GATE EE 2025 question paper with solutions

GATE EE 2025 answer key

General Aptitude Q.1 to Q.10

1. Q.1MCQ · 1 markOfficial answer: D

   General Aptitude › Grammar & vocabulary

   Fill the blanks in two passes. First use tense and grammar: "yesterday" needs past forms, and the bare verb "go" after the first blank needs a modal or "did". Then use meaning: illness stops you from being able to do something, so the modal of inability, "couldn't", fits better than the modal of refusal, "wouldn't".

   WATCH OUT FOR

   Picking (B) "wouldn't; wasn't" because it is also grammatical. "Wouldn't go" means Kavya refused to go, which is a choice. Being unwell is a reason she was unable to go, so "couldn't" is the right fit. Some students also pick (C) because it repeats "wasn't", but "wasn't go" is ungrammatical ("was" would need "going").

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2. Q.2MCQ · 1 markOfficial answer: A

   General Aptitude › Reasoning

   Name the relation in the first pair before you look at the options. Here Good and Evil are antonyms, so you need the antonym of Genuine, and that is Counterfeit (a fake made to pass as real).

   WATCH OUT FOR

   Three of the four options start with "Counter-", and that prefix can suggest "against" or "opposite". Choosing Counterpart or Counterfoil because of the prefix is the trap. Only Counterfeit actually means fake. Contraband also sounds wrong-doing and can tempt, but it means illegal goods, not false goods.

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3. Q.3MCQ · 1 markOfficial answer: A

   General Aptitude › Geometry

   Put the two intercepts straight into $x+py+q=0$. At $(-2,0)$ the equation becomes $-2+q=0$, so $q=2$. At $(0,4)$ it becomes $4p+q=0$, so $p=-\frac{1}{2}$. The coefficient of $x$ is fixed at 1, so the equation cannot be rescaled and $p$ and $q$ are unique.

   WATCH OUT FOR

   Seeing slope 2 and intercept 4 on the graph and choosing (D) $p=2;q=4$. Those numbers belong to $y=2x+4$, not to the form $x+py+q=0$. A second trap is a sign slip when solving for $x$. From $y=2x+4$ you get $x=\frac{1}{2}y-2$, which is $x-\frac{1}{2}y+2=0$, not $x+\frac{1}{2}y+\dots$. Getting that sign wrong points you to (C). Always check the answer by substituting one plotted point, such as $(-2,0)$.

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4. Q.4MCQ · 1 markOfficial answer: B

   General Aptitude › Reasoning

   Each Venn diagram shows one relationship. Disjoint circles inside a big circle: two categories that never share a member, both belonging to a larger class. Overlapping circles inside a big circle: two categories that can share members, both inside a larger class. Three mutually overlapping circles: three roles a person can hold in any combination. Nested circles: a chain in which each item is contained in the next.

   WATCH OUT FOR

   Swapping (P) and (Q) for rows (1) and (4). Zebra and Giraffe can never overlap, so they need the diagram with separate inner circles (Q). Pianist and Guitarist can overlap, because one person can play both, so they need the diagram with overlapping inner circles (P). Another slip is to draw Director, Producer, Actor as nested circles. None of these roles is a subset of another, so the circles must overlap (R).

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5. Q.5MCQ · 1 markOfficial answer: C

   General Aptitude › Numerical

   Write every term in the same base (here $27=3^3$) and then work only with exponents: multiply for a power of a power, and subtract for a quotient.

   WATCH OUT FOR

   Stopping at $3^{81-12}=3^{69}$ and forgetting the outer power $\frac{1}{3}$, which gives option (D). The other slip is to leave $27^4$ as it is, or to write it as $3^7$ by adding $3+4$ when the exponents should be multiplied to give $3^{12}$.

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6. Q.6MCQ · 2 marksOfficial answer: C

   General Aptitude › Reading

   Start with the one sentence that makes sense on its own and names the subject (Q, \"Tigers…\"). Then join the rest using pronoun reference (S's \"they\" means the tigers), the order of events in a hunt (search, then trail, then chase: S → R) and cause and effect (P's \"It is because\" explains R's \"may not catch them\").

   WATCH OUT FOR

   Opening with S because \"For several nights\" sounds like a story's first line (option A), or with P because it holds the vivid punchline (option D). Neither works. S's \"they\" has no noun before it to refer to, and P's \"It is because\" needs an earlier statement to explain. The opener must be the sentence that names \"Tigers\", which is Q.

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7. Q.7MCQ · 2 marksOfficial answer: C

   General Aptitude › Reasoning

   Get the rule from both examples: EF = 30 and HJ = 80 are the products of alphabet positions ($5\times 6$ and $8\times 10$). Then get the layout from the figure: each code sits in the sector diametrically opposite its letter pair. The layout says which pair (i) and (ii) belong to, and the rule gives their values.

   WATCH OUT FOR

   Assuming that (i) and (ii), which sit next to each other, are a letter pair and its code. On that reading option (A) looks right, because EH gives $5\times 8=40$, matching its (ii) = 40. The figure rules this out: EF is not next to 30, and HJ is not next to 80. Each code is diametrically opposite its pair, so (ii) must be FG's code, 42. Other slips are adding the positions instead of multiplying them ($5+6=11$, not 30) and miscounting J as 9.

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8. Q.8MCQ · 2 marksOfficial answer: B

   General Aptitude › Statistics & probability

   Sort each row before you read off the median. With 7 values, the median is the 4th value in sorted order. The mean is the sum divided by 7, and because both students have the same number of scores you can compare their means by comparing their sums. The difference between the maximum and minimum is the range.

   WATCH OUT FOR

   Reading the median from the unsorted table. The 4th column shows 45 for P and 56 for Q, and some students take those as the medians. The sorted median of P is actually 50. A second trap is marking the one statement you have checked and found true, even though the question asks for the INCORRECT one.

