GATE EE 2018 question paper with solutions

GATE EE 2018 answer key

General Aptitude GA Q.1 to Q.10

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

   General Aptitude › Grammar & vocabulary

   Homophones: tell them apart by what each blank needs grammatically. After "the" with nothing following, the blank needs a noun (course = path or track). Before a noun such as "sand", it needs an adjective (coarse = rough, large-grained).

   WATCH OUT FOR

   Picking by sound or by spelling habit, for example choosing (B) or (D) on the idea that the same word fills both blanks. Also mixing the two up and writing "coarse" for a route. Always ask whether the gap needs a noun or an adjective.

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

   General Aptitude › Grammar & vocabulary

   The semicolon clause explains the "misconception" that structure mirrors thought. In "the more ______ the structure, the more complicated the ideas", the blank must therefore be the structural equivalent of "complicated". That word is convoluted (twisted, hard to follow).

   WATCH OUT FOR

   Picking (B) simple or (C) clear because good writing should be simple. The question asks what the misconception says, not what good writing advice is, and a mirror needs matching qualities, not opposite ones. Another trap is (A) detailed: a detailed sentence has many parts, but it is not necessarily complicated or hard to follow.

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

   General Aptitude › Numerical

   If $r$ is a root of $f$, then $f(x-3)=0$ when $x-3=r$, that is, when $x=r+3$. Replacing $x$ by $x-3$ moves the graph 3 units to the right, so every root goes up by 3.

   WATCH OUT FOR

   Seeing the minus sign in $x-3$ and subtracting 3 from each root. That gives $-5,-3,0$, which is option (A). Another slip is to think the shift leaves the roots alone and pick option (B), $-2,0,3$. In $f(x-h)$ the graph moves right by $h$, so the roots go up by $h$. A quick check settles it: substitute the candidate $x$ into $x-3$ and see whether you get an original root.

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

   General Aptitude › Numerical

   Write the numerator in powers of the denominator (put $k+2=(k-3)+5$). This leaves a polynomial part plus a constant remainder over $(k-3)$, so the fraction is an integer exactly when $(k-3)$ divides that remainder, here 25.

   WATCH OUT FOR

   Checking only one or two values in an option. Option (A) passes for 4 and 8, but $k=18$ gives $\frac{400}{15}$, which is not an integer. Every value in the option has to work. A second slip is to expand $(k+2{)}^2$ and divide by $k$ instead of by $(k-3)$, which gives a wrong remainder.

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

   General Aptitude › Numerical

   In a composite function, evaluate the innermost calls first. Each result then becomes an argument of the outer function, in the same position: the first argument is $a$ and the second is $b$.

   WATCH OUT FOR

   Stopping at the inner value $F(1,3)=4$ and picking (B), or squaring at the outer level as well, e.g. computing $F(4,2)$ or $(4-2{)}^2$, or mixing up which function goes outside. Another slip is forgetting the absolute value, which only matters when the order of the arguments is reversed; here $4-2$ is already positive.

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

   General Aptitude › Numerical

   First choose which character of each type to use ($10\times 26\times 26$). Then count the positions: the numeral, the upper case letter and the lower case letter can sit in any order, which gives $3!$ arrangements. Every arrangement is a different password because the three characters are of different types.

   WATCH OUT FOR

   Stopping at $10\times 26\times 26=6760$ (option A), which silently fixes the order as numeral, then upper case, then lower case. Some students multiply by 2 instead of $3!=6$ and get 13,520 (option B). Others count every 3-character string over all 62 symbols, which gives $62^3$ and ignores the rule that the password needs one character of each type.

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

   General Aptitude › Reasoning

   Turn each letter into its alphabet position (A = 1, ..., Z = 26) and compare the pairs one by one. Both examples show the same shift, $+4$, for every letter with the order unchanged. Once the rule is fixed, the first letter alone decides the answer: D + 4 = H, and only option (C) starts with H.

   WATCH OUT FOR

   Picking (B) TLPH. It has the right shifted letters (H, L, P, T) but in reverse order. The code keeps the letters in order: the first letter of AMCF gives the first letter of EQGJ, and so on. A second trap is checking only one example. You need both examples to confirm the shift really is $+4$ for every position.

   Full solution →
8. GA Q.8MCQ · 2 marksOfficial answer: Marks to all

   General Aptitude › Statistics & probability

   A group of three children means three different children, so count combinations: favourable groups divided by $\left(\genfrac{}{}{0ex}{}{12}{3}\right)=220$. With three children, 'more girls than boys' means either 3 girls, or 2 girls and 1 boy.

   WATCH OUT FOR

   Treating the three picks as independent, with a girl each time at probability $\frac{5}{12}$, and using the binomial formula. This gives $\frac{325}{864}$ (option B), but it quietly lets the same child be picked twice. Another slip is to drop the all-girls case and count only the $70$ two-girl groups, which gives $\frac{70}{220}$.

   The question and its official answer →
9. GA Q.9MCQ · 2 marksOfficial answer: C

   General Aptitude › Numerical

   A fixed budget with a 25% higher price per marble buys $\frac{1}{1.25}=0.8$ times as many marbles, so 20% of the marbles go. Because they are removed in equal numbers from each of the four colours, each colour loses 5% of the original total. That leaves blue at $40-5=35$ out of $80$.

   WATCH OUT FOR

   Removing the same percentage from every colour. That leaves blue at 40%, which is not an option. The question says equal numbers, not equal proportions. Two other slips: dividing $35$ by the old total of $100$ to get 35% (option A, 35.75, is close to this), and treating a 25% price rise as a 25% cut in the number of marbles. The count falls to $100/1.25=80$, not $75$.

