GATE EE 2020 question paper with solutions

GATE EE 2020 answer key

General Aptitude GA Q.1 to Q.10

1. GA Q.1MCQ · 1 markOfficial answer: C

   General Aptitude › Grammar & vocabulary

   The verb agrees with the head subject. Phrases set off by commas ("including all its chapters") and add-on phrases ("as well as", "along with", "together with") do not change the subject's number. Only "and" makes a compound, plural subject.

   WATCH OUT FOR

   Matching the verb to the nearest noun. That gives "are" after "chapters" (plural) or "was" after "the instructor" (singular). Another mistake is treating "as well as" like "and". Here that happens to give the same plural verb, but with a singular first noun it would give the wrong one (e.g. "The instructor as well as the students IS ...").

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

   General Aptitude › Grammar & vocabulary

   Prohibit takes from + gerund: prohibit someone from doing something, or in the passive someone is prohibited from doing something. Unlike forbid, it does not take a to-infinitive.

   WATCH OUT FOR

   Carrying the forbid pattern (forbidden to park) over to prohibit and picking (A). Another mistake is picking (C) because prohibit parking is correct in the active voice, without noticing that the passive were prohibited, with people as the subject, needs from before the gerund.

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3. GA Q.3MCQ · 1 markOfficial answer: C

   General Aptitude › Reasoning

   In a word-formation analogy, first name exactly how the second word is made from the first. Here a negating prefix turns a word into its opposite (do → undo). Then use the prefix English actually uses with the new word. For \"trust\" that prefix is dis-, which gives \"distrust\".

   WATCH OUT FOR

   Copying the prefix instead of the relation. Students see un- in \"Undo\" and pick \"Untrust\", but English negates \"trust\" with dis-. Another trap is taking \"Entrust\" or \"Intrust\" because they look like prefixed forms of \"trust\". Neither one negates it: both mean to hand something over to someone's care.

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4. GA Q.4MCQ · 1 markOfficial answer: B

   General Aptitude › Grammar & vocabulary

   The blank needs a verb with no object that describes how markets react to bad political news. "Plunge", used intransitively, means to fall suddenly and steeply, which is exactly what markets do after a coup.

   WATCH OUT FOR

   Being thrown by the look-alike words. Plunged, plugged and probed all start with "p" and differ by a letter or two, and "poised" is a common word in market reports ("markets are poised to rise"). Read each option in the sentence for meaning and check whether it needs an object. Don't pick the word that just sounds like finance.

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5. GA Q.5MCQ · 1 markOfficial answer: C

   General Aptitude › Numerical

   Fix $Q$ in the team, then count the subsets of the other three people. There are $2^3=8$ of them. Drop the empty one, because a team of $Q$ alone has size 1. That leaves $7$.

   WATCH OUT FOR

   Two slips are common. The first is keeping the empty subset, which counts $Q$ alone as a team and gives 8 (option D). The second is leaving out the full team $PQRS$, which gives 6 (option B). A third error is counting every team of size 2 or more from all four people, which is $2^4-4-1=11$, and ignoring the condition that $Q$ must be a member.

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6. GA Q.6MCQ · 2 marksOfficial answer: B

   General Aptitude › Reading

   The passage sets its own unit: "four quarters (360 days)" makes one quarter 90 days. Three revisions, each taking off one quarter, bring the period from 4 quarters down to 1 quarter, which is 90 days.

   WATCH OUT FOR

   Reading "reduced by one quarter" as a 25% cut compounded three times, which gives 151.875 days and matches no option. Another slip is stopping after two revisions (180 days, option D), or making four cuts in place of the three that "thrice" means (0 days), or halving the period each time (45 days, option A).

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7. GA Q.7MCQ · 2 marksOfficial answer: C

   General Aptitude › Reasoning

   Two patterns run at once. Term length goes up by one (1, 2, 3, 4 letters, each term counting backwards), and the number of letters skipped between terms also goes up by one (2, 3, 4). So after P you skip O, N, M, L and start at K.

   WATCH OUT FOR

   Seeing only the four-letter length and picking the first backward run after P that looks right, such as NMLK (skips only O) or LKJI (skips O, N, M). Both miss the growing gap. Choosing JIHG means skipping one letter too many.

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8. GA Q.8MCQ · 2 marksOfficial answer: C

   General Aptitude › Numerical

   Count the occurrences position by position. "37" can be the first two, middle two or last two digits, which gives 100 + 90 + 90. Only the first and third positions can both hold "37" at once, and only in 3737. That number has two occurrences, and both are counted.

   WATCH OUT FOR

   Answering 279 (option B). That is the number of different numbers that contain "37": 3737 is counted once as a number even though it holds the group twice. The question asks how many times the group appears, so 3737 contributes 2 and the answer is 280. Another slip is allowing a leading 0 in the middle or last position (giving 100 instead of 90), which leads to answers such as 300 that are not among the options.

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9. GA Q.9MCQ · 2 marksOfficial answer: A

   General Aptitude › Geometry

   When C is the top of the semicircle ($OC\perp AB$), the triangles $AOC$ and $BOC$ are right-angled at $O$, with both legs equal to the radius $r$. So each chord is $r\sqrt{2}$ and the diameter is $2r$. The ratio does not depend on $r$ and equals $\frac{2r\sqrt{2}}{2r}=\sqrt{2}$.

   WATCH OUT FOR

   One mistake is to take each chord as equal to the radius, which gives $\frac{2r}{2r}=1$. Another is to use $\overline{AB}=r$ in place of $2r$, which gives $2\sqrt{2}$. A third is to add the chords as squares, since ${\overline{AC}}^2+{\overline{CB}}^2={\overline{AB}}^2$, and then pick option (C) $2$. The question adds the lengths, not their squares.