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9. Q.9MCQ · 2 marksOfficial answer: C

   General Aptitude › Geometry

   Integer centres form a unit cubic lattice, and the squared distance to any other centre is $l^2+m^2+n^2$. The nearest centre gives the smallest value of that sum ($1$, along an edge) and the next nearest gives the next value ($2$, along a face diagonal). So $x=1$, $y=\sqrt{2}$, and $\frac{y}{x}=\sqrt{2}$.

   WATCH OUT FOR

   Taking the next nearest centre as the second one along the same axis, $(2,0,0)$, which gives $y=2$ and option (D). The face-diagonal centre $(1,1,0)$ at $\sqrt{2}$ is closer. Other slips are jumping to the body diagonal $(1,1,1)$ at $\sqrt{3}$, which is the third-nearest, or inverting the ratio to get $\frac{1}{\sqrt{2}}$ (option B).

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10. Q.10MCQ · 2 marksOfficial answer: A

    General Aptitude › Geometry

    Because ST is parallel to QR, the basic proportionality theorem gives $PS:SQ=PT:TR=3:2$, so the diameters PS and PQ are in the ratio $3:5$. Semicircle areas go as the square of the diameter ($9:25$), and the shaded crescent is the difference, $25-9=16$ parts. Against the PS semicircle's 9 parts, that is $\frac{16}{9}$.

    WATCH OUT FOR

    Taking PQ as 3 + 2 but then comparing lengths instead of areas, which gives a ratio such as $\frac{5}{3}$ or $\frac{2}{3}$. Area scales with the square of the diameter. A second trap is reading $PS:SQ$ as $2:3$ because S looks close to Q in the figure. The figure is only representative, so the ratio has to come from the parallel line, not from how the drawing looks. Getting the ratio the wrong way round leads to $\frac{21}{4}$, which is not an option.

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Electrical Engineering Q.11 to Q.65

1. Q.11MCQ · 1 markOfficial answer: B

   Engineering Mathematics › Calculus

   Here $f$ depends on $x$ and $y$ only through $u=x+y$. Minimise the one-variable function $g(u)=(u-1{)}^2+u^2$ to get $u=\frac{1}{2}$. Every $(x,y)$ with $x+y=\frac{1}{2}$ gives that same minimum, so the minimizers make up a whole line.

   WATCH OUT FOR

   Setting $\partial f/\partial x=0$ and $\partial f/\partial y=0$ and expecting two independent equations with one solution, which leads to (D). The two equations are identical, so they fix only $x+y$. Another trap is reading the singular Hessian ($\det H=0$) as "no minimum exists" and picking (C). A zero determinant only means the test cannot decide, and completing the square shows $f\ge \frac{1}{2}$ with equality all along the line.

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2. Q.12MCQ · 1 markOfficial answer: B

   Engineering Mathematics › Linear algebra

   Write the scalar ${\mathbf{v}}_1^{T}A{\mathbf{v}}_2$ in two ways, once using $A{\mathbf{v}}_2={\lambda }_2{\mathbf{v}}_2$ and once using $A^T=A$ with $A{\mathbf{v}}_1={\lambda }_1{\mathbf{v}}_1$. Subtracting gives $({\lambda }_1-{\lambda }_2){\mathbf{v}}_1^T{\mathbf{v}}_2=0$, so the eigenvectors are orthogonal whenever ${\lambda }_1\ne {\lambda }_2$.

   WATCH OUT FOR

   Confusing linear independence with orthogonality. Eigenvectors for distinct eigenvalues are linearly independent for ANY matrix, but they are orthogonal (${\mathbf{v}}_1^T{\mathbf{v}}_2=0$) only because this matrix is symmetric. A second mistake is choosing (C) or (D): eigenvectors of distinct eigenvalues can never be multiples of each other, so neither ${\mathbf{v}}_1=-{\mathbf{v}}_2$ nor ${\mathbf{v}}_1={\mathbf{v}}_2$ can hold.

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3. Q.13MCQ · 1 markOfficial answer: B

   Engineering Mathematics › Linear algebra

   Compare $\rho (\mathbf{A})$ with $\rho ([\mathbf{A}|\mathbf{b}])$. If they differ, there is no solution. If they are equal to $n$, the solution is unique. If they are equal but less than $n$, there are infinitely many solutions. In this question row 2 is $-1$ times row 1 and $\mathbf{b}$ follows the same relation ($-1/3=-1\times 1/3$), so the dependent equation adds no new information and no contradiction.

   WATCH OUT FOR

   Seeing $\det \mathbf{A}=0$ and choosing 'no solution' (D) straight away. A singular matrix only rules out a unique solution. You still have to compare the rank of $\mathbf{A}$ with the rank of the augmented matrix. Option (C) is a trap: a linear system has exactly zero, one or infinitely many solutions, never a finite number greater than one.

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4. Q.14MCQ · 1 markOfficial answer: A

   Engineering Mathematics › Linear algebra

   Every eigenvalue of $P$ is $1$, and $(P-I{)}^2=0$, so the minimal polynomial is $(\lambda -1{)}^2$. Expanding $(P-I{)}^2=0$ gives $P^2=2P-I$ directly. For a $3\times 3$ matrix like this one, simply multiplying $P$ by itself and comparing with each option is just as quick, and it is a good check on the theory.

   WATCH OUT FOR

   Going only as far as Cayley-Hamilton, which gives the cubic $(P-I{)}^3=0$, and then trying to force a quadratic answer out of it without first checking that $(P-I{)}^2$ is already zero. The other slip is multiplying carelessly: a single sign error in the $(2,1)$ or $(2,2)$ entry of $P^2$ (true values $-2$ and $-1$) makes $P+I$ or $P$ look like a match.

