second common slip is assuming D3 can be off when S is OFF. With the switch open, the 2 A source's current into node R has no other way out, so D3 must conduct and clamp R at 40 V.

    Full solution →
35. Q.45NAT · 2 marksOfficial answer: 0.187 to 0.188

    Engineering Mathematics › Probability & statistics

    "Number of trials until the first head" follows a geometric distribution with mean $1/p$. From $E[X]=4$ you get $p=1/4$, and then $P(X=2)=(1-p)p$: one tail followed by one head.

    WATCH OUT FOR

    Reading the "expected number of trials" as the expected number of tails before the first head. That mean is $q/p$, so setting $q/p=4$ gives $p=1/5$ and $P=0.8\times 0.2=0.16$, which is wrong. The question counts every trial including the one that gives the head, so the mean is $1/p$. A second slip is writing $P(X=2)=p^2$ or $p$, which forgets that the first toss has to be a tail.

    Full solution →
36. Q.46NAT · 2 marksOfficial answer: 2

    Control Systems › State space

    For a stable system, the steady-state response to $\sin (\omega t)$ is $|G(j\omega )|\sin (\omega t+\angle G(j\omega ))$. It vanishes only at a frequency where $G(s)$ has a zero on the $j\omega$-axis. So find $G(s)=C(sI-A{)}^{-1}B+D$ and look for purely imaginary zeros. Here $G(s)=\frac{s^2+4}{(s+1{)}^2}$, whose zeros are at $s=\pm j2$.

    WATCH OUT FOR

    Dropping the feedthrough term $D=1$. Without it, $G(s)=\frac{3-2s}{(s+1{)}^2}$, whose only zero is the real $s=1.5$, and $|G(j\omega )|$ is never zero for $\omega >0$. Other slips: getting a sign wrong in the adjoint of $sI-A$ (it is $\left[\begin{array}{cc}s+2& 1\\ -1& s\end{array}\right]$), and not checking stability before using the frequency-response argument.

    Full solution →
37. Q.47NAT · 2 marksOfficial answer: 1.58 to 1.59

    Electrical Machines › Synchronous machines

    When armature resistance is included, the maximum mechanical power a synchronous motor develops is $P_{max}=\frac{EV}{|Z_s|}-\frac{E^{2}R}{|Z_s{|}^2}$, and it occurs at a load angle equal to the impedance angle, $\delta =\theta$. Setting this equal to the static stability limit gives a quadratic in $E$.

    WATCH OUT FOR

    A common mistake is to use the lossless formula $P_{max}=\frac{EV}{|Z_s|}$ (or $\frac{EV}{X_s}$) and drop the $-\frac{E^{2}R}{|Z_s{|}^2}$ term. That gives $0.79$, or $0.75$ with $X_s$. Here $R=0.1\text{p.u.}$ is a third of $X_s$ and cannot be ignored. Another mistake is to maximise the input power, $\frac{V^{2}R}{|Z_s{|}^2}+\frac{EV}{|Z_s|}$, which gives $0.47$. The static stability limit refers to the power developed at the shaft side of $E$.

    Full solution →
38. Q.48NAT · 2 marksOfficial answer: 760

    Electrical Machines › Induction machines

    Neglecting stator impedance, the low-slip torque approximation gives $T\propto \frac{V^{2}s}{f}$; since $V$ and $f$ are both scaled down by the same 20% (V/f held constant), torque becomes proportional to the slip frequency $sf$, so a constant-torque load keeps $sf$ unchanged across the frequency change.

    WATCH OUT FOR

    Assuming the slip stays at its rated value (0.04) and only recomputing the new synchronous speed with the reduced frequency, forgetting that a constant-torque load forces the slip to change (via constant slip-frequency $sf$) when both V and f drop together.

    Full solution →
39. Q.49NAT · 2 marksOfficial answer: 48

    Signals and Systems › Signals basics

    For a discrete-time sinusoid $\sin ({\Omega }_{0}n)$, write $\frac{{\Omega }_0}{2\pi }$ as a fraction $\frac{k}{N}$ in lowest terms; the period is the denominator $N$. For a sum of sinusoids, the period is the LCM of the individual periods. Constants, amplitudes and phase shifts play no part.