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10. GA Q.10MCQ · 2 marksOfficial answer: C

    General Aptitude › Data interpretation

    Work backwards from the chart value in two steps. A 20% rise means dividing by $1.2$ to get the 2014 revenue. A 10% profit on expenditure means revenue $=1.1\times$ expenditure, so you then divide by $1.1$. Together: $E_{2014}=\frac{45}{1.2\times 1.1}=\frac{45}{1.32}\approx 34.1$.

    WATCH OUT FOR

    There are three usual slips. (1) Taking 20% off the 2015 revenue ($45\times 0.8=36$) instead of dividing by $1.2$. (2) Treating the profit as 10% of revenue, which gives $E=0.9\times 37.5=33.75$, close to the distractor 33.7. (3) Using Q's 2015 expenditure bar ($35$), which gives option (D) 35.1 or something near it. The 2015 expenditure is not needed at all.

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Electrical Engineering Q.1 to Q.55

1. Q.1MCQ · 1 markOfficial answer: A or D

   Engineering Mathematics › Calculus

   $p(0)$ equals the constant term $d$, so $x=0$ is a root exactly when $d=0$. A real cubic always has at least one real root because non-real roots come in conjugate pairs. Repeated roots such as $(x-1{)}^3$ are allowed. For (D), making $c$ large and of the opposite sign to $a$ forces three sign changes and so three real roots. But no single fixed value of $c$ guarantees this whatever $a,b,d$ are, and that ambiguity is why D was also accepted.

   WATCH OUT FOR

   Picking (C) without noticing that a cubic with real coefficients must have a real root, or picking against (B) without remembering repeated roots such as $(x-1{)}^3$. Another mistake is to treat (D) as clearly true or clearly false: the answer depends on whether $c$ may be chosen after $a,b,d$ are known.

   The question and its official answer →
2. Q.2MCQ · 1 markOfficial answer: C

   Engineering Mathematics › Calculus

   The only property that holds for every allowed choice is symmetry: $\sin p\theta \cos q\theta$ is odd, so its integral over $[-\pi ,\pi ]$ is zero whatever $p$ and $q$ are. The sine-sine orthogonality results in (A), (B) and (D) need $m+n\ne 0$ (or $p+q\ne 0$), and (B) also needs a full period, so each one has a counter-example.

   WATCH OUT FOR

   Choosing (A) or (D) from the memorised orthogonality relation "different frequencies integrate to zero". The question only requires $m\ne n$ (or $p\ne q$) and non-zero values, so negative values are allowed. With $n=-m$ or $q=-p$ the product becomes $-{\sin }^2$, and its integral (or long-run average) is $-\frac{1}{2}$, not $0$.

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3. Q.3MCQ · 1 markOfficial answer: B

   Signals and Systems › Laplace

   A bounded, finite-duration signal gives a Laplace integral over a finite interval of a bounded integrand. That integral converges for every s, so the ROC is the whole s-plane. Poles cannot lie inside the ROC, so there are none. Any apparent pole, such as s = 0 in (1 - e^{-sT})/s, is a removable singularity.

   WATCH OUT FOR

   Seeing the factor 1/s in the rectangular-pulse transform (1 - e^{-sT})/s and concluding that finite-duration signals have a pole at s = 0. The numerator also vanishes there, so the limit is T and there is no pole. Two other errors: picking (A) because the Laplace transform 'generalises' the Fourier transform (a constant or two-sided signal can have an empty ROC), and picking (D) because only the delay factor e^{-st_0} is remembered and the 1/(s+a) factor is forgotten.

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4. Q.4MCQ · 1 markOfficial answer: D

   Signals and Systems › z-transform

   ${\left(\frac{1}{2}\right)}^{n}1[n]$ is a right-sided exponential with a single pole at $z=\frac{1}{2}$, so its ROC is $|z|>\frac{1}{2}$. Delaying it by $k>0$ multiplies the transform by $z^{-k}$. That factor can only change whether $z=0$ or $z=\infty$ belongs to the ROC, and $z=0$ is already outside it, so the ROC stays $|z|>\frac{1}{2}$.

   WATCH OUT FOR

   Inverting the pole and writing $|z|>2$: the pole sits at $z=a=\frac{1}{2}$, not at $1/a$. A second mistake is to think the factor $z^{-k}$, or the shift itself, turns the sequence left-sided and gives $|z|<\frac{1}{2}$. A delay by $k>0$ keeps the sequence causal. Some students also say $z^{-k}$ adds a pole of order $k$ at the origin. In fact $z/(z-\frac{1}{2})$ already has a zero there, so the pole has order $k-1$ (none when $k=1$). Either way the annulus does not change.

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5. Q.5MCQ · 1 markOfficial answer: C

   Engineering Mathematics › Complex analysis

   Residue theorem: ${\oint }_{C}fdz=2\pi i$ times the sum of the residues at the poles inside $C$. Here only the simple pole $z=0$ lies inside $|z|=1$, and its residue is ${\frac{z^2+1}{z-2}|}_{z=0}=-\frac{1}{2}$.

   WATCH OUT FOR

   Counting the pole at $z=2$ as well. It lies outside the unit circle, and adding its residue $\frac{5}{2}$ gives $2\pi i\cdot 2=4\pi i$, which is not an option. The other common slip is a sign error in the residue, $\frac{1}{0-2}=-\frac{1}{2}$ and not $+\frac{1}{2}$. Getting that sign wrong, or taking the contour clockwise, gives $\pi i$, the distractor (D).