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5. Q.15MCQ · 1 markOfficial answer: B

   Engineering Mathematics › Probability & statistics

   Conditioning on $X=1$ keeps only the $X=1$ entries of the joint PMF, rescaled by $P(X=1)$ so they add to 1. Then $E[Y\mid X=1]={\sum }_{y}yP(Y=y\mid X=1)$. Because $Y$ takes only the values 0 and 1, this equals $P(Y=1\mid X=1)$.

   WATCH OUT FOR

   Using the joint probability $P(X=1,Y=1)=\frac{1}{8}$ as the answer without dividing by $P(X=1)$ gives option (C). Taking the unconditional mean $E[Y]=P(Y=1)=\frac{1}{2}+\frac{1}{8}=\frac{5}{8}$ ignores the condition altogether, and $\frac{5}{8}$ is not among the options.

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6. Q.16MCQ · 1 markOfficial answer: C

   Analog Electronics › Amplifiers

   The sign of the gain depends on which terminal takes the input. At the gate (CS), $v_{gs}=+v_i$. At the source (CG), $v_{gs}=-v_i$. The output at the drain is always $-g_m{v}_{gs}{R}_D$, so the extra minus sign in CG cancels the inversion and gives a gain of the same size, $g_m{R}_D$, but positive.

   WATCH OUT FOR

   Students often assume every transistor amplifier with the output at the collector or drain is inverting, like the CE/CS stage, and choose (A). Driving the source makes $v_{gs}=-v_i$, and that second sign flip makes the common gate stage non-inverting. Another error is to confuse CG with the source follower (common drain). Both have positive gain, but the follower's gain is below 1 and its output is at the source, not the drain.

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7. Q.17MCQ · 1 markOfficial answer: D

   Analog Electronics › Op-amps

   With negative feedback, an ideal op-amp holds its two inputs at the same voltage (virtual short). Op-amp 2 is a follower that feeds $V_{out}$ back unchanged to the inverting input of op-amp 1, so op-amp 1 forces $V_{out}=V_{in}$. Finite-gain check: with gains $A_1$ and $A_2$, $v_{o2}=\frac{A_2}{1+A_2}{V}_{out}$ and $V_{out}=\frac{A_1{V}_{in}}{1+A_1\frac{A_2}{1+A_2}}$, which tends to $V_{in}$ as $A_1,A_2\to \infty$. The loop is a two-op-amp voltage follower.

   WATCH OUT FOR

   Being thrown by the two op-amps and guessing a difference or summing amplifier. Summing and difference amplifiers need a resistor network to weight and combine their inputs, and a logarithmic amplifier needs a diode or BJT in its feedback path. This circuit has none of these and only one input, $V_{in}$. Another slip is reading op-amp 2's feedback as going to its $+$ terminal, which would make it positive feedback. Its output actually loops to its own $-$ input, and $V_{out}$ goes to its $+$ input.

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8. Q.18MCQ · 1 markOfficial answer: B

   Networks (Electric Circuits) › Laws & node/mesh

   A current source in parallel with a resistor is a Norton source. KCL gives $I=I_s+V/R$, so the I-V characteristic is a straight line: the source sets the intercept on the current axis and the conductance $1/R$ sets the slope. The source's arrow direction relative to $I_{XY}$ decides the sign of the intercept.

   WATCH OUT FOR

   Picking (A) because the current source seems to fix the terminal current at 1 A. That is only true for a source in series with the resistor. Here the resistor is in parallel, so it draws an extra $V_{XY}/R$ and the line slopes upwards. A second slip is misreading the source direction as opposing $I_{XY}$, which gives $I_{XY}=V_{XY}/R-1\text{A}$ (intercept $-1\text{A}$). That matches no option and should send you back to check the sign.

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9. Q.19MCQ · 1 markOfficial answer: B

   Networks (Electric Circuits) › Laws & node/mesh

   A nullator fixes BOTH $v=0$ and $i=0$. In series, its $i=0$ passes to the resistor (KCL), which then has $v_R=0$ (Ohm's law), so the terminal voltage is also zero (KVL). A nullator in series with any element that has finite impedance is still a nullator.

   WATCH OUT FOR

   Seeing only the zero current and picking open circuit (C), or seeing only the zero voltage and picking short circuit (D). An open circuit leaves the voltage free and a short circuit leaves the current free. Here both are pinned to zero, which is a nullator, not either of the ideal limits.

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10. Q.20MCQ · 1 markOfficial answer: A

    Signals and Systems › LTI & convolution

    For an LTI system the response to $\delta [n]$ is the impulse response $h[n]$, and $y[n]={\sum }_{k}h[k]x[n-k]$. Each nonzero sample $h[k]$ adds the term $h[k]x[n-k]$, a copy of the input delayed by $k$. Here $h[0]=h[1]=h[2]=1$, so $y[n]=x[n]+x[n-1]+x[n-2]$.

    WATCH OUT FOR

    The most common slip is flipping the sign of the shift. $h[n]$ is nonzero at $n=0,1,2$, and students then write $x[n]+x[n+1]+x[n+2]$, which is option (C). In the convolution sum, $h[k]$ multiplies $x[n-k]$, so a nonzero $h[k]$ at positive $k$ means a delay ($x[n-k]$), not an advance. A quick check is to put $x[n]=\delta [n]$ into the option you picked and confirm that you get back the given $h[n]$.

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11. Q.21MCQ · 1 markOfficial answer: A

    Signals and Systems › Signals basics

    Sifting property: an impulse at $t_0$ inside an integral returns the value of the other factor at $t_0$. The impulse $\delta (t+3)$ sits at $t_0=-3$, not $+3$. Check that $t_0$ lies inside the rectangular window $u(t+4)-u(t-4)$, which is $1$ on $-4<t<4$.

    WATCH OUT FOR

    Three common slips. (1) Reading $\delta (t+3)$ as an impulse at $t=+3$. That happens to give the same value here because $t^2$ is even, so the wrong reasoning still lands on the right answer. (2) Dropping the leading minus sign and choosing $9$ (B). (3) Forgetting to square and choosing $\pm 3$ (C or D).

