GATE IN 2025 question paper with solutions

GATE IN 2025 answer key

General Aptitude Q.1 to Q.10

1. Q.1MCQ · 1 markOfficial answer: C

   General Aptitude › Grammar & vocabulary

   Let the contrast signal and the grammar make the choice. "Despite ... hesitation" plus "never wavered" asks for firm determination. The possessive "Rehman’s" in front and the infinitive "to contribute" after both call for a noun that takes "to + verb", and "resolve" (noun) is the only option that works.

   WATCH OUT FOR

   Picking "revolve" because it looks like "resolve" (it is a verb and means turning around a centre), or picking "ambivalence" because it sounds sophisticated. Ambivalence means about the same as hesitation, so it wipes out the contrast that "Despite" sets up.

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2. Q.2MCQ · 1 markOfficial answer: C

   General Aptitude › Reasoning

   Put the first pair's relation into one sentence ("a nest is the home of a bird") and then test every option in that same sentence for the second word ("a ___ is the home of a bee"). Only the option that fits the sentence exactly is correct.

   WATCH OUT FOR

   Picking Lair or Kennel only because each is some animal's home. That matches the general category (animal home) but not the specific animal. A kennel belongs to a dog and a lair to a wild beast. Only a hive belongs to a bee.

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3. Q.3MCQ · 1 markOfficial answer: A

   General Aptitude › Numerical

   $P{e}^x=Q{e}^{-x}$ for every $x$ means $P{e}^{2x}=Q$ for every $x$. A non-zero multiple of $e^{2x}$ is never constant, so the identity can only hold if $P=0$, and then $Q=0$ too. Put in two values of $x$ ($x=0$ gives $P=Q$, and any $x\ne 0$ then gives $P=0$).

   WATCH OUT FOR

   Using only $x=0$, getting $P=Q$, and then choosing (B) $P=Q=1$. The words "for all real values of $x$" are what force $P=Q=0$. Another trap is choosing (D) because $P=0$: once $Q=0$ as well, $\frac{P}{Q}$ is undefined, not zero.

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

   General Aptitude › Spatial

   In a cross-shaped net, two squares with exactly one square between them fold onto opposite faces. That gives the pairs open triangle / filled triangle, filled circle / open circle and blank / blank. Any cube drawing that shows two faces from the same pair side by side is impossible, and that rules out (B), (C) and (D).

   WATCH OUT FOR

   Picking (C) because blank faces look like they cannot contradict anything. The net has only two blank squares, and they are opposite each other (first and third in the middle row), so two blank faces can never share an edge. A second trap is assuming the two triangles are adjacent because they sit in the same row of the net. In fact they are two squares apart, so they are opposite faces.

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5. Q.5MCQ · 1 markOfficial answer: B

   General Aptitude › Numerical

   "For all values" means one counterexample kills an option. Test small primes, including $2$, the only even prime. Prove the option that survives from the definition: a product of two numbers each at least $2$ is always composite.

   WATCH OUT FOR

   Two traps here. For (A), students often think "the sum of two primes is even, so it is not prime", which forgets the prime $2$ ($2+3=5$). For (D), students often pick it because $2\times 3+1=7$ is prime, which half-remembers Euclid's argument. Euclid's argument only shows that $p_1{p}_2+1$ has a prime factor outside $\{p_1,p_2\}$, not that it is prime itself, and $3\times 5+1=16$ is a counterexample.

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6. Q.6MCQ · 2 marksOfficial answer: B

   General Aptitude › Reading

   A past counterfactual ("If I had known ..., I would not have ...") presents its if-clause as contrary to fact. So "had known" tells you the speaker did not know. "Even if" only adds that knowing would have made no difference to the outcome.

   WATCH OUT FOR

   Picking (C) because refusing to visit sounds hostile. The question asks for what follows from the sentence alone, and nothing in it mentions a past friendship. Another trap is reading "even if" as a sign that Ramya did know (option A). The past-perfect "had known" is what marks the condition as unreal.

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7. Q.7MCQ · 2 marksOfficial answer: B

   General Aptitude › Reasoning

   Each code is the word written backwards, with every letter then moved one place forward in the alphabet ($+1$). Find the rule from IMAGE, check it on FIELD, then apply it to BEACH.

   WATCH OUT FOR

   Comparing letters in the same position (I with F, M with H, and so on) gives unrelated shifts, and students then give up or guess. Another slip is shifting the letters without reversing the word. That turns BEACH into CFBDI, which is not among the options.

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8. Q.8MCQ · 2 marksOfficial answer: A

   General Aptitude › Data interpretation

   Test each candidate rule on the first step ($i=1$), then confirm the survivor on every later column. $X$ grows by adding the current input and $Y$ grows by multiplying by the current input. Both use $I(i)$, not the lagged $I(i-1)$.