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6. Q.6MCQ · 1 markOfficial answer: A

   Signals and Systems › Signals basics

   The average is a linear operation, so it scales by exactly $k$. The rms is the square root of a mean square: $k^2$ comes out of the integral and $\sqrt{k^2}=|k|$ comes out of the radical, so the rms scales by $|k|$. That equals $k$ for a non-negative gain. The rms is not additive, $\text{rms}(x_1+x_2)\ne \text{rms}(x_1)+\text{rms}(x_2)$ in general, but it does scale with a constant multiplier.

   WATCH OUT FOR

   Reasoning that the rms is 'nonlinear because of the squaring' and so cannot scale by $k$, which leads to (B). The rms fails additivity, not scaling: multiplying the signal by $k$ multiplies the rms by $|k|$. The opposite slip is forgetting that the rms is never negative, so $y_R=|k|x_R$ and not $k{x}_R$ when $k<0$. The question takes $k$ as a non-negative gain.

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7. Q.7MCQ · 1 markOfficial answer: A

   Electrical Machines › Synchronous machines

   For a cylindrical-rotor synchronous generator with negligible armature resistance, $E_A=V_A+jI{X}_s$. Writing this out for a load current at angle $\varphi$ from $V_A$ gives $|E_A{|}^2-V_A^2=I{X}_s(I{X}_s-2V_A\sin \varphi )$. For a lagging load ($\varphi <0$) this quantity is a sum of two positive terms and is always positive, so $E_A>V_A$ unconditionally. For a leading load ($\varphi >0$) the sign depends on the relative size of $I{X}_s$ and $2V_A\sin \varphi$, so no unconditional relation holds — an "always" claim for leading loads must be false.

   WATCH OUT FOR

   Over-generalizing the familiar rule of thumb "lagging loads reduce terminal voltage, leading loads raise it" into a universal statement for BOTH directions. The rule is unconditionally true only for lagging loads (Q); for leading loads (P), whether $V_A$ exceeds $E_A$ depends on the magnitude of the leading current relative to $2V_A\sin \varphi /X_s$, so treating P as always true (picking option D) is the typical error. A second common mistake is a sign slip when resolving $jI\angle \varphi$, which can flip which of P/Q comes out true.

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8. Q.8MCQ · 1 markOfficial answer: D
9. Q.9MCQ · 1 markOfficial answer: A
10. Q.10MCQ · 1 markOfficial answer: A

    Control Systems › Block diagrams & SFG

    Poles come only from the left-hand (output) side of the differential equation: its Laplace transform gives the characteristic polynomial $s^2+4$. The missing $s$ term means $\zeta =0$, so the poles sit on the imaginary axis at $\pm j{\omega }_n$ with ${\omega }_n=\sqrt{4}=2\text{rad/s}$. The input-side gain of 6 plays no part.

    WATCH OUT FOR

    One slip is solving $s^2=4$ in place of $s^2=-4$, which gives the real pair $\pm 2$ (option B). Another is reading the coefficient 4 as the pole location, which gives $\pm 4$ or $\pm 4j$ (options C and D), even though 4 is ${\omega }_n^2$, not ${\omega }_n$. A third is letting the right-hand-side gain of 6 affect the poles.

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11. Q.11MCQ · 1 markOfficial answer: C
12. Q.12MCQ · 1 markOfficial answer: Marks to all
13. Q.13MCQ · 1 markOfficial answer: C

    Control Systems › Block diagrams & SFG

    In a signal flow graph, a forward branch $d$ with a feedback branch $c$ around it forms a loop of gain $cd$. Branch gains simply multiply, with no minus sign, so the pair reduces to $\frac{d}{1-cd}$ and not $\frac{d}{1+cd}$. Removing a node such as $x_2$ never touches the branches between the nodes that remain, so $e$ from $x_3$ to $x_1$ survives unchanged.

    WATCH OUT FOR

    Not checking which way arrow $c$ points. If you read $c$ as a second forward branch alongside $d$, you add them and get $a(d+c)$, which is option (A). There are two smaller slips. One is using the block-diagram negative-feedback form $\frac{d}{1+cd}$. The other is putting $c$ in the numerator instead of $d$, which gives option (D). In a signal flow graph the numerator must be the forward branch, and the loop gain always goes in as $1-(\text{loop gain})$.

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14. Q.14MCQ · 1 markOfficial answer: A

    Analog Electronics › Amplifiers

    With $r_o$ neglected and the source at AC ground, a common-source stage has small-signal gain $A_v=-g_m{R}_D$. The DC supply is AC ground, so it sets the bias but never appears in the gain.

    WATCH OUT FOR

    Dropping the minus sign and choosing +2.44 (D). A common-source amplifier inverts the signal. Another mistake is trying to use the 10 V supply in the gain, or halving the result to about 1.22 (options B and C). The supply only fixes the DC bias, and nothing in the circuit halves $g_m{R}_D$.

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15. Q.15MCQ · 1 markOfficial answer: B

    Digital Electronics › Sequential

    An overlapping 3-bit sequence detector works like a sliding window. After each new bit, compare the last three bits received with the target (here $101$) and output 1 on a match. With 14 inputs there are $14-3+1=12$ complete windows, so there are 12 outputs. A detected sequence may reuse bits from the previous detection.

    WATCH OUT FOR

    The usual mistake is to treat the detector as non-overlapping, resetting after the match that ends at $b_{10}$. That misses the $101$ ending at $b_{12}$, because it reuses $b_{10}$, and gives $0,1,0,0,0,0,0,1,0,0,0,0$, which is not an option. Another mistake is to put each 1 at the first bit of the window instead of the last, or to expect 14 outputs rather than one per complete 3-bit window.