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12. Q.22MCQ · 1 markOfficial answer: B

    Control Systems › Time response

    For a first-order step response $y(t)=y_{ss}(1-e^{-t/T})$, normalise every reading by the final value $y_{ss}$ (here 3). Then $T=-t/\ln (1-y/y_{ss})$ at any point. Equivalently, $T$ is the time at which the output reaches 63.2% of $y_{ss}$.

    WATCH OUT FOR

    Forgetting that the final value is 3, not 1. Looking for an output of 0.632 puts $T$ between 0.6 s and 1.6 s and leads to option (A). The 63.2% mark is $0.632\times 3=1.896$. A second trap is treating $t=10\text{s}$ as the 2% settling time $4T$, which gives $T=2.5\text{s}$. In fact $2.98/3=99.3\%$, which is about $5T$.

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13. Q.23MCQ · 1 markOfficial answer: D

    Control Systems › Nyquist

    When the open loop is stable ($P=0$), the closed loop is stable if and only if the Nyquist plot does not encircle $-1$. The number of net encirclements equals the number of closed-loop RHP poles, $Z=N$. The direction of an encirclement only shows which way the contour was traversed. RHP zeros of $G(s)$, such as the one at $s=3$, are not counted in $P$ or $Z$.

    WATCH OUT FOR

    A common error is to fix a clockwise contour, put the anticlockwise encirclement into $N=P-Z$ as $N=+1$, get $Z=-1$, and then call the data contradictory and pick (A). The question never says which way the plot is traced, and 'anticlockwise' is only meaningful relative to that direction. What does not depend on the direction is that $-1$ is encircled once, and with $P=0$ that means one RHP closed-loop pole. Two other slips: counting the zero at $s=3$ as an open-loop RHP pole, and choosing 'marginally stable' even though the plot does not pass through $-1$.

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14. Q.24MCQ · 1 markOfficial answer: A
15. Q.25MCQ · 1 markOfficial answer: C

    Electromagnetics › Electrostatics

    For a uniformly charged sphere, Gauss's law gives $E_R=\rho R/(3{\epsilon }_0)$ inside, which grows linearly with $R$, and $E_R=Q/(4\pi {\epsilon }_0{R}^2)$ outside. The two agree at $R=a$. With $\rho =-3{\rho }_0$ the factor 3 cancels, leaving $E_R(a)=-{\rho }_{0}a/{\epsilon }_0$. The negative sign means the field points inward, towards the electron cloud.

    WATCH OUT FOR

    Dropping the minus sign because the cloud is described by positive constants ${\rho }_0$ and $a$. That gives the mirror-image graph, Fig. (i). A second slip is to forget the $\frac{4}{3}$ in the sphere's volume, or to use the cylinder result $\rho R/(2{\epsilon }_0)$. Either one gives the wrong peak magnitude and can pull you towards Fig. (iv).

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16. Q.26MCQ · 1 markOfficial answer: A

    Electrical Machines › Induction machines

    Since $Jd\omega /dt=T_{em}-T_L$, the transit time over a fixed speed range is minimized by keeping the developed torque $T_{em}$ at its largest feasible value at every speed — i.e., by riding the upper boundary of the allowed operating region.

    WATCH OUT FOR

    Assuming the shortest-time path is the one with the shortest-looking geometric route (e.g. the diagonal $O$–$P$–$S$ or $O$–$E$–$S$) rather than reasoning about which path maximizes accelerating torque $(T_{em}-T_L)$ at every intermediate speed.

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17. Q.27MCQ · 1 markOfficial answer: A
18. Q.28MSQ · 1 markOfficial answer: A, C

    Electrical Machines › Synchronous machines

    Synchronizing an alternator to an infinite bus/grid requires matching (i) voltage magnitude, (ii) frequency, (iii) phase sequence, and (iv) phase angle (zero at the instant of closing the breaker). A voltmeter (or double-voltmeter method) checks voltage magnitude equality, and a synchroscope directly displays the phase-angle/slip condition, telling the operator exactly when to close the breaker.

    WATCH OUT FOR

    Picking the wattmeter because it seems like a natural "power system" instrument — but real/reactive power flow is only meaningful after paralleling, not during the synchronizing check itself. Similarly, confusing the stroboscope (used for measuring rotational/mechanical speed by a flashing-light effect) with the synchroscope (an electrical instrument that indicates phase difference/slip between two AC systems) is a common name-similarity trap.

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19. Q.29MSQ · 1 markOfficial answer: A, C

    Control Systems › Root locus

    For $K\ge 0$, a real-axis point lies on the root locus exactly when an odd number of real open-loop poles and zeros lie to its right. A quick cross-check is to put the point into $G(s)=-1$: it is on the locus only if the $K$ that comes out is non-negative. Zeros count in the rule just as poles do, including the zero at the origin.

    WATCH OUT FOR

    Counting only the poles, or leaving out the zero at $s=0$. If you count only poles, you shade $(-7,-5)$ correctly but mark every point to the right of $-5$ as off the locus, so you miss $-1$. If you drop the zero at the origin, the parity flips in every segment, and you pick $-4$ and $-10$ instead. Another slip is to assume that the branches from $-5$ and $-7$ run straight along the real axis to the zeros. The segment $(-5,-2)$ is not on the locus, so the branches leave the axis between the poles and come back between the zeros.

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20. Q.30NAT · 1 markOfficial answer: 7

    Signals and Systems › Fourier series

    The quantity asked for is the average power of a periodic signal. By Parseval's theorem it equals the sum of the powers of the orthogonal Fourier components: $a_0^2$ for the DC term plus $A_k^2/2$ for each cosine of amplitude $A_k$. Equivalently it is $\sum |c_k{|}^2$ over the exponential coefficients. You never need to square $x(t)$ out and integrate.