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

    General Aptitude › Reasoning

    There are 5 trips and 4 rowers, and each person rows at least once, so exactly one person rows twice. Each person crosses forward exactly once more than they come back. P can row only when alone, which means only on a return. So P rows one return and crosses forward twice. Q can then only row a forward trip with R. Counting the six forward seats (who fills them) leaves R as the only person who can take the second return, and R also has to row P across once.

    WATCH OUT FOR

    Two errors are common. The first is to assume that P, who can row only alone, rows both returns. That puts P on every forward trip and leaves no trip where Q can row with R. The second is to build one schedule where S rows twice without checking condition (iv): if S makes the second return, R's only crossing is the trip Q rows, so R never rows. A full case check also has to rule out R rowing both of P's crossings. That would need R on three forward trips, which the seat count does not allow.

    Full solution →

Electrical Engineering Q.1 to Q.55

1. Q.1MCQ · 1 markOfficial answer: A

   Electrical Machines › Transformers

   Total per-unit impedance drop $e_z$ splits into a resistive part $e_r$ and reactive part $e_x$ via $e_z^2=e_r^2+e_x^2$ (right-triangle relation, since resistance drop and reactance drop are in phase quadrature relative to the load current phasor). For lagging power factor, the approximate percentage regulation is $\%\text{Reg}\approx e_r\cos \varphi +e_x\sin \varphi$ (a second-order correction term is negligible here and does not change the answer).

   WATCH OUT FOR

   Using the plus/minus sign wrong for the reactive term (lagging power factor requires the reactive-drop term to be added, not subtracted — subtraction applies to leading pf), or mis-splitting the 5% total impedance drop into resistive/reactive components using an arithmetic sum instead of $\sqrt{e_z^2-e_r^2}$.

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

   Electrical Machines › Synchronous machines

   In a salient-pole synchronous machine, $X_d\ne X_q$, so besides the excitation (fundamental) power/torque term proportional to $\sin \delta$, there is a reluctance power/torque term proportional to $\sin 2\delta$ arising purely from the rotor saliency ($X_d>X_q$). This reluctance term is maximum when $\sin 2\delta =1$, i.e. at $\delta =45^{\circ }$ electrical — independent of excitation.

   WATCH OUT FOR

   Confusing the reluctance-torque maximum (at $\delta =45^{\circ }$) with the excitation-torque maximum (at $\delta =90^{\circ }$), or assuming the total developed torque and the reluctance-torque component peak at the same angle.

   Full solution →
3. Q.3MCQ · 1 markOfficial answer: C
4. Q.4MCQ · 1 markOfficial answer: B
5. Q.5MCQ · 1 markOfficial answer: D

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

   In the two-wattmeter method on a balanced load, the power factor depends only on the ratio of the two readings: $\tan \varphi =\frac{\sqrt{3}(W_1-W_2)}{W_1+W_2}$. When one reading is half the other and both are positive, $\tan \varphi =\frac{1}{\sqrt{3}}$, so $\varphi =30^{\circ }$ and the power factor is 0.866. Points worth remembering: equal readings give a power factor of 1, a ratio of 2 : 1 gives 0.866, one reading at zero gives 0.5, and one negative reading means the power factor is below 0.5.

   WATCH OUT FOR

   Leaving out the $\sqrt{3}$ gives $\tan \varphi =\frac{1}{3}$ and a power factor of about 0.949, which is not among the options. Another slip is treating the smaller reading as negative even though the question says both are positive. That gives $\tan \varphi =\frac{\sqrt{3}(1.5)}{0.5}=3\sqrt{3}$ and a power factor of about 0.189. Some students also put the sum on top and the difference below, which gives $\tan \varphi =3\sqrt{3}$ as well.

   Full solution →
6. Q.6MCQ · 1 markOfficial answer: B
7. Q.7MCQ · 1 markOfficial answer: B

   Networks (Electric Circuits) › Laws & node/mesh

   In a connected graph, the number of independent loops equals the number of links: $L=b-n+1$. A tree uses $n-1$ branches and each remaining branch closes one fundamental loop, so $b=L+n-1$.

   WATCH OUT FOR

   Using $L=b-n$, which leaves out the $+1$ and gives 13 (option C), or confusing independent loops with meshes or node equations ($n-1=7$). Another mistake is adding $L+n=13$ without subtracting 1 for the tree.

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

   Networks (Electric Circuits) › AC, power & resonance

   When sources of equal magnitude are joined by a mostly inductive branch ($R\ll \omega L$), real power flows from the source whose voltage leads to the one that lags, $P\approx \frac{V^2\sin \delta }{X}$. Compare each source only with the source it is actually wired to. Here $v_1$ and $v_3$ are each wired to $v_2$, and $v_2$ leads both of them.

   WATCH OUT FOR

   Picking (A) because all three magnitudes are equal, when it is the phase difference that drives real power through a reactance. Another mistake is picking (B): comparing $v_3$ with $v_1$ (it leads $v_1$ by $5^{\circ }$) and calling it a supplier. But $v_3$ is connected only to $v_2$, which leads it by $5^{\circ }$, so $v_3$ absorbs power.

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

   Control Systems › Time response

   The damping ratio of a second-order system comes from its denominator alone: for $s^2+bs+c$, $\zeta =\frac{b}{2\sqrt{c}}$. Equivalently, compare $b^2$ with $4c$. If $b^2<4c$ the poles are complex (underdamped), if $b^2=4c$ they are repeated (critically damped), and if $b^2>4c$ they are real and distinct (overdamped). The numerator plays no part.