    WATCH OUT FOR

    Using the continuous-time formula $\frac{2\pi }{{\Omega }_0}$ gives $\frac{16}{15}$ for the first term. That is not an integer, so it cannot be a discrete-time period; you must go on to the smallest integer $N=16$. A second mistake is to cancel the $\pi$ wrongly and take $N_1=8$ from the denominator of $\frac{15\pi }{8}$, which gives $\text{LCM}(8,6)=24$ instead of 48.

    Full solution →
40. Q.50NAT · 2 marksOfficial answer: 0.037 to 0.039

    Signals and Systems › DTFT & DFT

    When one period of a periodic signal is a finite-length signal $x[n]$ padded with zeros, its DTFS coefficients are the DTFT of $x[n]$ sampled at $N$ equally spaced frequencies and divided by $N$: $a_k=\frac{1}{N}X\left(\frac{2\pi k}{N}\right)$. Here $x[n]=\{\frac{1}{2},1,\frac{1}{2}\}$ fits inside $0\le n\le 4$, so nothing overlaps (no time-domain aliasing) and $a_3=\frac{1}{5}X\left(\frac{6\pi }{5}\right)$.

    WATCH OUT FOR

    Forgetting the $\frac{1}{N}$ factor, which gives $|X(6\pi /5)|=0.191$ instead of $0.038$. Another mistake is using $\cos (6\pi /5)=+0.809$ (the angle $216^{\circ }$ is in the third quadrant, so its cosine is negative), which gives $0.362$. Students also sometimes sample at $\Omega =2\pi k$ or $\pi k/N$ instead of $2\pi k/N$.

    Full solution →
41. Q.51NAT · 2 marksOfficial answer: -12.1 to -11.9

    Networks (Electric Circuits) › Theorems

    With $a$-$b$ open, $V_{th}$ is just the open-circuit voltage at node A. Because no current leaves through terminal $a$, the current through the dependent source equals the controlling current $i_x$. That turns the circuit into one KCL equation at node B with one unknown, $i_x$. The resulting transfer $V_{th}/I=12.5$ is purely real, so the Thevenin voltage is in phase with $\sin 1000t$.

    WATCH OUT FOR

    Working out $\sin (5^{\circ })$ with the calculator in degree mode gives $+1.09\text{V}$. The argument $1000t=5$ is in radians, so the answer is $-11.99\text{V}$. Other slips: writing the dependent-source KVL with the polarity reversed ($V_B=V_A+4i_x$), and assuming the source path carries none of the current or all of it, when with $a$-$b$ open it carries exactly $i_x$.

    Full solution →
42. Q.52NAT · 2 marksOfficial answer: 237 to 239

    Networks (Electric Circuits) › Two-port

    A resistor placed across the top terminals 1 and 2 of a common-ground two-port is itself a two-port connected in parallel with it. Add its y-matrix, $\frac{1}{R}\left[\begin{array}{cc}1& -1\\ -1& 1\end{array}\right]$, to the given one. Then apply the two terminations: $V_1$ fixed by the source and $I_2=-V_2/R_L$ at the load.

    WATCH OUT FOR

    The most common mistake is to ignore the 3 Ω bridging resistor and use the given y-parameters alone. That gives $V_2=300/7\text{V}$ and about $306.12\text{W}$. Another mistake is to write the load condition as $I_2=+V_2/R_L$, forgetting that $I_2$ is defined into the port. That gives $V_2=340/7\text{V}$ and about $393.20\text{W}$. Some students also give the bridging resistor $y_{12}=+\frac{1}{3}\text{S}$ instead of $-\frac{1}{3}\text{S}$.

    Full solution →
43. Q.53NAT · 2 marksOfficial answer: 2150 to 2250

    Electrical Machines › Transformers

    The a-b winding is open-circuited, so its rated induced EMF (200 V) together with the given $B_{max}$ (the flux density at which the transformer is being operated) fixes its number of turns via $E=4.44fN{\Phi }_{max}$. The core's magnetic permeance (${\mu }_0{\mu }_{r}A/l$) then gives the magnetizing inductance referred to that same winding as $N^2$ times the permeance. The total self-inductance of a real winding is always (its own leakage inductance) + (its own magnetizing inductance referred to it) — the leakage impedance data for the c-d side is a distractor for this particular sub-question.