    WATCH OUT FOR

    The usual slips: (1) squaring the amplitudes without halving, which gives $1+3\times 4=13$; (2) halving the DC term as well, which gives $0.5+6=6.5$; (3) adding the amplitudes before squaring, which gives $(1+2+2+2{)}^2=49$, the peak value of $x^2(t)$ at $t=0$ and not its average. The cross terms vanish because the harmonics are orthogonal, and the DC term contributes its full square.

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21. Q.31NAT · 1 markOfficial answer: 5

    Electrical and Electronic Measurements › Errors & standards

    For resistors in parallel, the relative error in the result is a weighted average of the individual relative errors, with weights $R_2/(R_1+R_2)$ and $R_1/(R_1+R_2)$. The weights add up to 1, so when every resistor has the same maximum percentage error, the combination has that same maximum percentage error. Another way to see it: conductances add, so $G=G_1+G_2$, and a sum of terms that each carry 5% error also carries at most 5%.

    WATCH OUT FOR

    Using the product/quotient rule on $R_1{R}_2/(R_1+R_2)$ as if the numerator and denominator were independent: 5% + 5% for the product plus 5% for the sum gives 15%. The same resistors appear in both, so their errors partly cancel. Simply adding the two errors to get 10% is also wrong. A parallel combination never has a larger percentage error than the worst of its resistors.

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22. Q.32NAT · 1 markOfficial answer: 0.975 to 0.985
23. Q.33NAT · 1 markOfficial answer: 0.380 to 0.400
24. Q.34NAT · 1 markOfficial answer: 90
25. Q.35NAT · 1 markOfficial answer: 66.25 to 66.75

    Electrical Machines › Synchronous machines

    For a synchronous machine with P poles, a mechanical angle of rotor shift corresponds to an electrical (torque) angle of δ_elec = (P/2)·δ_mech. With |V| = |E| (no-load emf equals terminal voltage) and negligible armature resistance, the current is found from the phasor difference V − E = jXsIa, whose magnitude for two equal phasors separated by δ is 2V sin(δ/2).

    WATCH OUT FOR

    Treating the given 2° as the electrical torque angle directly, instead of converting it from mechanical to electrical degrees using δ_electrical = (P/2) × δ_mechanical for a 4-pole machine — this halves the actual current found.

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

    Engineering Mathematics › Probability & statistics

    To find the density of a function of a random variable, work out the CDF first and then differentiate: $F_Y(y)=P(X^2\le y)=P(X\le \sqrt{y})=\sqrt{y}$ on $(0,1]$, so $P_Y(y)=\frac{1}{2\sqrt{y}}$. For a monotone map, the shortcut $P_Y(y)=P_X(x)\left|\frac{dx}{dy}\right|$ gives the same answer.

    WATCH OUT FOR

    Assuming $Y$ is uniform because $X$ is uniform (option B). Squaring squeezes values near 0 closer together, so $Y$ piles up near 0 and its density grows without bound as $y\to 0^{+}$. A second trap is checking only that an option integrates to 1. All four options do, so that test can't decide the answer. A third is multiplying by $\left|\frac{dy}{dx}\right|$ where the formula needs $\left|\frac{dx}{dy}\right|$.

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27. Q.37MCQ · 2 marksOfficial answer: A

    Digital Electronics › Boolean logic

    Adjacency, $(A+y)(A+\overline{y})=A$: when two sum terms differ only in one variable, that variable drops out. Use it twice here, first on $x$ and then on $v$, to get $f=\overline{u}+\overline{w}=\overline{uw}$. As a check, the four maxterms are exactly the inputs where $u=w=1$.

    WATCH OUT FOR

    Two mistakes are common. One is to drop the complements and pick (B) ($u+w$), forgetting that every factor has $\overline{u}$ and $\overline{w}$, never $u$ or $w$. The other is to keep the wrong pair of variables and pick (C) or (D). Only $v$ and $x$ appear both complemented and uncomplemented across the factors, so those are the variables that cancel. The variables that are complemented in every factor, $u$ and $w$, are the ones that stay.

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28. Q.38MCQ · 2 marksOfficial answer: C

    Analog Electronics › Amplifiers

    Looking into the drain of a MOSFET whose source has a resistance $R_S$ to ground gives $R=r_{ds}+R_S+g_m{r}_{ds}{R}_S$. In a cascode, $R_S$ is the $r_{ds}$ of the lower transistor, so $R=2r_{ds}+g_m{r}_{ds}^2$. The ideal current-source load is an open circuit in parallel with this, so it leaves the result unchanged.

    WATCH OUT FOR

    Choosing (D) because the load is an ideal current source. The infinite resistance is in PARALLEL with the cascode at the output node, not in series, so it drops out and leaves the finite cascode resistance. Choosing (A) means treating $M_2$ as nothing more than $r_{ds}$ and missing the $g_m{r}_{ds}^2$ boost that comes from its source being degenerated by $M_1$. Choosing (B) means treating $M_2$ as a source follower or diode ($1/g_m$), but here the output is taken at its drain.

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29. Q.39MCQ · 2 marksOfficial answer: C

    Networks (Electric Circuits) › Transients

    When the time constant is much longer than the half-period ($RC\gg T/2$), an RC network with the output across C acts as an integrator. The square wave in becomes a triangular wave out, with zero average and peak $\tanh (T/4RC)\approx T/(4RC)$, so the larger $RC$ (or the higher the frequency), the smaller the triangle.

    WATCH OUT FOR

    Choosing (A) because 'integrator gives a triangle' and forgetting the amplitude. The integrator's output amplitude is about $T/(4RC)$, which is tiny here. A full $\pm 1\text{ V}$ swing would need $RC\ll T$, and in that case the output across C is square, not triangular. The other trap is picking (D): that is the differentiated waveform, taken across R with $RC\ll T$.