    Full solution →
16. Q.16NAT · 1 markOfficial answer: 0.8774 to 0.8952

    Engineering Mathematics › Differential equations

    $\frac{dy}{dx}+y=2x$ is a first-order linear ODE. Use the integrating factor $e^x$ to get $y=2x-2+C{e}^{-x}$. Then $y(0)=1$ gives $C=3$, and $e^{-\ln 2}=\frac{1}{2}$ makes the evaluation exact before you round.

    WATCH OUT FOR

    Using the initial condition before adding the constant $C$, or dropping $C$ after integrating by parts. Getting the sign of the $-e^x$ term wrong in $\int x{e}^{x}dx=(x-1)e^x$. Writing $e^{-\ln 2}=2$ instead of $\frac{1}{2}$ gives $2\ln 2+4\approx 5.386$.

    Full solution →
17. Q.17NAT · 1 markOfficial answer: 4.10 to 4.40

    Electrical Machines › Induction machines

    Synchronous speed $N_s=120f/P$ is fixed by supply frequency and pole count and does not change with load; only the slip changes between no-load and full-load. Speed regulation is defined the same way as voltage regulation — as a percentage change measured relative to the full-load (rated operating) value, i.e. $(N_{NL}-N_{FL})/N_{FL}\times 100$.

    WATCH OUT FOR

    Using the synchronous speed $N_s$ in the denominator instead of the full-load speed $N_{FL}$ — that gives $0.04/1\times 100=4.00\%$, which is outside the accepted key range because it silently redefines "regulation" against the wrong reference speed.

    Full solution →
18. Q.18NAT · 1 markOfficial answer: 1.700 to 1.750

    Networks (Electric Circuits) › AC, power & resonance

    In AC circuits KCL holds for phasors, not for meter readings. The current through A1 is the vector sum $1\angle 10^{\circ }+1\angle 70^{\circ }$. The two currents are $60^{\circ }$ apart, so the sum has magnitude $\sqrt{1+1+2\cos 60^{\circ }}=\sqrt{3}$.

    WATCH OUT FOR

    Adding the ammeter readings as plain numbers ($1+1=2\text{A}$) ignores the $60^{\circ }$ phase difference between the branch currents. Another slip is to use the $60^{\circ }$ difference directly in $2\cos \theta$ instead of the half-angle, which gives $2\cos 60^{\circ }=1\text{A}$. The correct result is $2\cos 30^{\circ }=\sqrt{3}\text{A}$.

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19. Q.19NAT · 1 markOfficial answer: 13.80 to 14.20

    Networks (Electric Circuits) › Theorems

    Only two facts are needed. An ideal voltage source fixes its node voltage no matter what resistor sits across it, and a resistor that carries no current (the open-circuited output $3\Omega$) drops no voltage. The whole $5\text{A}$ therefore flows through the $2\Omega$, so $V_{TH}=4+5\times 2=14\text{V}$.

    WATCH OUT FOR

    Letting the left $3\Omega$ carry part of the $5\text{A}$ (treating it as a divider with the $2\Omega$, or using source transformation on it). It is directly across an ideal voltage source, so it has no effect on $V_A$ or $V_B$. Another slip is to count a voltage drop across the output $3\Omega$. With the terminals open it carries no current, so it drops nothing.

    Full solution →
20. Q.20NAT · 1 markOfficial answer: 3
21. Q.21NAT · 1 markOfficial answer: 162.41 to 162.59

    Electrical Machines › Transformers

    No-load (core) loss splits into hysteresis loss (∝ f · B_m^n) and eddy current loss (∝ f² · B_m²). Since E = 4.44 f N B_m A, the flux density B_m depends only on the ratio V/f. Whenever V/f is held at the same value across different test conditions, B_m stays constant, so the two loss components reduce to simple functions of frequency alone: P_h ∝ f and P_e ∝ f². This lets two known (V, f, loss) data points fix P_h and P_e at a reference condition, from which the loss at any other same-(V/f) condition follows by scaling.

    WATCH OUT FOR

    Forgetting to check that V/f is actually the same in all three conditions before dropping B_m from the proportionality — if V/f changed, B_m would change too and P_h ∝ f alone (or P_e ∝ f² alone) would no longer hold. A second common slip is mixing up which loss scales as f versus f² (swapping hysteresis and eddy current), or using absolute frequencies in the ratio instead of the ratio relative to the condition where P_h, P_e were computed.

    Full solution →
22. Q.22NAT · 1 markOfficial answer: 19.90 to 20.20
23. Q.23NAT · 1 markOfficial answer: 123.00 to 135.00
24. Q.24NAT · 1 markOfficial answer: 2

    Control Systems › Nyquist

    The plot is of $1+G(s)$, so encirclements are counted around the origin, not around $-1$. With a clockwise contour, $N=Z-P$: the open-loop pole at $s=+b$ gives $P=1$, the one clockwise encirclement gives $N=+1$, so $Z=2$. The zeros of $1+G(s)$ are the closed-loop poles, and here they cannot cancel against anything because the numerator of $G(s)$ is just the constant $K$.

    WATCH OUT FOR

    Using $N=P-Z$, which counts anticlockwise encirclements as positive, gives $Z=P-N=0$ and wrongly calls the system stable. The other common slip is to treat $G(s)$ as having no RHP poles and write $Z=N=1$, missing the pole at $s=+b$ that comes from the factor $(s-b)$.

    Full solution →
25. Q.25NAT · 1 markOfficial answer: 0
26. Q.26MCQ · 2 marksOfficial answer: A

    Engineering Mathematics › Calculus

    For a continuous function on a closed interval, the absolute extrema can only occur at interior critical points or at the endpoints, so evaluate $y$ at all of them and compare. Here $y=3{\left(x+\frac{1}{2}\right)}^2+\frac{1}{4}$ is an upward parabola whose vertex $x=-\frac{1}{2}$ lies inside $[-2,0]$. The vertex gives the minimum, and the endpoint farther from the vertex gives the maximum.