   WATCH OUT FOR

   Checking only one column, or only one of the two outputs. Option (D) has the right rule for $X$, so it looks correct until you work out $Y(1)=Y(0)I(0)=400$, which is not $-80$. A student who checks only $i=1$ against the wrong row can also be fooled, because $X(0)=Y(0)=I(0)=20$ all start from the same value.

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9. Q.9MCQ · 2 marksOfficial answer: D

   General Aptitude › Geometry

   Triangle PLS is half of the rectangle, because it stands on side PL and its apex S lies on the opposite side MN. It is also half of the square, because it stands on side PS and its apex L lies on the opposite side QR. So the rectangle and the square have the same area, 2 × 2 = 4 cm², wherever L lies on QR.

   WATCH OUT FOR

   Three slips are common. (1) Measuring the figure, or assuming L is exactly the midpoint of QR, and then working out PL and LM with Pythagoras. That is slow, and a rounding slip leads to 2√2 or 8. (2) Taking triangle PLS as the whole rectangle instead of half of it, which gives 2 (option B). (3) Doubling twice and getting 8 (option C). The note says the figure is only representative: the area does not depend on where L sits.

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

    General Aptitude › Reasoning

    Treat the zones as vertices and the bridges as edges. Joining $n$ zones with the fewest bridges means a spanning tree: exactly $n-1$ bridges that reach every zone and form no cycle. For 5 zones that is 4 bridges, so check every option for all three conditions: the count, reaching every zone, and having no cycle.

    WATCH OUT FOR

    Choosing (B) because it has the right number of bridges (4) without checking that it reaches Z4. Q, S and T form a loop, so one bridge is wasted. The other trap is choosing (D) because it does connect every zone, even though it uses 5 bridges when 4 are enough.

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Instrumentation Engineering Q.11 to Q.65

1. Q.11MCQ · 1 markOfficial answer: A

   Engineering Mathematics › Linear algebra

   Each entry factors as $a_{ij}=-\beta (-1{)}^i(-1{)}^j$, so $A=-\beta u{u}^T$ with $u_i=(-1{)}^i$. A non-zero matrix that is a scalar times an outer product $u{u}^T$ has rank exactly 1: every row is a multiple ($\pm 1$) of the first row.

   WATCH OUT FOR

   Treating the checkerboard of $\beta$ and $-\beta$ as two independent row types (odd rows and even rows) and answering 2. An even row is exactly $-1$ times an odd row, so the two types are dependent. The other trap is thinking the size $2n$ or the condition $n>2$ affects the answer. Neither does: the rank is 1 for any size and any non-zero $\beta$.

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2. Q.12MCQ · 1 markOfficial answer: A

   Engineering Mathematics › Differential equations

   Separate the variables and integrate, then read the conic from the signs of the squared terms. The same signs give an ellipse (a circle if the coefficients are also equal). Opposite signs give a hyperbola. Only one squared variable gives a parabola.

   WATCH OUT FOR

   Losing or flipping the sign. $\frac{dy}{dx}=-9\frac{x}{y}$ gives $y^2+9x^2=C$, which is an ellipse, and students who mix up the two equations pick (C). Some also see only the $x^2$ and $y^2$ terms and answer "circle" without checking the signs and coefficients.

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3. Q.13MCQ · 1 markOfficial answer: D
4. Q.14MCQ · 1 markOfficial answer: C
5. Q.15MCQ · 1 markOfficial answer: D
6. Q.16MCQ · 1 markOfficial answer: B

   Signals and Systems › Signals basics

   Factor out the average frequency: $e^{j{\omega }_{1}t}+e^{j{\omega }_{2}t}=2\cos \left(\frac{{\omega }_1-{\omega }_2}{2}t\right)e^{j\frac{{\omega }_1+{\omega }_2}{2}t}$. The unit-magnitude phasor has no effect on $|x(t)|$, so the magnitude is $2|\cos (2t)|$. Taking the absolute value halves the period of the cosine.

   WATCH OUT FOR

   There are three common slips. One is to give the period of $x(t)$ itself: $\gcd (9,5)=1$ rad/s gives $2\pi$, which is not an option, and students then guess. Another is to stop at the period of $\cos (2t)$, which is $\pi$ (option A), and forget that the absolute value halves it. The third is to read the answer off one exponent and pick $\pi /9$ or $\pi /5$.