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

    Signals and Systems › LTI & convolution

    The area under a convolution is the product of the areas of the two signals, $\int ydt=\left(\int x_{1}dt\right)\left(\int x_{2}dt\right)$, which is the same statement as $Y(0)=X_1(0)X_2(0)$. So you never need to work out $y(t)$ itself.

    WATCH OUT FOR

    Adding the two areas instead of multiplying them: $1.5+1=2.5$ is exactly option (B). A second trap is counting only the rectangle part of $x_1(t)$ and taking its area as 1, which drops the triangle of area $\frac{1}{2}$ on $[1,2]$. Working out the full piecewise convolution also reaches 1.5, but it takes much longer and leaves more room for algebra slips.

    Full solution →
31. Q.41MCQ · 2 marksOfficial answer: B

    Control Systems › Routh–Hurwitz

    The forward path has a zero in the right half-plane at $s=+1$. As $K$ increases, one root-locus branch leaves the pole at $s=-1$ and heads for that zero, crossing the origin at $K=2$. For a second-order characteristic polynomial the Routh condition is simply that all coefficients are positive, here $3+K>0$ and $2-K>0$.

    WATCH OUT FOR

    Writing the characteristic equation as $(s+1)(s+2)+K(s+1)$ or $+K(1-s)$, which loses the sign of the right-half-plane zero. The constant term is $2-K$, not $2+K$. That sign slip makes the system look stable for every $K>0$ and pushes students towards (A) or (D). The other trap is picking (C) because "higher gain destabilises": the loop is still stable for $1<K<2$, so the threshold is 2, not 1.

    Full solution →
32. Q.42MCQ · 2 marksOfficial answer: B

    Signals and Systems › LTI & convolution

    A continuous-time LTI system is BIBO stable if and only if ${\int }_{-\infty }^{\infty }|h(t)|dt<\infty$. An impulse response of $h(t)=u(t)$ makes the system an ideal integrator. Its integral diverges, so the system is unstable, and a bounded step input produces an unbounded ramp.

    WATCH OUT FOR

    Calling the system stable because its impulse response $u(t)$ is bounded (this leads to A). Stability needs $h(t)$ to be absolutely integrable, and being bounded is not enough. A second mistake is to pick (D) by reading 'unstable' as 'every output is unbounded'. Instability only means that at least one bounded input gives an unbounded output. Many inputs, such as $\delta (t)$ or a finite pulse, still give bounded outputs.

    Full solution →
33. Q.43MCQ · 2 marksOfficial answer: A

    Electrical Machines › Transformers

    Match the primary star-winding current on each leg to the co-labelled secondary delta-winding current by ampere-turn balance (N1 I_primary = N2 I_secondary, in phase), then get the delta's external line current from the phasor difference of the two adjacent internal winding currents that meet at that terminal (I_ad = I_a - I_c), and finally compare magnitude and angle with the primary line current.

    WATCH OUT FOR

    Two common errors: (1) taking the delta line current as the sum (I_a + I_c) or as a single winding current (just I_a) instead of the correct phasor difference of the two windings meeting at that corner, which changes both the √3 factor and the ±30° shift; (2) getting the sign of the 30° shift backwards by mislabelling which winding is "incoming" vs "outgoing" at the corner, turning a lag into a lead (confusing options A and B, or C and D).

    Full solution →
34. Q.44MCQ · 2 marksOfficial answer: A

    Electrical Machines › DC machines

    In a DC series motor with a linear magnetic circuit, flux is proportional to armature current (φ ∝ Ia), so electromagnetic torque T ∝ Ia², and back-EMF E ∝ φN ∝ Ia·N. Combining the load's T ∝ N³ law with the motor's T ∝ Ia² law, and then using the circuit equation V = E + Ia·R, pins down the external resistance needed for a given speed change.

    WATCH OUT FOR

    Assuming back-EMF stays proportional to speed alone (E ∝ N) as in a shunt/separately excited motor, forgetting that in a series motor with linear magnetic circuit, flux itself changes with armature current, so E ∝ φN ∝ Ia·N — missing this collapses the whole ratio calculation.

    Full solution →
35. Q.45MCQ · 2 marksOfficial answer: C

    Electrical Machines › Induction machines

    See keyIdea field above.

    WATCH OUT FOR

    Assuming $s_{maxT}=R_r^{\prime }/X_{20}$ must be evaluated at the rated 50 Hz reactance (giving $s_{maxT}=1/4$, not 1) and then either stopping there or wrongly forcing slip = 1 into the rated-frequency torque expression instead of solving for the frequency at which the reactance shrinks enough to make $s_{maxT}=1$; a second common slip is treating $R_r^{\prime }$ as scaling with frequency along with the reactance, which cancels the frequency dependence and gives no solution.

    Full solution →
36. Q.46MCQ · 2 marksOfficial answer: A
37. Q.47MCQ · 2 marksOfficial answer: A
38. Q.48MCQ · 2 marksOfficial answer: A
39. Q.49MCQ · 2 marksOfficial answer: B
40. Q.50MCQ · 2 marksOfficial answer: C

    Electromagnetics › Electrostatics

    Look at how the air-dielectric boundary lies relative to the field. Here the boundary is a radial plane, so the radial field $\mathbf{E}$ runs along it. Both regions then have the same voltage across them and the same $\mathbf{E}$, so they add as parallel capacitors, each scaled by its share of the angle: $C=\sum ({\theta }_i/2\pi ){\epsilon }_{ri}{C}_0$. When the boundary is a cylinder $\rho =c$ instead, it is normal to $\mathbf{E}$, $\mathbf{D}$ is continuous across it, and the layers add in series.