    WATCH OUT FOR

    Checking only the endpoints gives $y(-2)=7$ and $y(0)=1$, which leads to option (B) "7 and 1". The interior critical point $x=-\frac{1}{2}$, where $y=\frac{1}{4}$, gets missed. Option (C) comes from confusing the $x$-values (the endpoint $-2$, the vertex $-\frac{1}{2}$) with the values of $y$.

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

    Engineering Mathematics › Vector calculus

    A field defined on all of space is conservative exactly when its curl is zero, so it is the gradient of a scalar potential. Set each of the three curl components to zero. Each one fixes one constant: the $x$-component gives $k_3$, the $y$-component gives $k_1$ and the $z$-component gives $k_2$. Then check by building the potential $\varphi =5xy-4xz+3yz$.

    WATCH OUT FOR

    The usual slip is the sign or order in the $y$-component of the curl. It is $\partial F_x/\partial z-\partial F_z/\partial x$, and reversing it gives $k_1=-4$, which matches no option. The other mistake is testing divergence (solenoidal) instead of curl. Here $\nabla \cdot \mathbf{F}=0$ for every choice of $k_1,k_2,k_3$, because no component depends on its own coordinate, so a divergence test says nothing. Option (D) is a trap: with every $k=0$ the curl is $(-3,4,-5)\ne 0$.

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

    Electrical Machines › DC machines

    For a dc shunt motor, back EMF $E_b=V-I_a{R}_a-V_{brush}$, and speed is proportional to $E_b/\varphi$ (since $E_b\propto N\varphi$). Field current stays fixed (rated voltage across a fixed field resistance), so armature current at each condition is found by subtracting field current from line current (no-load) or total line current (loaded).

    WATCH OUT FOR

    Using the given no-load armature current directly (correct) but then forgetting to subtract the field current from the loaded total line current to get loaded armature current — using $I_a=50\text{A}$ instead of 47.5 A. Also common: forgetting the 2 V total brush drop (1 V/brush × 2 brushes), or applying the flux reduction the wrong way (multiplying instead of dividing by 0.95 in the speed ratio).

    Full solution →
29. Q.29MCQ · 2 marksOfficial answer: C
30. Q.30MCQ · 2 marksOfficial answer: D

    Networks (Electric Circuits) › Laws & node/mesh

    For any fixed resistive load, KCL at the diode node shows that raising the shunt resistance $R_2$ raises the node voltage and so the load current, and lowering the series resistance $R_1$ lowers the current lost in D and $R_2$, which also raises the load current. Since this is true for every load, it is also true for the maximum power. The ideal is $R_1\to 0$, $R_2\to \infty$, just like a good current source or a good solar cell.

    WATCH OUT FOR

    Applying the matched-load rule ($R_L$ equal to the source resistance) to the internal resistors, which leads to choosing "matched" or small values of $R_2$. $R_1$ and $R_2$ are parts of the source, not the load, and every watt they dissipate is lost. A second trap is getting stuck on the stated sign $I<0$: the printed arrows show $I$ leaving the $+$ terminal, so a resistive load actually gives $V$ and $I$ the same sign. The design conclusion does not depend on which sign convention is used.

    The question and its official answer →
31. Q.31MCQ · 2 marksOfficial answer: B

    Analog Electronics › Diode circuits

    At a reverse bias of only about one $\eta V_T$, keep the full Shockley equation $I=I_S\left(e^{V/(\eta V_T)}-1\right)$. The reverse current saturates toward $-I_S$, so raising its magnitude by 50% (from $0.632I_S$ to $0.948I_S$) takes a lot more reverse voltage, and $V$ ends up about three times the original bias.

    WATCH OUT FOR

    Dropping the $-1$ term and writing $I\approx I_S{e}^{V/(\eta V_T)}$. That approximation holds only for forward bias well above $\eta V_T$. Using it here gives $V=-0.03+0.03\ln 1.5\approx -0.018\text{V}\approx -0.02\text{V}$, which is distractor (A). It moves the voltage toward zero even though the reverse current is supposed to grow. A second slip is using $V_T=26\text{mV}$ without multiplying by the ideality factor.

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

    Networks (Electric Circuits) › Laws & node/mesh

    This supply has no linear Thevenin equivalent. It is current-limited below 10 V and voltage-limited up to 4 A, so the rule $R_L=R_{th}$ does not apply. Delivered power $P=VI$ rises along the current-limited branch ($P=16R$) and falls along the voltage-limited branch ($P=100/R$), so the maximum is at the knee of the V–I characteristic, $(10\text{V},4\text{A})$, which the load reaches when $R=10/4=2.5\Omega$.

    WATCH OUT FOR

    Applying the linear maximum-power-transfer rule $R_L=R_{th}$. An ideal current source has $R_{th}=\infty$, which suggests an open load, and an ideal voltage source has $R_{th}=0$, which suggests a short. Both loads deliver zero power here. A second trap is the matched-load habit of taking half the source voltage (5 V), which gives option (D). But 4 A through 2.5 Ω is 10 V, not 5 V, so (D) breaks Ohm's law.

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

    Electromagnetics › Electrostatics

    Free charge comes from the divergence of $\mathbf{D}$, not of $\mathbf{E}$: in a uniform linear dielectric, ${\rho }_v=\nabla \cdot \mathbf{D}={\epsilon }_0{\epsilon }_r\nabla \cdot \mathbf{E}$. In cylindrical coordinates the radial term is $\frac{1}{r}\frac{\partial (r{E}_r)}{\partial r}$, and a $\phi$-component that depends only on $r$ adds nothing to the divergence.