   WATCH OUT FOR

   Taking $\zeta$ to be the coefficient of $s$ divided by $2$, forgetting the division by ${\omega }_n$. That gives $\zeta =2.5$, $5$ and $9$, which would wrongly call all three systems overdamped. A second mistake is to rank the systems by the size of the $s$ coefficient alone. R has the largest $s$ coefficient, but you only know it is overdamped once you check $18^2=324>4\times 35=140$.

   Full solution →
10. Q.10MCQ · 1 markOfficial answer: D

    Electromagnetics › Electrostatics

    A point charge above a grounded conducting plane gives the same field (above the plane) as the charge plus an equal and opposite image charge placed at its mirror point. At a point between the real charge and the plane, both the real charge (pushing away) and the negative image (pulling toward) give a field along −z, so the two magnitudes add.

    WATCH OUT FOR

    Leaving out the image charge gives −899.18 V/m (option B). Getting the direction wrong gives a positive value (option A or C). Another mistake is to subtract the image contribution instead of adding it. Directly below a positive charge, and above a negative image, both fields point along −z.

    Full solution →
11. Q.11MCQ · 1 markOfficial answer: D

    Engineering Mathematics › Calculus

    $f(x)=x|x|$. Its derivative is $f^{\prime }(x)=2|x|$, which is continuous at $0$ but has a corner there. So $f$ has a continuous first derivative, but no second derivative exists at $0$: the one-sided second derivatives are $+2$ and $-2$. When a function is defined in pieces, test each level of smoothness at the joint separately with one-sided limits.

    WATCH OUT FOR

    Mixing up $x|x|$ with $|x|$ and choosing (B). Another mistake is to see that the branch derivatives $2x$ and $-2x$ come from different formulas and conclude that $f^{\prime }$ jumps, which gives (C). In fact both branches give $0$ at $x=0$, so $f^{\prime }$ is continuous there. What fails is the next level: the slopes of $f^{\prime }$ are $+2$ and $-2$.

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

    Engineering Mathematics › Vector calculus

    Directional derivative = gradient at the point dotted with the UNIT vector in the given direction: $D_{\mathbf{p}}\Phi =\nabla \Phi \cdot \mathbf{p}/|\mathbf{p}|$. Here $\nabla \Phi (2,-1,1)=(5,-3,2)$ and $|\mathbf{p}|=3$, giving $3/3=1$.

    WATCH OUT FOR

    Dotting the gradient with $\mathbf{p}$ itself instead of the unit vector gives $3$, which is not an option. The other trap is differentiating the cyclic terms carelessly: $z{x}^2$ contributes $2zx$ to $\partial \Phi /\partial x$ and $x^2$ to $\partial \Phi /\partial z$. Also watch the sign of $y=-1$ in $2xy$ and $2yz$.

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

    Engineering Mathematics › Complex analysis

    Only poles inside the contour count. $|z+2|=1$ encloses $z=-2$ but not $z=2$, so the integral is $2\pi i$ times the residue at $z=-2$, which is $\frac{z+1}{z-2}$ evaluated at $z=-2$, that is $\frac{1}{4}$.

    WATCH OUT FOR

    Three slips are common here. The first is counting the pole at $z=2$ as well: it lies at distance 4 from the centre, outside the unit circle. The second is losing a sign in $\frac{-1}{-4}$, which gives $-\frac{\pi i}{2}$ (option C). The third is forgetting the residue altogether and answering $2\pi i$ (option B).

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

    Digital Electronics › Boolean logic

    A two-level NAND-NAND circuit carries out an AND-OR (sum-of-products) function. By De Morgan's theorem, the output gate's NAND turns into an OR of the complemented inputs, and those complements cancel the bubbles of the first-level gates. So each first-level NAND gives one product term, and the output is the OR of those terms: $AB+CD$.

    WATCH OUT FOR

    Treating each bubble as a separate inversion of the whole circuit and ending up with $\overline{AB+CD}$, or applying De Morgan to the inner NANDs as well, which gives $(\overline{A}+\overline{B})$-type terms and leads to option (B) or (C). Only the output NAND needs De Morgan. Its complements cancel the first-level bubbles, so the inner products $AB$ and $CD$ come through unchanged.

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

    Analog Electronics › Op-amps

    The $R_1$–$R_2$ network makes a non-inverting amplifier, so $V_o=v_{in}(1+R_1/R_2)$ is larger than $v_{in}$. The only path for the input current is through $Z$ to the output. Since the output sits above the input, current is pushed back into the source, so the input impedance comes out negative: $-Z{R}_2/R_1$. This is a negative impedance converter.

    WATCH OUT FOR

    Answering $Z$ (option C) by treating $Z$ as if it simply loads the source, which ignores the fact that the far end of $Z$ is the amplified output $V_o$, not ground. Another slip is writing the current through $Z$ as $(V_o-v_{in})/Z$ while still calling it $i_{in}$. That drops the minus sign, the key feature of the result.

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

    Signals and Systems › LTI & convolution

    An eigenfunction passes through the system unchanged in shape: $T\{x(t)\}=kx(t)$, where $k$ is a constant (possibly complex) that does not depend on $t$. For an LTI system, $e^{j{\omega }_{0}t}$ is an eigenfunction and its eigenvalue is $H({\omega }_0)$.

    WATCH OUT FOR

    Picking (D) because the frequency response is involved. For the input $e^{j{\omega }_{0}t}$, $H({\omega }_0)$ is itself the eigenvalue, and the output is $H({\omega }_0)x(t)$, not $kH(\omega )$. The output always has to contain the input signal. Picking (B) or (C) is the other mistake: both multiply the input by a factor that changes with time, which is modulation, not scaling.