    WATCH OUT FOR

    The common mistake is treating the two regions as series layers. That is right only when the dielectric forms a coaxial shell (a boundary at constant $\rho$), not when it fills an angular sector. The other trap is answering ${\epsilon }_r=5$ or $10$: $5$ assumes filling half the gap scales $C_0$ by ${\epsilon }_r$, and $10$ forgets that the air half still adds $C_0/2$. The correct equation is $\frac{1}{2}(1+{\epsilon }_r)=5$.

    Full solution →
41. Q.51MSQ · 2 marksOfficial answer: B, D

    Electromagnetics › Electrostatics

    For a purely radial field in spherical coordinates, $\nabla \cdot \mathbf{f}=\frac{1}{R^2}\frac{d}{dR}(R^2{f}_R)$. With $f_R=R^{-n}$ this gives $(2-n)R^{-(n+1)}$. That expression can lose its $R$-dependence in two ways: the power of $R$ becomes zero ($n=-1$), or the coefficient becomes zero ($n=2$, the inverse-square field, which has zero divergence).

    WATCH OUT FOR

    Setting only the power of $R$ to zero and marking just (B). Students forget that a divergence of zero is also independent of $R$, so they miss $n=2$, the inverse-square (Coulomb) field. A second trap is using the Cartesian-style derivative $\partial f_R/\partial R$ without the $R^2$ factor. That gives $-n{R}^{-n-1}$ and wrongly picks $n=-1$ and $n=0$.

    Full solution →
42. Q.52MSQ · 2 marksOfficial answer: A, D

    Networks (Electric Circuits) › Coupled circuits

    $W=\frac{1}{2}{L}_1{I}_1^2+\frac{1}{2}{L}_2{I}_2^2+M{I}_1{I}_2$ is a quadratic in the ratio $x=I_1/I_2$, with its minimum at $x=-M/L_1$. When $k=1$, $M^2=L_1{L}_2$ turns $W$ into a perfect square, so the minimum energy is exactly zero at $x=-\sqrt{L_2/L_1}$. Both $-M/L_1$ and $-L_2/M$ reduce to that value.

    WATCH OUT FOR

    Stopping at $-M/L_1$ and marking only (A), without using $k=1$ to see that $-L_2/M$ is the same number. Also common: minimising with respect to $I_2$ (which gives $I_2/I_1=-M/L_2$) and then marking $-M/L_2$ as a value of $I_1/I_2$, when it is really the reciprocal ratio. A third slip is dropping the sign of the mutual term, which moves the minimum to a positive ratio.

    Full solution →
43. Q.53NAT · 2 marksOfficial answer: 0 to 1

    Engineering Mathematics › Vector calculus

    "Rate of change in a direction" means the directional derivative: the gradient dotted with the unit vector pointing from the given point towards the target point. Here $\nabla V(0,0)=(1,0)$ and $\hat{u}=(1,2)/\sqrt{5}$, which gives $1/\sqrt{5}$.

    WATCH OUT FOR

    Skipping the normalisation and computing $\nabla V\cdot (1,2)=1$, or answering with the maximum rate $|\nabla V(0,0)|=1$ (the rate along the gradient, not towards $(1,2)$). A directional derivative is a rate per unit distance, so the direction vector has to be a unit vector. The correct value is $1/\sqrt{5}\approx 0.447$, which rounds to 0. The key's range of 0 to 1 happens to accept 1 as well, but only 0 follows from the definition.

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

    Engineering Mathematics › Differential equations

    This is a chain of two integrators, $x_1\leftarrow 2\int x_2$ and $x_2\leftarrow \int r$. A unit step fed through it gives a ramp for $x_2$ and a parabola for $x_1$. Integrate twice and add the initial condition at each stage: $x_2=t$, and $x_1=1+t^2$.

    WATCH OUT FOR

    Dropping the initial condition $x_1(0)=1$ gives $x_1(1)=1$. This is the zero-state response only, and it misses the zero-input part $1/s$. Another common slip is forgetting the factor $2$ in ${\dot{x}}_1=2x_2$, which gives $x_1=1+t^2/2$, that is $1.5$ at $t=1$.

    Full solution →
45. Q.55NAT · 2 marksOfficial answer: 0

    Engineering Mathematics › Complex analysis

    Before you compute anything, look for singularities inside the contour. $f(z)=jz$ is entire, so Cauchy's integral theorem gives ${\oint }_{C}f(z)dz=0$. The orientation (clockwise) and the constant $j$ only multiply that zero.

    WATCH OUT FOR

    Mixing up $\oint zdz$ with $\oint \frac{dz}{z}$ and writing $\pm 2\pi j$ (or $\pm 2\pi$ after multiplying by $j$). Only a pole inside the contour gives a nonzero result, and $jz$ has none. A second slip is to think that the clockwise orientation must make the answer nonzero and negative. Reversing direction only flips the sign, and zero stays zero.

    Full solution →
46. Q.56NAT · 2 marksOfficial answer: 2

    Analog Electronics › Op-amps

    This is the two-op-amp instrumentation amplifier. The first stage scales $V_a$ by $(1+R_2/R_1)$. The second stage outputs $(1+R_4/R_3)V_b-(R_4/R_3)V_1$. When $R_4/R_3=R_1/R_2$ (here all resistors are equal), the common-mode part cancels and $V_{out}=(1+R_4/R_3)(V_b-V_a)=2V_{in}$.

    WATCH OUT FOR

    Treating the second stage as a plain inverting amplifier, $V_{out}=-V_1=-2V_a$, and forgetting that its + input carries $V_b$. That gives a gain of −2 or 1 instead of 2. A second mistake is taking op-amp 1's gain as 1 (a follower) when its − input actually sits at the midpoint of a 10 kΩ/10 kΩ divider.