    WATCH OUT FOR

    Using the leakage reactance value of the c-d winding (or adding both leakage inductances) instead of only the a-b winding's own leakage inductance when forming $L_{aa}$; or forgetting that $B_{max}$ pins down the actual peak flux (via the a-b EMF equation) needed to get $N_{ab}$, and instead trying to solve for turns ratio from incomplete data.

    Full solution →
44. Q.54NAT · 2 marksOfficial answer: 1.56 to 1.58

    Networks (Electric Circuits) › Transients

    If the post-switching forced response is DC, the capacitor shows no transient only when you switch at the moment its pre-switching voltage already equals the new final value. Here that means the AC steady-state voltage $5\sqrt{2}\sin (1000t-\pi /4)$ has to equal $+5\text{V}$. The time constant plays no part.

    WATCH OUT FOR

    One slip is to include the battery in the pre-switching circuit, but $\overline{K}$ shorts that branch. Another is to forget the $-45^{\circ }$ phase of the parallel RC impedance and solve $5\sqrt{2}\sin 1000t_1=5$, which gives $0.785\text{ms}$. A third is to misread the battery's polarity. Taking $v_C(\infty )=-5\text{V}$ leads to $4.71\text{ms}$.

    Full solution →
45. Q.55NAT · 2 marksOfficial answer: 7.1 to 7.5
46. Q.56NAT · 2 marksOfficial answer: 8

    Digital Electronics › Sequential

    A state that does not change with the clock is a fixed point of the next-state equations: for every D flip-flop, $Q^{+}=D$ must equal the present $Q$. Write the D inputs in terms of $Q_1,Q_2$, solve $Q_1^{+}=Q_1$ and $Q_2^{+}=Q_2$, then work out the node values A, B, C, D from that state.

    WATCH OUT FOR

    Reading A as $Q_1$ instead of $\overline{Q_1}$ (A is wired to the $\overline{Q}$ output). That changes the equations to $Q_2^{+}=\overline{Q_1}\oplus Q_2$, which gives the fixed state $11$ and the wrong answer $(1011{)}_2=11$. Two other traps: answering with the flip-flop state $Q_1{Q}_2=00$ (giving 0) when the question asks for the four node values [ABCD], and forgetting that a NOR with both inputs tied together is just an inverter, so $B=\overline{A}$.

    Full solution →
47. Q.57NAT · 2 marksOfficial answer: 2

    Digital Electronics › Processors & memory

    Do the 8-bit addition once, then read each flag off it separately. Carry is an unsigned idea (carry out of bit 7). Auxiliary carry is the carry out of bit 3. Overflow is a signed idea: it equals the carry into bit 7 XOR the carry out of bit 7. Parity follows whatever convention the question states, here 0 for an even number of 1s.

    WATCH OUT FOR

    Treating overflow as the same thing as carry. Here there is no carry ($166\le 255$) but there is signed overflow ($+96+70=166>+127$), so $O=1$ while $C=0$. A second trap is using the 8085/8051 convention (parity flag set for even parity). This question defines P as 0 for even, so four 1s gives $P=0$, not 1.

    Full solution →
48. Q.58NAT · 2 marksOfficial answer: 39 to 40.5
49. Q.59NAT · 2 marksOfficial answer: 245 to 255

    Analog Electronics › Biasing

    With $V_{BE}=0$, the Zener is connected straight across the 10 $\Omega$ emitter resistor and is reverse biased by $10I_E$. Assume one Zener state, solve the emitter-follower bias with the emitter resistance reflected into the base loop as $(\beta +1)(10+20)\Omega$, then check the assumption. Here breakdown gives a negative Zener current, so the Zener is off. That leaves $V_Z=2.5\text{V}<5\text{V}$ and $I_{20}=I_E=250\text{mA}$.

    WATCH OUT FOR

    Seeing a 5 V Zener and assuming it is in breakdown. That puts 5 V across the 10 Ω resistor, so $I_E=0.5\text{A}$ and $I_{20}\approx 496.44\text{mA}$ (or 500 mA if the Zener current is ignored), and the Zener current comes out negative, which is impossible. Always check a Zener's state after solving. Other slips: leaving out the $(\beta +1)$ factor on the 20 Ω resistor in the base loop, and reporting $I_C$ instead of $I_E$ as the current in the 20 Ω resistor.