    Full solution →
17. Q.17NAT · 1 markOfficial answer: 5.5

    Engineering Mathematics › Linear algebra

    For a $2\times 2$ matrix, trace = ${\lambda }_1+{\lambda }_2$ and determinant = ${\lambda }_1{\lambda }_2$, while $\text{tr}({\mathbf{A}}^2)={\lambda }_1^2+{\lambda }_2^2$. The identity $({\lambda }_1+{\lambda }_2{)}^2={\lambda }_1^2+{\lambda }_2^2+2{\lambda }_1{\lambda }_2$ therefore gives $\det (\mathbf{A})=\frac{1}{2}\left[(\text{tr}\mathbf{A}{)}^2-\text{tr}({\mathbf{A}}^2)\right]$. You get the same formula by taking the trace of the Cayley–Hamilton equation ${\mathbf{A}}^2-(\text{tr}\mathbf{A})\mathbf{A}+(\det \mathbf{A})\mathbf{I}=0$.

    WATCH OUT FOR

    Assuming $\text{tr}({\mathbf{A}}^2)=[\text{tr}(\mathbf{A}){]}^2$ (this would give 16, not 5), or dropping the factor 2 and answering $16-5=11$. Another trap is to try to find real eigenvalues, get a negative discriminant ($16-22=-6$) and decide the data are inconsistent. The eigenvalues are simply the complex pair $2\pm j\sqrt{1.5}$, and the sum and product relations still hold for them.

    Full solution →
18. Q.18NAT · 1 markOfficial answer: 0.49 to 0.51

    Engineering Mathematics › Calculus

    $x-[x]$ is the fractional part of $x$: a sawtooth with period 1 that rises from 0 towards 1 on every unit interval. Split the integral where $[x]$ jumps (here at $x=1$), or note that $[0.25,1.25]$ is exactly one period long, so the integral equals ${\int }_0^{1}xdx=\frac{1}{2}$. Geometrically, it is the area of a triangle with base 1 and height 1.

    WATCH OUT FOR

    Treating $[x]$ as 0 across the whole range and integrating $x$ from 0.25 to 1.25, which gives 0.75. Another slip is reading $[x]$ as rounding to the nearest integer rather than the floor. On $[1,1.25]$ the integrand is $x-1$, not $x$.

    Full solution →
19. Q.19NAT · 1 markOfficial answer: 0.45 to 0.55

    Networks (Electric Circuits) › Two-port

    $h_{11}$ is the input impedance with the output shorted. Set $V_2=0$ so that node B is grounded, write one KCL equation at the central node with the controlled source current $2I_1$ included, and then $V_1=1\cdot I_1+V_A$. You cannot simplify the network with series/parallel reduction because the controlled source depends on $I_1$. The dependent source lowers the input impedance from the passive value $1+(1\parallel 1)=1.5\Omega$ to $0.5\Omega$.

    WATCH OUT FOR

    Ignoring the dependent source, or treating it as disappearing when port 2 is shorted, and reporting the passive value $1+(1\parallel 1)=1.5\Omega$. With $V_2=0$ the source is not removed. It still takes $2I_1$ out of node A (into the short), so it must stay in the KCL equation. A second slip is getting its direction wrong (taking $2I_1$ as entering node A). That gives $V_A=1.5I_1$ and $h_{11}=2.5\Omega$.

    Full solution →
20. Q.20NAT · 1 markOfficial answer: 7.0
21. Q.21NAT · 1 markOfficial answer: 73.0 to 74.0
22. Q.22NAT · 1 markOfficial answer: 3500.0
23. Q.23NAT · 1 markOfficial answer: 16.90 to 17.70
24. Q.24NAT · 1 markOfficial answer: 600.0

    Electrical Machines › DC machines

    At zero speed the back-emf is zero, so the entire applied voltage drops across armature resistance, fixing the stall current and hence the torque constant $K\varphi$ (from $T=K\varphi I_a$). At no load (no mechanical losses, so no-load torque = 0), the armature current must be zero, so the full new armature voltage appears as back-emf; speed then follows from $E=K\varphi \omega$.

    WATCH OUT FOR

    Assuming the armature current found from the stall condition (2000 A) is reused at the new voltage, or forgetting that at no-load with no mechanical losses the torque — and hence $I_a$ — must be exactly zero, which is what makes $E=V_a$ at no load.

    Full solution →
25. Q.25NAT · 1 markOfficial answer: 0.65 to 0.69

    Control Systems › Time response

    In a stable unity-feedback type-0 system, the unit-step steady-state error is $e_{ss}=\frac{1}{1+K_p}$, where $K_p=G(0)$ is the DC gain of the open-loop transfer function. You can check it quickly: the closed-loop DC gain is $\frac{G(0)}{1+G(0)}$, and the error is 1 minus that.

    WATCH OUT FOR

    Writing $e_{ss}=\frac{1}{K_p}=2$ (the formula for ramp error with $K_v$, or a slip in the step formula). Another error is to give the steady-state output, $\frac{1}{3}\approx 0.33$, as the error. A third is to skip the stability check. The final value theorem only holds when the closed loop is stable, and here that means confirming $s^2+3s+3$ has both roots in the left half-plane.

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

    Electromagnetics › Magnetostatics

    With the secondary open, the primary behaves as a toroidal inductor with $L=\frac{\mu N_P^2(\pi r^2)}{2\pi R}=\frac{\mu N_P^2{r}^2}{2R}$. Its voltage and current amplitudes satisfy $V=\omega LI$. Solving this for $R$ gives the answer.