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7. Q.17MCQ · 1 markOfficial answer: A

   Digital Electronics › Combinational

   Treat the MUXes one at a time. The 4:1 MUX with inputs 0, 1, 1, 0 gives $P\oplus Q$. The 2:1 MUX, selected by R, passes that result when $R=0$ and a constant 1 when $R=1$. That is the OR of R with $P\oplus Q$, because $R+\overline{R}X=R+X$.

   WATCH OUT FOR

   Writing the 2:1 MUX output as $R(P\oplus Q)$, which is option (B). That swaps the roles of the two data inputs. The first MUX's output is on input 0, which is chosen when $R=0$, and the constant 1 is on input 1. A second mistake is reading the 0, 1, 1, 0 pattern as XNOR, which leads to option (C).

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8. Q.18MCQ · 1 markOfficial answer: A

   Digital Electronics › Number systems

   An $n$-bit 2's complement register can hold only values from $-2^{n-1}$ to $2^{n-1}-1$. When two positive numbers add up to more than $2^{n-1}-1$, the sum wraps around. The stored pattern is the true sum minus $2^n$, here $10-16=-6$. You can spot the overflow because two operands with sign bit 0 give a result with sign bit 1.

   WATCH OUT FOR

   Reading $1010$ as unsigned and answering $+10$, which is not an option, or treating it as sign-magnitude and getting $-2$. A related slip is choosing $-8$ because it is the most negative 4-bit value. Some students also flip the bits and forget to add 1, getting $0101=5$ and choosing $-5$ (C). The MSB carries weight $-8$, so $1010=-8+2=-6$.

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9. Q.19MCQ · 1 markOfficial answer: A

   Electromagnetics › Electrostatics

   Gauss's law for $\mathbf{D}$ uses only the free charge enclosed. For an infinite sheet, $\mathbf{D}=\frac{{\rho }_s}{2}{\hat{a}}_n$ whatever the medium and whatever the distance from the sheet. The permittivity changes only $\mathbf{E}=\mathbf{D}/({ϵ}_0{ϵ}_r)$.

   WATCH OUT FOR

   Dividing by ${ϵ}_r$ to get $0.5\hat{c}$ (option D), or by $2{ϵ}_r$ again to get $0.25\hat{c}$ (option B). Both mix up $\mathbf{D}$ with $\mathbf{E}$, and only $\mathbf{E}$ depends on the medium. A third mistake is to forget that the flux leaves through both faces of the pillbox and write $D={\rho }_s=10$ (option C). $D={\rho }_s$ is the result for a conductor surface, where the field on the inside is zero.

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

    Analog Electronics › Op-amps

    The source feeds two branches. The lower one ($R_3$, $R_4$) is a plain divider because the + input draws no current, and it fixes $V_{+}$. The virtual short copies $V_{+}$ onto the − input, so the current in $R_1$ is set by $V-V_{+}$ alone. $I$ is the sum of the two branch currents, and $R_2$ never enters it.

    WATCH OUT FOR

    Two common slips. The first is treating the − input as virtual ground (true only when the + input is grounded), which gives $I_1=V/R_1$ and the wrong answer. The second is letting $R_2$ into $V/I$ (as option (B) does), when all $R_2$ does is set the output voltage. Only the lower branch loads the source as a simple series resistance $R_3+R_4$, so option (D) misses the current through $R_1$.

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

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

    A wattmeter always reads a little more than the load power, because it also counts the loss in one of its own coils. With the pressure coil on the source side (Topology-A) it adds the current coil's copper loss $I_L^2{R}_C$. The relative error is $I_L{R}_C/V_L$, which is small when the current-coil drop is small next to the load voltage ($V_L\gg V_C$). With the pressure coil on the load side (Topology-B) it adds the pressure-coil loss $V_L^2/R_P$. The relative error is $I_P/I_L$, which is small when the load current is much larger than the pressure-coil current ($I_L\gg I_P$). Topology-A suits high-resistance, low-current loads, and Topology-B suits low-resistance, high-current loads.

    WATCH OUT FOR

    Mixing up which coil's loss belongs to which connection. In Topology-A the pressure-coil current is drawn directly from the source and never flows through the current coil, so $I_P$ does not matter there. What matters is the drop $V_C$ that adds to the voltage the pressure coil sees. In Topology-B the pressure coil sees exactly $V_L$, so $V_C$ does not matter there. What matters is the current $I_P$ that the current coil carries on top of $I_L$. Swapping the two gives option (D). Another error is treating the coils as pure resistors and adding their voltages as plain numbers. Only the real part $\text{Re}(V_p{I}_c^{\ast })$ counts, and that gives the loss term $I_L^2{R}_C$ whatever the coil's reactance.