    WATCH OUT FOR

    Using ${\rho }_v={\epsilon }_0\nabla \cdot \mathbf{E}$ and leaving out ${\epsilon }_r$ gives $4{\epsilon }_0$ (option B), which is the total (free plus bound) charge density, not the free charge. Another slip is to take the radial term as $\partial E_r/\partial r=2$ and forget the $\frac{1}{r}\frac{\partial (r\cdot )}{\partial r}$ form, which gives $2.25\times 2=4.5{\epsilon }_0$, an answer that is not among the options.

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

    Control Systems › Time response

    If armature inductance is negligible, a PMDC motor behaves as a first-order system $\frac{K}{\tau s+1}$. Its DC gain is the final speed divided by the step voltage ($10/5=2$). Its time constant is the time taken to reach $63.2\%$ of the final value ($0.5\text{s}$).

    WATCH OUT FOR

    Using the final speed ($10\text{rad/s}$) as the gain and forgetting to divide by the $5\text{V}$ input, which gives option (A). Another mistake is reading the pole form $\frac{K}{s+a}$ as if $a$ were the time constant. In fact $\tau =1/a$ and the DC gain is $K/a$, so $\frac{2}{s+0.5}$ has $\tau =2\text{s}$ and DC gain $4$.

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

    Control Systems › Routh–Hurwitz

    The order of the closed-loop system is the degree of $1+G(s)H(s)=0$ after clearing fractions, and that degree includes the pole in the feedback block. If any coefficient of the characteristic polynomial is missing (here the $s^1$ term), the system cannot be stable. A Routh array then counts the right-half-plane poles: two sign changes here.

    WATCH OUT FOR

    Counting only the forward-path poles ($s^2(s+1)$, which gives 3rd order) and ignoring the pole of the feedback block at $s=-20$. A second mistake is to treat the missing $s^1$ coefficient as a sign of marginal stability. The Routh array shows two sign changes, which means two poles in the right half-plane, so the system is truly unstable.

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

    Control Systems › Routh–Hurwitz

    For a unity negative feedback loop, the closed-loop characteristic polynomial is the open-loop denominator plus the open-loop numerator. Build the Routh array for it. A full row of zeros marks poles on the $j\omega$ axis, and the auxiliary polynomial from the row above gives their location.

    WATCH OUT FOR

    Running Routh on the open-loop denominator $s^3+2s^2+2s+K$ alone. Its $s^1$ entry is $(4-K)/2$, which gives $K=4$, and that is not among the options. The numerator $s^2+s+1$ has to be added to the denominator first. A second slip is getting the $s^1$ entry as $(9+K+1)/3$ by using the wrong sign in the cross-multiplication.

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

    Signals and Systems › LTI & convolution

    The magnitude of a convolution is at most the convolution of the magnitudes, $|x\ast h|(t)\le (|x|\ast |h|)(t)$, at every $t$. This is the triangle inequality applied to the convolution integral. Since a real number is at most its magnitude, $y(t)\le |y(t)|\le z(t)$ everywhere, for any $x$ and $h$.

    WATCH OUT FOR

    Testing a single pair of signals and reading the options as claims about that pair. With $x=u(t)$ and $h=-u(t)$, $z(t)\le y(t)$ holds for $t<0$ and fails for $t>0$, which tempts students into (B). The question asks what is true for EVERY $x$ and $h$, and one pair can never establish that. Some students also pick (D) without seeing that its 'but not all' clause makes it false, because the inequality holds at every instant.

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

    Signals and Systems › Laplace

    Read each property straight from the pole-zero pattern. Zeros that mirror the poles about the $j\omega$ axis ($z=-p^{\ast }$) give an all-pass. Complex poles without their conjugates give complex coefficients, so the system is complex. A causal system with a right-half-plane pole has ROC $\text{Re}(s)>2$, which misses the $j\omega$ axis, so it is unstable.

    WATCH OUT FOR

    Taking the symmetric-looking pattern as proof the system is real. The poles $2-j$ and $-2+j$ are symmetric about the origin ($p$ and $-p$), not about the real axis. Symmetry about the origin only makes $H$ a function of $s^2$; it does not make the coefficients real. A second trap is assuming an all-pass is automatically stable. All-pass describes only $|H(j\omega )|$. Causality combined with the right-half-plane pole at $2-j$ makes the system unstable, which rules out (A).

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

    Signals and Systems › LTI & convolution

    A recursive term $ay[n-1]$ with a genuine pole at $z=a\ne 0$ (no pole-zero cancellation) makes the impulse response infinitely long for every LTI system that obeys the equation. The general solution is the causal response plus $K{a}^n$, and it is non-zero at infinitely many $n$ whatever $K$ is. A difference equation does not fix the ROC or the initial condition $y[-1]$, so it guarantees neither causality nor that the input alone determines the output.

    WATCH OUT FOR

    Choosing (D) by quietly assuming initial rest ($y[-1]=0$). That is exactly the causality assumption the question withholds, the same reason (B) is false. With $x[n]=\delta [n]$, the causal and anticausal systems that obey this equation give different $y[n]$ for $n>0$. Another mistake is rejecting (C) with degenerate coefficients ($a=0$ or $b_1=a{b}_0$), which remove the recursion the question is about.

    The question and its official answer →
40. Q.40MCQ · 2 marksOfficial answer: D

    Electromagnetics › Electrostatics

    The flux out of a closed surface is the volume integral of $\nabla \cdot \mathbf{D}$ (divergence theorem). In cylindrical coordinates the radial term is $\frac{1}{r}\frac{\partial (r{D}_r)}{\partial r}$, and the extra $r$ in $dV=rdrd\varphi dz$ cancels the $\frac{1}{r}$. The $\varphi$ component carries no flux through any face of a cylinder centred on the z-axis.