    WATCH OUT FOR

    There are three easy slips. (1) Using the mutual inductance, which gives $N_P{N}_S$ as in option (B), when it is the PRIMARY voltage and current that are given. The self-inductance goes as $N_P^2$. (2) Inverting the relation to $R\propto V/I$ as in options (A) and (C). A larger radius means a longer magnetic path and so a SMALLER inductance, so $R$ must grow with $I/V$. (3) Dropping the factor 2 that comes from $\pi r^2/(2\pi R)$.

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

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

    For a shunt to divide the current the same way at every frequency, the shunt branch must have the same $L/R$ time constant as the meter coil. The impedance ratio is then a pure resistance ratio, so the DC formula $R_{sh}=R_m/(m-1)$ still holds.

    WATCH OUT FOR

    One common slip is to use $R_{sh}=R_m/m=5\text{m}\Omega$ instead of $R_m/(m-1)$. The meter carries 1 A of the 10 A, so the shunt carries only 9 A. Another is to think frequency independence means the shunt must be purely resistive. A non-inductive shunt makes the reading depend on frequency, because the meter's reactance $\omega L_m$ grows with $\omega$. The shunt needs the same time constant as the meter, not zero inductance.

    Full solution →
28. Q.28MCQ · 2 marksOfficial answer: A
29. Q.29MCQ · 2 marksOfficial answer: C
30. Q.30MCQ · 2 marksOfficial answer: C

    Engineering Mathematics › Numerical methods (IN)

    Newton-Raphson on $f(\delta )=P(\delta )-0.8$ gives ${\delta }_1={\delta }_0-f({\delta }_0)/f^{\prime }({\delta }_0)$. Differentiating the sine terms gives $f^{\prime }$ per radian, so the correction comes out in radians and has to be converted to degrees before you subtract it from $30^{\circ }$.

    WATCH OUT FOR

    There are two traps. (1) Subtracting the radian correction $0.02195$ straight from $30$ gives $29.98^{\circ }$. (2) Dropping the chain-rule factor, i.e. writing $\frac{d}{d\delta }(0.15\sin 2\delta )=0.15\cos 2\delta$ instead of $0.3\cos 2\delta$, gives about $28.67^{\circ }$. Neither result is among the options.

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

    Networks (Electric Circuits) › Transients

    A DC step applied to a lossless series LC circuit makes the capacitor voltage oscillate about the source voltage. Here $v(t)=5-5\cos t$, so the voltage has amplitude 5 V around a centre of 5 V. The current $i(t)=5\sin t$ is 90 degrees out of phase with that oscillation, with amplitude $V_s/\sqrt{L/C}$. Plotting two quadrature sinusoids against each other gives an ellipse centred at $(V_s,0)$. It is a circle here only because $\sqrt{L/C}=1\Omega$, which makes both amplitudes equal to 5.

    WATCH OUT FOR

    One mistake is assuming the capacitor charges only up to 5 V, as it would in an RC circuit. With no resistance, it overshoots to 10 V, so the circle is centred at $v=5$, not at the origin. Another is mixing up the axes (the vertical axis is $i$) and choosing the circle centred at $(0,5)$. A third is a polarity slip that gives the centre $(-5,0)$. $v(t)=5(1-\cos t)$ is never negative, so every point of the locus has $v\ge 0$.

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

    Networks (Electric Circuits) › AC, power & resonance

    An infinite ladder looks the same after you remove its first section. So $Z_{eq}=Z_1+(Z_2\parallel Z_{eq})$, which is a quadratic in $Z_{eq}$. Of its two roots, keep the one a physical ladder can actually have. Here every element is inductive, so $Z_{eq}$ must be a positive reactance.

    WATCH OUT FOR

    One mistake is to stop after one section and answer $j9+j4=j13\Omega$ (option C); that is only the first step of the iteration. Another is to take the wrong quadratic root ($-j3\Omega$), or to slip a sign in $j\cdot j=-1$ and land on $-j12\Omega$ (option B). A third is to forget that the shunt branch is a series L–C pair and must be combined first, giving $j5-j1=j4\Omega$.

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

    Control Systems › State space

    Break the system into its modes. The sum $x_1+x_2$ is conserved (the $\lambda =0$ mode) and the difference $x_2-x_1$ decays as $e^{-2t}$ (the $\lambda =-2$ mode). The states therefore meet at the average of their initial values. The system is marginally stable, not asymptotically stable: it stays bounded but does not return to zero.

    WATCH OUT FOR

    Reading $\lambda =0$ as instability and choosing (D). A simple eigenvalue at the origin gives a constant mode, not a growing one; only a repeated eigenvalue at zero (with a Jordan block) would give a term that grows like $t$. The other mistake is assuming the ordering $x_1<x_2$ survives to the limit and choosing (A). The gap stays positive for every finite $t$ but goes to zero as $t\to \infty$.

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

    Control Systems › Routh–Hurwitz

    Routh–Hurwitz with an all-zero row: the zero row at $s^3$ means an even auxiliary polynomial $A(s)=s^4+6s^2+25$ divides $P(s)$. Replace the zero row with $dA/ds$ and continue. The sign changes in the first column count the RHP roots (4). Then check whether any roots of $A(s)$ lie on the $j\omega$ axis (none here; they are $\pm 1\pm j2$). LHP count = total − RHP − $j\omega$ = 7 − 4 − 0 = 3.

    WATCH OUT FOR

    Seeing a zero row and assuming it always means roots on the $j\omega$ axis. That makes a student subtract four imaginary-axis roots, or count $A(s)$'s roots as marginal, and so pick a wrong option. An even polynomial's roots are symmetric about the origin: they can be real pairs $\pm a$, imaginary pairs $\pm j\omega$, or complex quadruples $\pm \sigma \pm j\omega$. Here $s^2=-3\pm j4$ gives the quadruple $\pm 1\pm j2$, so none are on the axis. A second trap: people read "4 sign changes" as 4 LHP roots (option B). The sign changes count right-half-plane roots.