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12. Q.22MCQ · 1 markOfficial answer: D
13. Q.23MCQ · 1 markOfficial answer: C

    Electrical and Electronic Measurements › Digital instruments

    For a DVM spec of $\pm (\text{\% of reading}+n\text{digits})$, first find the range the reading sits on. On that range one digit is one count of the last displayed place. Change the digit term into volts, add it to the percentage-of-reading term, then divide by the reading. For a $3\frac{1}{2}$ digit meter, $10\text{V}$ sits on the $19.99\text{V}$ range, so $1\text{digit}=0.01\text{V}$.

    WATCH OUT FOR

    There are three common mistakes. The first is to answer $\pm 0.5\%$ and forget the digit term, but that is not one of the options. The second is to add the "1" as $1\%$, which gives $\pm 1.5\%$ (option B). The third is to take one digit as $0.1\text{V}$ by assuming the wrong range, which gives $0.05+0.1=0.15\text{V}$, or $1.5\%$. The digit term has to be turned into volts using the resolution of the range actually in use, which here is $0.01\text{V}$ on the $19.99\text{V}$ range.

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14. Q.24MCQ · 1 markOfficial answer: A

    Networks (Electric Circuits) › Coupled circuits

    A coupled pair with a common terminal is equivalent to a T of uncoupled inductors. When both terminal currents enter dotted ends (mutual terms $+M$), the arms are $L_a-M$ and $L_b-M$ and the shunt is $+M$. When one current enters a dot and the other does not (mutual terms $-M$), the arms are $L_a+M$ and $L_b+M$ and the shunt is $-M$. Here that gives $3+4=7\text{H}$, $2+4=6\text{H}$ and $-4\text{H}$.

    WATCH OUT FOR

    Ignoring where the dots are and always using $L_3=+M$. That gives $L_3=4\text{H}$, $L_1=3-4=-1\text{H}$ and $L_2=2-4=-2\text{H}$, which is exactly distractor (B). Here the 2 H coil's dot is at its bottom (common) end, so $i_2$ enters the undotted terminal and the mutual terms are negative. A second trap is rejecting option (A) because it contains a negative inductance. Negative elements are allowed in an equivalent circuit.

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15. Q.25MCQ · 1 markOfficial answer: A

    Networks (Electric Circuits) › AC, power & resonance

    Every branch sees the same voltage $V_s$, so every current is $V_s$ times an admittance. The angle between any two of these currents is therefore the angle between the two admittances, $\angle (Y_b+Y_c)-\angle (Y_a+Y_b+Y_c)$. The $45^{\circ }$ of the source cancels out.

    WATCH OUT FOR

    Picking $33.69^{\circ }$ (options C or D). That is the angle of $I_s$ measured from $V_s$, not from $I$. Other slips are leaving out $Y_a=-j0.2\text{S}$ when forming $I_s$, and reading the sign the wrong way round: $I$ is at the larger angle, so it leads $I_s$.

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

    Electrical and Electronic Measurements › Time, frequency & scopes

    A 10X passive probe is a series resistor $R_p=9R_{in}$ that forms a 10:1 divider with the oscilloscope's input resistance. At the tip, $R_p$ and $R_{in}$ are in series, so the tip resistance is $R_p+R_{in}=10R_{in}$. The attenuation factor and the factor by which the input resistance rises are the same number.

    WATCH OUT FOR

    One mistake is to combine $R_p=9\text{M}\Omega$ and $R_{in}=1\text{M}\Omega$ in parallel, which gives $0.9\text{M}\Omega$ (option A). The other is to report only the probe resistor, $9\text{M}\Omega$, or to guess $R_{in}(1+1/10)$. The current from the tip passes through $R_p$ and then through $R_{in}$, so the two resistors are in series and the total is $10\text{M}\Omega$.

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17. Q.27MCQ · 1 markOfficial answer: D

    Control Systems › Routh–Hurwitz

    Because of the leading $1+$, you first have to rewrite $G(s)$ as a single fraction. Its zeros are then the roots of $D(s)+(2s-1)=s^3+5s^2+5s+21$. Apply the Routh criterion to that combined numerator, not to the denominator. For a cubic $s^3+a_2{s}^2+a_{1}s+a_0$ with all coefficients positive, every root is in the left half-plane exactly when $a_2{a}_1>a_0$. Here $5\cdot 5=25>21$, so all three zeros are in the LHP.

    WATCH OUT FOR

    Reading the zeros off the $2s-1$ term gives a single zero at $s=+\frac{1}{2}$, which leads to the wrong answer (A) 0. A second mistake is running the Routh test on the denominator $s^3+5s^2+3s+22$. That polynomial gives the poles, not the zeros: its $s^1$ entry is $\frac{15-22}{5}<0$, so it has two right-half-plane roots, and this can mislead you into picking (B) 1.