    WATCH OUT FOR

    Taking $\frac{\partial D_r}{\partial r}=0$ (as if in Cartesian coordinates), which drops the radial term and gives $-270\pi$, which is not an option. Leaving out the Jacobian $r$ in $dV$ is another route to a wrong answer. Other slips are giving the top face the wrong sign (its outward normal is $+{\hat{a}}_z$ and $D_z=-15r$ there, so its flux is negative) and counting flux from $D_{\varphi }$, which is tangential to every face.

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

    Control Systems › Bode & margins

    Use the dc gain $H(0)$ to find the unknown pole. Then set $|H(j\omega ){|}^2=1$, which is a quadratic in ${\omega }^2$. The zeros at $\pm j10$ make a notch at $\omega =10$, so the gain falls from 5 to 0 and then rises again, and it crosses unity once on each side of the notch. The smaller crossing is the answer.

    WATCH OUT FOR

    Picking 78.13 (option C) or 122.87 (option D). Those are the roots for ${\omega }^2$, not $\omega$, and the options are set up to catch anyone who forgets the square root. Another trap is a sign slip in $H(0)=100/(-p)$ that gives $p=+20$. That pole is unstable and contradicts the problem statement, though it gives the same magnitude, so check the sign against stability.

    Full solution →
42. Q.42MCQ · 2 marksOfficial answer: Marks to all

    Engineering Mathematics › Linear algebra

    $A=L{L}^T$ is the Cholesky form. Solve for $L$ column by column. A negative leading minor ($2\cdot 2-3^2=-5$) means $A$ is not positive definite, so some diagonal entry of $L$ has to be imaginary. Here that is $a_{22}^2=\det A_2/a_{11}^2=-\frac{5}{2}$, and the imaginary part carries into $a_{23}$. Signs never change the realness pattern, so the count is fixed.

    WATCH OUT FOR

    Factoring in the wrong order ($A=L^{T}L$, i.e. starting from $a_{33}^2=7$). This gives a different factor with only $a_{11}$ imaginary and leads to option A (5) or C (8). The question prints $L$ times $L^T$, which puts the imaginary entries at $a_{22}$ and $a_{23}$ instead. Another slip is to forget that $a_{23}=(1-a_{12}{a}_{13})/a_{22}$ picks up the $j$ from $a_{22}$.

    The question and its official answer →
43. Q.43MCQ · 2 marksOfficial answer: A

    Electrical Machines › Synchronous machines

    Saturated $Z_s$ = rated phase voltage on the actual OCC divided by the SCC current at the same field current. Unsaturated $Z_s$ = voltage on the air-gap (straight) part of the OCC divided by the SCC current at the same field current. In a delta-connected machine, phase voltage equals line voltage.

    WATCH OUT FOR

    Dividing 400 by $\sqrt{3}$ as if the machine were star-connected. The machine is delta-connected, so the phase voltage is 400 V. A second trap: the OCC's straight part does not pass through the origin (it starts at 10 V). So 'air-gap voltage over short-circuit current' changes with the field current it is read at, and different readings give 2.0, 2.025, 2.1 or 2.2.

    The question and its official answer →
44. Q.44NAT · 2 marksOfficial answer: -3.05 to -2.95

    Engineering Mathematics › Vector calculus

    For a line integral along $y=x^2$, write $\mathbf{F}\cdot d\mathbf{l}=F_{x}dx+F_{y}dy$. Then replace $y$ and $dy$ with $x^2$ and $2xdx$ and integrate in $x$ from the start point to the end point shown by the arrows. Here $\nabla \times \mathbf{F}=-2{\mathbf{a}}_z\ne 0$, so no potential function exists. The integral cannot be found from the endpoints alone and must be taken along the given curve.

    WATCH OUT FOR

    Integrating only over the right branch ($x$ from $0$ to $2$), which gives $-8/3\approx -2.67$. Other common slips: reversing the direction of travel, which gives $+3$; dropping the minus sign on the ${\mathbf{a}}_{y}x$ term, so the integrand becomes $3x^{2}dx$ and the answer $9$; and treating the field as conservative by taking a difference between the endpoints.

    Full solution →
45. Q.45NAT · 2 marksOfficial answer: 54.00 to 56.00

    Networks (Electric Circuits) › Transients

    A capacitor charging towards $+10\text{V}$ can only pass through zero if its initial voltage is negative. Setting $v_C(t_0)=0$ in $v_C(t)=10+(V_0-10)e^{-t/\tau }$ gives $V_0=10(1-e^{t_0/\tau })$. Work out $V_0$ for both crossing times and compare them. The ratio simplifies to $e^{0.2}/(1+e^{0.2})$.

    WATCH OUT FOR

    Assuming the capacitor starts uncharged (then it never crosses 0 V), or putting $e^{-t_0/\tau }$ in place of $e^{t_0/\tau }$ when solving for $V_0$. Another mistake is dividing the change by the new voltage $V_0^{\prime }$ instead of the original $V_0$, which gives about 122% instead of 54.98%.

    Full solution →
46. Q.46NAT · 2 marksOfficial answer: 123.00 to 127.00

    Networks (Electric Circuits) › AC, power & resonance

    The terminal voltage is fixed, so the active current tracks $P$ and the reactive current tracks $Q$. The reactor leaves the active current at $90\text{A}$ and doubles the lagging reactive current from $43.59\text{A}$ to $87.18\text{A}$. The new armature current is these two added in quadrature.