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

    Engineering Mathematics › Complex analysis

    Both double poles, $z=1$ and $z=2$, lie inside $|z|=4$. Their residues are $+4$ and $-4$, which cancel. There is a shortcut: when a contour encloses every singularity of a rational function whose denominator degree is at least 2 more than its numerator degree (here 4 against 2), the integral is $0$, because the function has no $1/z$ term at infinity.

    WATCH OUT FOR

    Using the simple-pole formula $\lim (z-a)f(z)$ at what are actually double poles, since the whole denominator is squared. A second mistake is a sign error in the quotient rule, for example writing $(z-2{)}^3$ where it should be $(z-2{)}^2$, which gives residues that do not cancel and leads to $\pm 1$ or $-2$.

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

    Digital Electronics › Sequential

    A shift register with feedback is a counter. Its division ratio is the length of the state cycle, which you find by stepping through the next-state equations $Q_1^{+}=(Q_2{Q}_3{)}^{\prime }$, $Q_2^{+}=Q_1$, $Q_3^{+}=Q_2$. Here the cycle has 5 states. To rule out lock-out, check where every unused state goes. All three unused states here lead into the main cycle, so the counter starts itself and divides by 5.

    WATCH OUT FOR

    Seeing three flip-flops and assuming divide-by-3 (a ring counter) or divide-by-7/8. Another trap is picking (D) without tracing the unused states. $000$, $010$ and $101$ all feed into the 5-state cycle, so there is no disjoint loop. A third slip is reading the NAND inputs as $Q_1$ and $Q_3$ instead of $Q_2$ and $Q_3$.

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

    Digital Electronics › Boolean logic

    Set the don't care $m_1$ to 1 where it makes a group bigger. Here it completes two quads: the row $A=0$ gives $\overline{A}$, and the columns with $C=1$ give $C$. Before accepting the answer, check that $F$ is 0 exactly at $m_4$ and $m_6$.

    WATCH OUT FOR

    Leaving the don't care $m_1$ as 0. The 1s then break into smaller groups, such as $\overline{A}\overline{C}+\overline{A}B+C$ or something close to option (A), and it looks as if no option simplifies far enough. Another slip is to use B's weight for A and get the minterm numbers wrong, which puts the 0s in the wrong cells and makes (C) look right.

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

    Signals and Systems › Signals basics

    Sample at $t=nT$, list the finite set of samples (including the endpoints $t=\pm 1$, which are inside the closed interval), and add their squares: $E={\sum }_n|x[n]{|}^2$. Here $x_1[n]+x_2[n]=1$ for $|n|\le 4$. Both sequences take the values $0,0.25,0.5,0.75,1$, but $x_1[n]$ reaches the peak value $1$ at two samples ($n=\pm 4$) and $x_2[n]$ at only one ($n=0$).

    WATCH OUT FOR

    Choosing (C) by reasoning from the continuous-time signals. Their energies are equal, $\int |x_1(t){|}^{2}dt=\int |x_2(t){|}^{2}dt=\frac{2}{3}$, but sampling does not keep that equality: the samples at the interval edges $t=\pm 1$ are where $x_1$ is largest and $x_2$ is zero, so $x_1[n]$ gets two peak samples and $x_2[n]$ only one. Dropping the endpoint samples $n=\pm 4$ (treating the interval as open) gives the same kind of miscount, with $E_1=1.75$ against $E_2=2.75$, which points to (B).

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

    Signals and Systems › Signals basics

    A sum of shifted ramps $(t-a)u(t-a)$ is a piecewise-linear signal: at each breakpoint the slope changes by that ramp's coefficient. Sketch the waveform first, then integrate $x^2(t)$ one piece at a time. A linear edge that rises or falls by $A$ over a width $T$ contributes $A^{2}T/3$, and a flat segment of height $A$ and width $T$ contributes $A^{2}T$.

    WATCH OUT FOR

    Students often find the energy of each ramp term separately and add the results. Each ramp on its own has infinite energy, and energy does not add across overlapping terms because $x^2$ has cross terms. Another slip is to take the energy as the area under $x(t)$, which is $2$, not $5/3$. A third is to mix up the signs at $t=3$ and $t=4$, which gives a waveform that never returns to zero.

    Full solution →
40. Q.40NAT · 2 marksOfficial answer: 9.5 to 10.5

    Signals and Systems › Fourier transform

    A squared first-order factor $\frac{1}{(a+j\omega {)}^2}$ in frequency is $t{e}^{-at}u(t)$ in time. You can get it from frequency differentiation ($tx(t)\leftrightarrow jdX/d\omega$) or from convolving $e^{-at}u(t)$ with itself. After that, the logarithm simply undoes the exponential.

    WATCH OUT FOR

    Writing $x(t)=e^{-10t}u(t)$ or $e^{-20t}u(t)$ and forgetting the factor $t$ that the squared denominator brings. Luckily it is 1 at $t=1$, so the answer comes out as 10 anyway, but the $e^{-20t}$ slip gives 20, which is wrong. Another trap is using $1/(a-j\omega )$, the anticausal pair, and getting a signal that is zero at $t=1$, whose logarithm does not exist.