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18. Q.28MCQ · 1 markOfficial answer: B

    Control Systems › Nyquist

    The argument principle counts encirclements in the same direction as the s-plane contour is traversed. This contour is counterclockwise, so the counterclockwise encirclements of the origin by $1+GH$ number $Z-P$. Stability needs $Z=0$, and with $P=2$ that gives $-2$ counterclockwise encirclements, which is 2 clockwise ones. For $1+GH$ the critical point is the origin, and for $GH$ it is $-1+j0$.

    WATCH OUT FOR

    Picking (C) by habit. The memorised rule 'P counterclockwise encirclements of $-1+j0$' assumes the usual clockwise Nyquist contour. The contour here is counterclockwise, so every encirclement direction flips. A second trap is using the wrong critical point: $-1+j0$ goes with $G(s)H(s)$ and the origin goes with $1+G(s)H(s)$. That rules out (A) and (D).

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19. Q.29MCQ · 1 markOfficial answer: B

    Digital Electronics › Boolean logic

    Factor first, then complement: $X=\overline{A}(\overline{B}+\overline{C})$, and De Morgan applied twice gives $\overline{X}=A+BC$. De Morgan turns a complemented product into a sum of complements, and a complemented sum into a product of complements.

    WATCH OUT FOR

    Complementing each literal but keeping the AND/OR operators, which gives $AB+AC$. The other slip is doing De Morgan only halfway and writing $A+\overline{B}+\overline{C}$ or $\overline{A}+\overline{B}+\overline{C}$ (option A). Both $\cdot$ and $+$ have to swap when you complement.

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20. Q.30MCQ · 1 markOfficial answer: A

    Electromagnetics › Magnetostatics

    At $t=0^{+}$ the bar is stationary, so the motional emf $Blv$ is zero and the current is set by $R_{int}$ alone: $I=V/R_{int}$. The force on the bar is $\mathbf{F}=I\mathbf{l}\times \mathbf{B}$, where $\mathbf{l}$ points along the current in the bar. Its magnitude is $BIl$ and the cross product gives its direction.

    WATCH OUT FOR

    Three mistakes are common. (1) Reporting the current, 120, as the force, which gives option (C). (2) Taking the current in the bar as upward, or reversing the order of the cross product ($\mathbf{B}\times \mathbf{l}$), which flips the direction and gives option (B). (3) Including a back emf, or concluding the force is zero because the bar is not yet moving. At the instant of closing, $v=0$, so there is no back emf, but the current, and therefore the force, is already at its maximum.

    Full solution →
21. Q.31MSQ · 1 markOfficial answer: B, C, D

    Analog Electronics › Op-amps

    An op-amp circuit is linear only when its output is a fixed linear combination of its inputs (superposition and homogeneity both hold). Positive feedback with saturation (Schmitt trigger), an exponential device in the feedback path (log amplifier) and a diode that conducts for one polarity only (precision rectifier) all break this. A purely resistive negative-feedback network (instrumentation amplifier) keeps it.

    WATCH OUT FOR

    Calling the precision rectifier linear because the op-amp makes it 'precise' (no 0.7 V drop). The op-amp removes the diode's cut-in voltage but does not change the piecewise transfer characteristic, so the circuit is still non-linear. Another mistake is thinking every circuit built with negative feedback is linear: the log amplifier uses negative feedback but has an exponential device in its feedback path.

    Full solution →
22. Q.32NAT · 1 markOfficial answer: 2

    Engineering Mathematics › Linear algebra

    If you are given an eigenvector but not its eigenvalue, write out $Av=\lambda v$ one row at a time. A row with no unknown in it gives you $\lambda$ straight away, and the other row then gives the unknown entry. The scale $\alpha$ cancels, so set $\alpha =1$.

    WATCH OUT FOR

    Assuming a value for $\lambda$, or trying to solve the characteristic equation with $x$ still unknown, when $Av=\lambda v$ gives it directly. Another slip is multiplying $v$ into the columns rather than the rows, which gives $x+2(-4)$ in the wrong place or a sign error in $-1-2=-3$. Also do not keep $\alpha$ as a variable: it cancels from both sides, and it cannot fix $x$.

    Full solution →
23. Q.33NAT · 1 markOfficial answer: 2

    Engineering Mathematics › Complex analysis

    For a proper rational function with simple poles, each residue is the partial-fraction coefficient: cover up the factor $(z-a)$ and put $z=a$ in what is left. As a quick check, the finite residues of a rational function add up to minus the residue at infinity. That equals the coefficient of $1/z$ in the large-$|z|$ expansion, here the leading-coefficient ratio $2/1=2$.