    WATCH OUT FOR

    Doubling the whole current to 200 A, or doubling the angle $\varphi$ instead of $Q$ (which gives $100\times 0.9/\cos (2\varphi )=146.9\text{A}$). Another slip is adding the reactor's current to the load current by simple arithmetic instead of in quadrature. Only the reactive component doubles, and the active component stays at $90\text{A}$.

    Full solution →
47. Q.47NAT · 2 marksOfficial answer: 1.11 to 1.13
48. Q.48NAT · 2 marksOfficial answer: Marks to all

    Electrical Machines › Transformers

    In a single-loop core, every winding links the same flux, so an ideal winding's voltage is fixed at $\pm j\omega N\Phi$. Ideal voltage sources that are not equal up to sign cannot all be connected across equal-turn windings on one flux path at the same time.

    WATCH OUT FOR

    Treating each voltage-fed winding as setting up its own flux, $V_i/(j\omega N)$, and adding these fluxes. That adds fluxes as if they were MMFs. When the windings are connected together the common flux cannot equal that sum, and this is what produces the 46 V answer. A second mistake is taking the balanced sum $V_A+V_B+V_C=0$ to prove that X reads 0 V without checking whether the circuit has a solution at all.

    The question and its official answer →
49. Q.49NAT · 2 marksOfficial answer: 1.01 to 1.03

    Electrical Machines › Synchronous machines

    Flux linkages (and hence the emf behind the subtransient reactance) cannot change instantaneously at the instant of fault, so $E^{\prime \prime }$ is computed using the prefault load current together with the subtransient reactance, evaluated at the prefault terminal voltage.

    WATCH OUT FOR

    Using the subtransient reactance to recompute a "prefault" current, or trying to use the given internal emf $E=1+j0.7$ directly as $E^{\prime \prime }$, instead of first finding the actual prefault line current from the steady-state relation $E=V+j{X}_{d}I$ and then applying that same current with $X_d^{\prime \prime }$.

    Full solution →
50. Q.50NAT · 2 marksOfficial answer: 4.95 to 5.05
51. Q.51NAT · 2 marksOfficial answer: 4.51 to 5.10
52. Q.52NAT · 2 marksOfficial answer: 11.50 to 12.00

    Electrical and Electronic Measurements › Errors & standards

    Propagate the uncertainty through the full diode law $I=I_s\left(e^{V/V_T}-1\right)$. At $V=20\text{mV}$ the bias is smaller than $V_T=29\text{mV}$, so the $-1$ term matters. It multiplies both relative sensitivities by $e^{V/V_T}/(e^{V/V_T}-1)\approx 2.007$. The independent contributions from $\delta V$ and $\delta V_T$ then combine as a root-sum-square.

    WATCH OUT FOR

    The usual mistake is to use $I\approx I_s{e}^{V/V_T}$, which is only valid for $V\gg V_T$. At $20\text{mV}$ it gives $5.87\%$ (root-sum-square) or $8.20\%$ (linear sum), both far from the key. Two other slips: adding the two contributions linearly with the full equation gives $16.47\%$, and taking half the source resolution ($0.5\text{mV}$) as $\delta V$ gives $10.15\%$.

    Full solution →
53. Q.53NAT · 2 marksOfficial answer: Marks to all

    Analog Electronics › Op-amps

    Write the output as a closed-form function of the uncertain parameter. Here that is V_out = 9/(3 + 2αT) − 0.2, from a divider followed by a non-inverting stage with an offset. Differentiate with respect to α and multiply by Δα to get the limiting error. Always check the operating point first: in this question the printed α makes the thermistor resistance negative above 25 °C.

    WATCH OUT FOR

    Using the tolerance in percent, Δα = 0.25 instead of 0.0025 per °C, which inflates the error a hundredfold. Quoting the full spread (about 0.17 V) instead of the ± error. Not noticing that the printed α gives a negative thermistor resistance at 150 °C, which is why the question could not be keyed.

    Full solution →
54. Q.54NAT · 2 marksOfficial answer: 210

    Digital Electronics › Processors & memory

    In the 8085, $M$ always means the memory byte at the address in $HL$. Follow $HL$ through each $\text{INX H}$: the first one runs before $\text{ADD M}$, so the byte added is the one at $2002\text{H}$ ($\text{B1H}$), and $98\text{H}$ at $2001\text{H}$ is never read. The second $\text{INX H}$ moves $HL$ to $2003\text{H}$, and the sum is stored there.

    WATCH OUT FOR

    Adding $98\text{H}$ as well, or adding it instead of $\text{B1H}$, because the question names both locations. $98\text{H}+\text{B1H}+21\text{H}$ overflows 8 bits and gives the wrong answer. The other common slip is writing the answer in hex ($\text{D2}$), or converting $\text{D2H}$ as $12\times 16+2=194$ by mistaking the value of the digit D, which is 13.

    Full solution →
55. Q.55NAT · 2 marksOfficial answer: 138.10 to 139.20

    Electromagnetics › Magnetostatics

    The wire's field $B={\mu }_{0}I/(2\pi \rho )$ changes only with distance from the wire. Integrate it over the loop in strips parallel to the wire to get $\Phi =\frac{{\mu }_{0}Ia}{2\pi }\ln \frac{d+a}{d}$. Then $M=\Phi /I$ depends only on geometry, so the 2 A current is a distractor.

    WATCH OUT FOR

    Taking $B$ at one distance (the near side, the far side or the centre, 1.5 m) and multiplying by the area $a^2$ instead of integrating across $1\text{m}\le \rho \le 2\text{m}$. At the centre this gives $2\times 10^{-7}/1.5=133.33\text{nH}$, outside the key range. The other common slips are multiplying by $I=2$ A, which gives flux rather than inductance, and using ${\log }_{10}$ where the natural log is needed.

    Full solution →