    Full solution →
41. Q.41NAT · 2 marksOfficial answer: 89.0 to 91.5

    Analog Electronics › Biasing

    Because $V_{BE}$ is constant, the base-emitter junction has zero incremental resistance ($r_{\pi }=0$). Looking into the emitter you therefore see only the base resistance divided by $(\beta +1)$, which is $10\text{k}\Omega /101=99.01\Omega$. The collector branch is a current source and drops out. $R_{Th}$ is this value in parallel with the $1\text{k}\Omega$ across $a$–$b$. Dividing the open-circuit voltage by the short-circuit current gives the same answer, because the circuit is linear while the BJT stays active.

    WATCH OUT FOR

    Common mistakes: (1) Taking $R_{Th}=1\text{k}\Omega$ and ignoring the transistor. (2) Multiplying $R_B$ by $(\beta +1)$ instead of dividing: that is the rule for looking into the base, not the emitter. (3) Including the $10\Omega$ collector resistor, which sits behind an ideal current source ($r_o$ neglected). (4) Adding a nonzero $r_{\pi }=V_T/I_B$, when the question says $V_{BE}$ is constant.

    Full solution →
42. Q.42NAT · 2 marksOfficial answer: 0.0

    Engineering Mathematics › Vector calculus

    If $\partial P/\partial y=\partial Q/\partial x$ on a simply connected region, then $Pdx+Qdy$ is the exact differential of a potential $\varphi$. The line integral is then $\varphi (\text{end})-\varphi (\text{start})$, whatever the path. Here $\varphi =xy(x+y)$ is zero on both coordinate axes, and both endpoints lie on an axis.

    WATCH OUT FOR

    Parametrising the arc as $x=3\cos t$, $y=3\sin t$ and grinding through the trigonometric integral, which wastes time and invites sign or limit errors. Another mistake is applying Green's theorem to the open arc as if it were a closed curve. The exactness test takes seconds and shows the answer is simply $\varphi (0,3)-\varphi (3,0)=0$.

    Full solution →
43. Q.43NAT · 2 marksOfficial answer: 11.5 to 12.5

    Engineering Mathematics › Calculus

    For the absolute maximum on a closed interval, compare $f$ at every critical point inside the interval AND at both endpoints. A local maximum found with $f^{\prime \prime }<0$ is only a candidate. Here the cubic has a positive leading coefficient, so it is rising again after its local minimum at $x=1$ and ends higher at $x=2$ than at its local peak.

    WATCH OUT FOR

    Setting $f^{\prime }(x)=0$, finding the local maximum at $x=5/9$ with $f^{\prime \prime }<0$, and answering $f(5/9)\approx 7.1$ without checking the endpoints. The absolute maximum on $[0,2]$ is at the endpoint $x=2$, where $f(2)=12$.

    Full solution →
44. Q.44NAT · 2 marksOfficial answer: 0.9 to 1.1

    Engineering Mathematics › Linear algebra

    The polynomial in $B$ differs from $A$'s characteristic polynomial ${\lambda }^3-{\lambda }^2-4\lambda +4$ only in the constant term. By Cayley-Hamilton that characteristic polynomial evaluated at $A$ is the zero matrix, so $B=0+I=I$ without computing any powers of $A$. The same thing shows up in the eigenvalues: $p({\lambda }_i)=1$ for every eigenvalue ${\lambda }_i$ of $A$.

    WATCH OUT FOR

    Students often take $\det B$ as $\det (A^3)-\det (A^2)-4\det A+5$. The determinant is not linear, so that is wrong. Other slips: getting a sign wrong when expanding $\det (A-\lambda I)$ (for example, writing the eigenvalue $-2$ as $+2$), or computing $A^2$ and $A^3$ by hand and making an arithmetic error. Also, $\det (4A)$ for a $3\times 3$ matrix is $4^3\det A$, not $4\det A$.

    Full solution →
45. Q.45NAT · 2 marksOfficial answer: 2.50 to 2.55

    Electromagnetics › Electrostatics

    A slab parallel to the plates that covers the whole plate area splits the gap into layers stacked along the field. The same $\mathbf{D}$ passes through every layer, so the layer capacitors add in series. Each half-thickness layer has twice the capacitance of the full gap filled with the same medium, which gives $C=\frac{2{ϵ}_r{C}_0}{{ϵ}_r+1}$.

    WATCH OUT FOR

    Treating the two layers as capacitors in parallel, as if the slab covered half the plate area over the full gap. That gives $C=\frac{C_0(1+{ϵ}_r)}{2}$ and ${ϵ}_r\approx 1.87$. A second mistake is forgetting that each layer is only $d/2$ thick, so its capacitance is $2C_0$ or $2{ϵ}_r{C}_0$, not $C_0$ or ${ϵ}_r{C}_0$.

    Full solution →
46. Q.46NAT · 2 marksOfficial answer: 8.0

    Control Systems › Time response

    Only one number on the graph matters: the steady-state value. For a stable unity-feedback loop, $y(\infty )=T(0)=\frac{G(0)H(0)}{1+G(0)H(0)}$ for a unit step. Here $G(0)H(0)=K/2$, so $0.8=\frac{K/2}{1+K/2}$ fixes $K$. The overshoot and ringing are there to distract you, or at most to confirm the answer afterwards.

    WATCH OUT FOR

    1) Reading the 1.2 peak and trying to extract $K$ from percent overshoot with second-order formulas. The system is third order, and the overshoot has to be measured from 0.8, not from 1. 2) Taking $G(0)H(0)=K$ instead of $K/2$, forgetting the factor $(0+2)=2$, which gives $K=4$. 3) Assuming the response settles at 1, the usual unit-step reflex. A type-0 loop always leaves a steady-state error. 4) Skipping the stability check. The final value theorem is only valid if $K$ is in the stable range, here $K<18$.