    WATCH OUT FOR

    Using $z=1$ where the pole is at $z=0$, or getting the sign wrong in $z-1$ at $z=0$, so the residue there comes out $+1$ instead of $-1$ and the sum becomes $4$. Another slip is to include the residue at infinity ($-2$) in the sum and answer $0$. The question asks only for the singular points, which are $z=0$ and $z=1$.

    Full solution →
24. Q.34NAT · 1 markOfficial answer: 125

    Digital Electronics › ADC & DAC

    In a dual-slope ADC the run-down time is $T_2=T_1{V}_{in}/|V_{ref}|$, independent of $R$ and $C$. The time to measure one input is the fixed run-up time plus that run-down time: $T_{conv}=T_1+T_2$.

    WATCH OUT FOR

    Giving only the run-down time, $T_2=25\text{ms}$, which is the part proportional to the input. The measurement also includes the fixed 100 ms integration of the input, so the answer is 125 ms. A second slip is carrying the minus sign of $V_{ref}$ into the ratio and getting a negative time. The reference is negative only so that it opposes the input, and the magnitude is what sets the slope.

    Full solution →
25. Q.35NAT · 1 markOfficial answer: 3.90 to 4.10

    Analog Electronics › Biasing

    The Zener fixes the base at $4.7\text{V}$, so the emitter sits one $V_{BE}$ lower at $4.0\text{V}$. That fixes $I_E=V_E/R_E$. With very high $\beta$, $I_C=I_E$, and the LED in the collector carries this current whatever its forward drop is. The circuit is a Zener-referenced constant-current sink.

    WATCH OUT FOR

    One mistake is to find the current through $R_1$ instead, $(12-4.7)/2.2\text{k}\Omega \approx 3.32\text{mA}$. That is the Zener current, not the LED current. Another is to forget to subtract $V_{BE}$ and get $4.7\text{V}/1\text{k}\Omega =4.70\text{mA}$. A third is to try to use an LED forward voltage. It is not given and does not affect the answer while the BJT stays active.

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

    Engineering Mathematics › Vector calculus

    Read $F_{1}dydz+F_{2}dzdx+F_{3}dxdy$ over a closed surface as the outward flux of $\vec{F}=(F_1,F_2,F_3)$. Then use the divergence theorem. Here $\nabla \cdot \vec{F}=4$ is constant, so the flux is just $4\times$ the volume of the ball of radius 3.

    WATCH OUT FOR

    Using the surface area of the sphere ($4\pi r^2=36\pi$) where the enclosed volume ($\frac{4}{3}\pi r^3=36\pi$) is needed. For $r=3$ these happen to be the same number, which hides the error, and it gives the wrong answer for any other radius. Other slips: taking the radius as 9 instead of $\sqrt{9}=3$ (which gives a far larger value), writing $\partial (2x+z)/\partial y=1$ (divergence 5, so $180\pi$, which is not an option), and dropping one of the two 2's to get $72\pi$ (option A).

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

    Engineering Mathematics › Numerical methods (IN)

    To find $1/a$ with Newton-Raphson, apply it to $f(x)=\frac{1}{x}-a$. That gives the division-free iteration $x_{n+1}=x_n(2-a{x}_n)$. Its relative error $e_n=1-a{x}_n$ squares at every step ($e_{n+1}=e_n^2$), so the method converges exactly when $0<x_0<2/a$. Here $2/1.6=1.25$, and that boundary value is one of the options.

    WATCH OUT FOR

    Using $f(x)=1.6x-1$ instead. That function is linear, so Newton-Raphson reaches the root in one step from any guess and none of the options would fail. A second mistake is treating the boundary $x_0=2/a=1.25$ as safe because it 'looks close' to 0.625. At the boundary $|e_0|=1$, the first step sends $x_1$ to exactly 0 and the iteration sticks there. A third is picking 0.55 or 1.15 because they look far from the root: both lie inside $(0,1.25)$ and both converge.

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

    Engineering Mathematics › Calculus

    Over one full period $[-\pi ,\pi ]$, an even power of cosine equals four times its integral over $[0,\pi /2]$. Wallis' formula then gives ${\int }_{-\pi }^{\pi }{\cos }^{n}xdx=2\pi \cdot \frac{(n-1)(n-3)\cdots 1}{n(n-2)\cdots 2}$ for even $n$, which is $\frac{5\pi }{8}$ for $n=6$ and $\frac{3\pi }{4}$ for $n=4$. Integrate each term separately and add.