    Full solution →
47. Q.57NAT · 2 marksOfficial answer: 4.0 to 4.2

    Networks (Electric Circuits) › Transients

    The capacitor does not start uncharged. Before the switch closes, the 3 A source drives 75 V across the 25 Ω resistor, and that voltage opposes the 100 V source, so $v_C(0)=100-75=25$ V. Closing S shorts out the whole parallel group. That leaves a plain series RC loop (100 V, 5 F, 2 Ω) with $v_C(\infty )=100$ V and $\tau =10$ s. Then use $v_C(t)=v_C(\infty )+[v_C(0^{+})-v_C(\infty )]e^{-t/\tau }$.

    WATCH OUT FOR

    Assuming the capacitor starts at 0 V gives $v_C=100(1-e^{-t/10})$ and $t=10\ln 2\approx 6.9$ s, which is wrong. Another trap is taking $v_C(0)$ as the 75 V across the parallel group rather than 100 − 75 = 25 V. A third is leaving the 25 Ω resistor in the time constant after S closes, when the switch has shorted it out.

    Full solution →
48. Q.58NAT · 2 marksOfficial answer: 78.0 to 81.0

    Electrical and Electronic Measurements › V, I, power & energy

    Three-voltmeter method: the series resistor voltage is in phase with the current, so the supply voltage is the phasor sum of $V_R$ and $V_L$ with angle $\varphi$ between them. The cosine rule gives $V_L\cos \varphi$, and with $I=V_R/R$ the load power reduces to $P=(V_S^2-V_R^2-V_L^2)/(2R)$. Only the three readings and $R$ are needed.

    WATCH OUT FOR

    Adding the voltages arithmetically ($9+199=208\ne 200$) and concluding the data are inconsistent, when they are phasors at an angle. Other mistakes: dividing by $2V_R$ or by $R$ instead of $2R$ (giving $17.7\text{W}$ or $159\text{W}$), including the resistor's own loss $V_R^2/R=40.5\text{W}$ in the load power, or using a minus sign in the cosine rule, which gives a negative $\cos \varphi$.

    Full solution →
49. Q.59NAT · 2 marksOfficial answer: -2.00 to -1.94, or 1.94 to 2.00

    Electrical and Electronic Measurements › Bridges

    Near balance, only the divider that contains the changed arm moves the detector voltage. With $E=\frac{VS}{R+S}-\frac{VQ}{P+Q}$, the sensitivity is $\left|\frac{dE}{dR}\right|=\frac{VS}{(R+S{)}^2}$. This is the same as the standard small-change result $\Delta E=\frac{VRS}{(R+S{)}^2}\cdot \frac{\Delta R}{R}$.

    WATCH OUT FOR

    There are three common slips. The first is using the equal-arm approximation $\Delta E\approx \frac{V}{4}\frac{\Delta R}{R}$, which gives $50\text{mV}/\Omega$; it only holds when all four arms are equal, and here $R/S=0.01$. The second is pairing $R$ with the wrong arm, for example using $(R+P)$ instead of $(R+S)$; $R$ and $S$ are the two arms in series across the battery, meeting at the left detector terminal. The third is quoting the answer in $\text{V}/\Omega$ (0.00196) when the question asks for $\text{mV}/\Omega$.

    Full solution →
50. Q.60NAT · 2 marksOfficial answer: 1.7 to 1.9
51. Q.61NAT · 2 marksOfficial answer: 200

    Signals and Systems › Fourier transform

    The value of a signal at $t=0$ is the total area under its Fourier transform divided by $2\pi$: $h(0)=\frac{1}{2\pi }\int H(j\omega )d\omega$. For an ideal rectangular low-pass response of height 1 and cutoff ${\omega }_c$, that area is $2{\omega }_c$, so $h(0)={\omega }_c/\pi$. Here ${\omega }_c=200\pi$ gives $200$.

    WATCH OUT FOR

    Leaving out the $\frac{1}{2\pi }$ factor gives $400\pi \approx 1257$. Taking $200\pi$ as a frequency in hertz, or using only the positive half of the passband (width $200\pi$ instead of $400\pi$), gives $h(0)=100$. The passband runs from $-200\pi$ to $+200\pi$ rad/s, and both halves count.

    Full solution →
52. Q.62NAT · 2 marksOfficial answer: 3

    Control Systems › State space

    The poles of a state-space system are the eigenvalues of $A$, the roots of $\det (sI-A)=0$. For a companion (controllable canonical) matrix, the last row $[-a_0-a_1]$ gives the characteristic polynomial $s^2+a_{1}s+a_0$ directly. Check $C(sI-A{)}^{-1}B$ for cancellation before treating every eigenvalue as a transfer-function pole.

    WATCH OUT FOR

    Adding the poles themselves, $(-1)+(-2)=-3$, which is the trace of $A$, instead of adding their magnitudes. Another common slip is a sign error in the last row: reading the polynomial as $s^2-3s-2$ instead of $s^2+3s+2$.

    Full solution →
53. Q.63NAT · 2 marksOfficial answer: 8.5 to 10.0
54. Q.64NAT · 2 marksOfficial answer: 216
55. Q.65NAT · 2 marksOfficial answer: 0.1

    Control Systems › Compensators & PID

    At $\omega =10\text{rad/s}$ the plant has a phase of exactly $-180^{\circ }$ and a magnitude of $1/\sqrt{2}$. A $45^{\circ }$ phase margin therefore needs the PD zero to add exactly $45^{\circ }$ there, so $10K_D=1$. With that value, $|1+j|=\sqrt{2}$ cancels the plant's $1/\sqrt{2}$, so $10\text{rad/s}$ really is the gain crossover. The question is self-consistent, and the magnitude check confirms the answer.

    WATCH OUT FOR

    Using only one pole's phase ($-45^{\circ }$ in place of $-90^{\circ }$ for the squared factor $(s+10{)}^2$), which puts the plant phase at $-135^{\circ }$ and leads to the wrong conclusion that no controller phase is needed. Another mistake is computing $|(j10+10{)}^2|$ as $|j10+10|=10\sqrt{2}$ instead of $200$, which spoils the magnitude check.

    Full solution →