    Full solution →
50. Q.60NAT · 2 marksOfficial answer: 7.1 to 7.3
51. Q.61NAT · 2 marksOfficial answer: 2.49 to 2.51

    Electromagnetics › Magnetostatics

    An infinite sheet of surface current $\mathbf{K}$ produces a uniform field $\mathbf{H}=\frac{1}{2}\mathbf{K}\times {\hat{a}}_n$ on each side. Its magnitude is $K/2$ however far the point is from the sheet. The position of the point only sets the side of the sheet, and so the sign of the direction.

    WATCH OUT FOR

    A common error is to treat the sheet like a line current and let the field fall off with distance, for example by using the distance $\sqrt{3}$ from the origin or $1$ from the plane. That gives a wrong magnitude. Another is to use $H=K$ (the field between two opposite sheets, or inside a solenoid) instead of $K/2$. Getting ${\hat{a}}_n$ backwards only flips the sign of ${\hat{a}}_y$. It does not change the magnitude.

    Full solution →
52. Q.62NAT · 2 marksOfficial answer: 9 to 10

    Engineering Mathematics › Vector calculus

    For $\vec{F}=-y\hat{i}+x\hat{j}$, Green's theorem gives ${\oint }_C(xdy-ydx)=2\times \text{(enclosed area)}$. The problem therefore reduces to the area of the cardioid $r=1+\cos \theta$, which is $\frac{1}{2}{\int }_0^{2\pi }{r}^{2}d\theta =\frac{3\pi }{2}$. The same result follows directly from $xdy-ydx=r^{2}d\theta$.

    WATCH OUT FOR

    The usual slip is reporting the area $3\pi /2\approx 4.71$ instead of twice the area. The curl term is $1-(-1)=2$, not 1. The second most common error is forgetting the factor $\frac{1}{2}$ in the polar area formula, which gives $6\pi$. Other errors: dropping the ${\cos }^2\theta$ term, which gives $2\pi$, or integrating only over $0$ to $\pi$, which covers half the curve.

    Full solution →
53. Q.63NAT · 2 marksOfficial answer: 80

    Signals and Systems › Sampling

    First split the product into its two tones (60 Hz and 120 Hz). Sampling at f_s copies each tone f_0 to |k f_s ± f_0|. The reconstruction filter keeps every copy below its cut-off, and that includes any alias that folds down into the passband.

    WATCH OUT FOR

    Three answers come up often and are wrong. 90 Hz comes from reading the tones straight off the product (90 Hz and 30 Hz) without applying product-to-sum. 60 Hz comes from assuming the filter just deletes the 120 Hz tone. 120 Hz comes from ignoring aliasing. The 120 Hz tone is not simply removed: it lands at 80 Hz, inside the passband, and survives.

    Full solution →
54. Q.64NAT · 2 marksOfficial answer: 414 to 416

    Networks (Electric Circuits) › Three-phase (EE)

    Convert the balanced delta load to an equivalent star ($Z_Y=Z_{\Delta }/3=5-j5\Omega$) and solve a single-phase circuit: a stiff source, then $j10\Omega$ of line reactance, then the star-phase load. The divider ratio $(5-j5)/(5+j5)=-j$ has magnitude 1. So the load line voltage has the same magnitude as the supply (415 V) and lags it by $90^{\circ }$.

    WATCH OUT FOR

    Putting the line inductance in series with the delta branch impedance without converting to star first, i.e. $j10+15-j15$. Line elements carry line current, so they can only be combined with the per-phase star equivalent $Z_{\Delta }/3$. The second trap is assuming a series line inductance must lower the load voltage. With a capacitive load there can be no drop at all, or even a rise. Here the magnitude is unchanged.

    Full solution →
55. Q.65NAT · 2 marksOfficial answer: 10

    Engineering Mathematics › Vector calculus

    The maximum directional derivative of a scalar function at a point is the magnitude of its gradient there, and it is reached in the direction of the gradient: ${\max }_{\hat{u}}{D}_{\hat{u}}f=|\nabla f|$.

    WATCH OUT FOR

    Dropping the cross term $x_1{x}_2$ when differentiating. It adds $x_2$ to $\partial f/\partial x_1$ and $x_1$ to $\partial f/\partial x_2$. Leaving it out gives $\nabla f(1,1)=5\hat{i}+7\hat{j}$ and a magnitude of $\sqrt{74}\approx 9$ instead of $10$. The other common slip is adding the components ($6+8=14$) where the square root of the sum of squares is needed.

    Full solution →