    WATCH OUT FOR

    Using the $[0,\pi /2]$ Wallis value directly, without multiplying by 4 for the full interval $[-\pi ,\pi ]$, gives $\frac{5\pi }{32}+\frac{3\pi }{16}=\frac{11\pi }{32}$, which is not an option. Doubling only once (for evenness) gives $\frac{11\pi }{16}$. Picking (B) $\frac{5\pi }{8}$ means only the ${\cos }^{6}x$ term was evaluated and the ${\cos }^{4}x$ term was dropped.

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

    Signals and Systems › LTI & convolution

    For an LTI system at initial rest, the response to $\delta [n-p]$ is the impulse response delayed by $p$. So $y[p+1]=h[1]$, and one step of the recursion $h[n]=\frac{1}{\alpha }h[n-1]+\delta [n]$ gives $h[0]=1$, $h[1]=1/\alpha$.

    WATCH OUT FOR

    Reading the answer at the wrong index. $y[p]=1$ (option B) is the value at the instant the impulse arrives, but the question asks for one sample later, $y[p+1]=1/\alpha$. Going two samples later gives $1/{\alpha }^2$ (option D), and forgetting that the output only starts at $n=p$ gives 0 (option A). A second trap is thinking that $p>10$ or $\alpha >1$ changes the value. Neither does.

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

    Digital Electronics › Sequential

    Write the D-input equations and trace the state sequence. The output frequency is the clock frequency divided by the number of clock periods before F repeats. With Control = 1 the OR gate is forced to 1 and FF2 becomes a toggle flip-flop (divide by 2). With Control = 0 the feedback from $\overline{Q_2}$ gives the 3-state cycle 00 -> 10 -> 11 (divide by 3). The unused state 01 falls into that cycle.

    WATCH OUT FOR

    Assuming two flip-flops always make a mod-4 counter and dividing by 4 (3 MHz, options C and D), without tracing the states. Another is swapping the order the question asks for (Control = 0 first), which gives option (B).

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

    Signals and Systems › Fourier series

    The input is multiplied by the ±1 square wave twice. Since $s^2(t)=1$, it comes back at baseband with gain $100$. The offset enters between the two multipliers, so it is multiplied only once. Because $s(t)$ has no DC term, the offset ends up near the odd harmonics $f_s,3f_s,\dots$. The output filter's cutoff $\pi /T$ equals $f_s/2$ in hertz. It must pass the $50\text{Hz}$ signal band, which gives $f_s\ge 100\text{Hz}$. At that frequency the offset band ($90$ to $110\text{Hz}$ and up) lies outside the passband.

    WATCH OUT FOR

    One mistake is to take the fundamental amplitude $4/\pi$ of $s(t)$ and conclude that the offset reaches the output with gain $100\cdot \frac{4}{\pi }\cdot \frac{1}{2}=\frac{200}{\pi }$ (options B and D). That component sits at $f_s\pm 10\text{Hz}$, above the filter cutoff $f_s/2$, so it never reaches $v_{out}$. Another mistake is to read the cutoff $\pi /T$ as a frequency in hertz, or never to check it against the signal band. At $f_s=80\text{Hz}$ the cutoff is only $40\text{Hz}$, so part of the $50\text{Hz}$ signal band is cut off (options C and D).

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

    Digital Electronics › Processors & memory

    Address lines that go straight into a memory chip change freely across its range. The lines that feed the chip-select logic are fixed at the values that make the active-low $\overline{CS}$ equal to 0. Here $\overline{CS}=A_{15}+A_{14}^{\prime }$ is 0 only for $A_{15}{A}_{14}=01$. The 14 free lines $A_{13}\dots A_0$ then run from all 0s to all 1s, giving $\text{4000 H}$ to $\text{7FFF H}$.

    WATCH OUT FOR

    Treating $\overline{CS}$ as active high, or forgetting the inverter on $A_{14}$. Both errors give $A_{15}{A}_{14}=00$ or $10$, which lands on a wrong block such as $\text{8000 H}$ to $\text{BFFF H}$ (option D). Another trap is placing the RAM straight after the 4 KB ROM at $\text{1000 H}$ (option A). The decoder, not the ROM's size, decides where the RAM sits, and a 16 KB block decoded on $A_{15}{A}_{14}$ always starts on a 16 KB boundary.

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

    Digital Electronics › ADC & DAC

    In a binary-weighted-resistor DAC, the op-amp's inverting input is a virtual ground. Each closed branch therefore runs from $V_{ref}$ to $0\text{V}$ and draws $V_{ref}/R$ on its own. The reference source sees the closed branches in parallel, and their currents add in the feedback resistor, so $I=V_{ref}/R_{in}$.