    Full solution →
47. Q.47NAT · 2 marksOfficial answer: 5.70 to 5.85

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

    When the star point is tied to the neutral, each branch current is its own phase voltage divided by its own impedance, and the neutral carries their phasor sum. The j10 and -j10 branches each draw 10 A. Their quadrature parts cancel and leave a real current of $-10\sqrt{3}\text{A}$. Phase a must therefore draw $+10\sqrt{3}\angle 0^{\circ }\text{A}$, which is in phase with $V_{an}$, so a resistor can supply it: $R=100/(10\sqrt{3})$.

    WATCH OUT FOR

    The usual slip is swapping the reactances between phases, i.e. putting the capacitor on b and the inductor on c. Then $I_b+I_c=+17.32\text{A}$, which needs $R=-5.77\Omega$. A negative value like this shows the phase assignment is wrong, but candidates often just enter its magnitude without noticing. Other slips are forgetting that dividing by $-j10$ adds $+90^{\circ }$ to the angle, and using line voltages ($100\sqrt{3}\text{V}$) when a neutral-connected star has phase voltages across its branches.

    Full solution →
48. Q.48NAT · 2 marksOfficial answer: 10.0

    Networks (Electric Circuits) › AC, power & resonance

    When voltage and current contain components at different frequencies (here DC and $\omega$), the average power is the sum of the powers at each frequency, because products of components at different frequencies average to zero. At DC, $P=V_0{I}_0$. At $\omega$, $P=\frac{1}{2}{V}_m{I}_m\cos ({\theta }_v-{\theta }_i)$, and only after both signals are written as cosines with positive amplitudes.

    WATCH OUT FOR

    Taking the voltage angle as $+60^{\circ }$ and ignoring the minus sign in $-10\cos (\omega t+60^{\circ })$. That gives $P_{\omega }=+2.5X$ and $X=-10$, a sign error. The minus sign is a $180^{\circ }$ phase shift, so the angle difference is $120^{\circ }$, not $60^{\circ }$. A second mistake is to leave out the DC power $5\times 5=25\text{W}$, or to add cross terms such as $5\cdot X\cos \omega t$, which average to zero.

    Full solution →
49. Q.49NAT · 2 marksOfficial answer: 90.0
50. Q.50NAT · 2 marksOfficial answer: 11.50 to 13.50
51. Q.51NAT · 2 marksOfficial answer: 97.20 to 97.55

    Electrical Machines › Transformers

    For a Δ/Y three-phase transformer, phase voltage equals line voltage on the Δ side but line voltage/√3 on the Y side; full-load phase currents on each side then follow from rated kVA = 3 × V_phase × I_phase for that side. Total copper loss is the sum of I²R loss in all three phases of both windings, and efficiency at any power factor is (output)/(output + copper loss + iron loss), with output = rated kVA × cosφ at full load.

    WATCH OUT FOR

    Using the line voltage as the phase voltage on the Y (secondary) side (forgetting to divide by √3), or computing copper loss for only one phase instead of multiplying by 3, both of which give a wrong current/loss and shift the efficiency outside the key range.

    Full solution →
52. Q.52NAT · 2 marksOfficial answer: 823 to 827

    Electrical Machines › DC machines

    In a linearly-saturated (unsaturated, linear) DC series motor, flux is proportional to armature current, so torque is proportional to the square of armature current. Equating motor torque to the new load torque gives the new armature current directly; back-EMF at each operating point then follows from V − IaR, and since Eb ∝ φN ∝ IaN, the ratio of back-EMFs gives the new speed.

    WATCH OUT FOR

    Assuming speed is inversely proportional to current alone (as in a shunt motor with constant flux) instead of properly forming the ratio Eb1/Eb2 = (N1 Ia1)/(N2 Ia2), which correctly accounts for flux also changing with armature current in a series motor.

    Full solution →
53. Q.53NAT · 2 marksOfficial answer: 12.0
54. Q.54NAT · 2 marksOfficial answer: 0.104 to 0.112

    Electrical Machines › Induction machines

    At no-load the slip s → 0, so in the forward-field branch of the double-revolving-field equivalent circuit the rotor term $R_2^{\prime }/(2s)$ tends to infinity (open circuit), leaving only the magnetizing reactance $j{X}_M/2$ in the forward branch; in the backward-field branch, $2-s\to 2$, so it stays a finite parallel combination that must be computed and added to the series stator and forward-branch impedances to get the total no-load impedance and hence the power factor.

    WATCH OUT FOR

    Forgetting that the backward-field branch does NOT vanish or simplify away at no load (only the forward branch's rotor resistance term goes to infinity because of s → 0 in the denominator; the backward branch's denominator is $2-s\to 2$, a finite number), so students either drop the backward branch entirely (giving a much higher, wrong power factor) or incorrectly send $2-s\to 0$ as well.

    Full solution →
55. Q.55NAT · 2 marksOfficial answer: 50.0

    Networks (Electric Circuits) › AC, power & resonance

    Because the inductor carries the terminal current, "i in phase with v" means the whole input impedance is purely resistive. First reduce the parallel R-C branch to its series form, $50-j50\Omega$. The inductor's $+j\omega L$ has to cancel the $-j50$, so the resistive part left over, $50\Omega$, is $|Z|$. You don't need to find L to get the answer.

    WATCH OUT FOR

    A common mistake is to resonate L with the capacitor alone, using ${\omega }^{2}LC=1$, which gives $L=1\text{H}$. That ignores the parallel resistor. With $L=1\text{H}$ you get $Z=50+j50\Omega$ and $|Z|=70.71\Omega$, and there the current is not in phase with the voltage. Another mistake is to quote $|Z_p|=|50-j50|=70.71\Omega$ as the answer, which leaves out the inductor.

    Full solution →