    WATCH OUT FOR

    There are three common mistakes. (1) Reading the bits in the wrong order, as if $D_0=1$, $D_1=1$, $D_2=0$. That closes the 12 k$\Omega$ and 6 k$\Omega$ branches and gives 4 k$\Omega$ and 1.5 mA, which is not an option. (2) Taking $I$ as the current in the top 12 k$\Omega$ branch only, because the arrow is drawn at that node. This leads towards option (B), yet that branch is open and carries nothing. $I$ is the total current into the summing node. (3) Treating the op-amp input as an open circuit, which gives $R_{in}=\infty$ (options C and D). The inverting input draws no current, but it is held at 0 V, so each closed branch still draws current from the source.

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

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

    In the two-wattmeter method on a balanced load, the readings add up to the total three-phase power, and they are $V_L{I}_L\cos (30^{\circ }\mp \varphi )$. At a power factor of $\cos 30^{\circ }$, one meter reads the full $V_L{I}_L$ and the other reads half of it. The two readings are then in a 2 : 1 ratio and add up to $3\times$ the per-phase power.

    WATCH OUT FOR

    Taking the 30 kW as the total power instead of the per-phase power. That gives readings of 20 kW and 10 kW, which is not an option, or pushes you towards (A) or (B), whose readings also add up to only 30 kW. A second mistake is splitting the total power equally, as in (D). That is only true at unity power factor. At 30 degrees lagging the two meters read differently.

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

    Networks (Electric Circuits) › Theorems

    Each half of the bridge is a voltage divider fed by an ideal source, so it reduces to its own Thevenin equivalent measured from GND: $V_H=12\text{V}$ behind $2\text{k}\Omega$ and $V_L=6\text{V}$ behind $2\text{k}\Omega$. Figure (b) splits these into a common-mode part and a differential part. $V_C=\frac{V_H+V_L}{2}=9\text{V}$ is the common-mode voltage, and $V_1=V_2=\frac{V_H-V_L}{2}=3\text{V}$ is half the bridge's differential output, which is why both are 3 V.

    WATCH OUT FOR

    Using only the resistor from the output node to GND as the Thevenin resistance. That gives $R_1=6\text{k}\Omega$ and $R_2=3\text{k}\Omega$ (option A). When you find the Thevenin resistance, the ideal 18 V source becomes a short, so each output sees both of its arms in parallel: $3\parallel 6=2\text{k}\Omega$. A second trap is getting the sign of $V_2$ wrong. $V_L=V_C-V_2$, not $V_C+V_2$, because the $+$ terminal of $V_2$ faces the middle node.

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

    Electrical Machines › Induction machines

    In the linear region of the induction-motor torque-speed curve (from synchronous speed down to 95% of synchronous speed), torque is directly proportional to slip speed (N_s - N). Doubling the load torque doubles the slip speed.

    WATCH OUT FOR

    Forgetting to verify that the doubled-torque operating point (2940 rpm) still lies inside the linear segment (between 3000 rpm and 2850 rpm) before accepting it.

    Full solution →
37. Q.47MCQ · 2 marksOfficial answer: B

    Control Systems › Compensators & PID

    Use the root-locus angle condition at the desired pole. Every open-loop pole at the origin contributes $-\angle s_d$, measured from the positive real axis. The compensator has to supply whatever angle is still needed to bring the total to an odd multiple of $180^{\circ }$. Here that is $-180^{\circ }-(-2\times 120^{\circ })=60^{\circ }$.

    WATCH OUT FOR

    One mistake is to measure the pole angle from the negative real axis ($60^{\circ }$) instead of from the positive real axis ($120^{\circ }$). The plant phase then comes out as $-120^{\circ }$, which wrongly calls for $-60^{\circ }$ (a lag). Another is to count the double pole at the origin only once. A third is to pick $120^{\circ }$ (D) because that is $\angle s_d$ itself, not the deficit the compensator has to make up.

    Full solution →
38. Q.48MCQ · 2 marksOfficial answer: A

    Signals and Systems › Fourier transform

    Cross-correlation is convolution with the second signal time-reversed: $R_{xy}(\tau )=x(\tau )\ast y(-\tau )$. If the second signal is already $g(-t)$, the two reversals cancel, $h=f\ast g$, and the convolution property gives $H(j\omega )=F(j\omega )G(j\omega )$.

    WATCH OUT FOR

    Using the correlation theorem as $F(j\omega )G^{\ast }(j\omega )=F(j\omega )G(-j\omega )$ with $G$ instead of the transform of $g(-t)$. That gives option (C), which is the answer for correlating $f$ with $g(t)$ and ignores the time reversal the question builds in. Swapping the order of the two signals gives option (B).

    Full solution →
39. Q.49MSQ · 2 marksOfficial answer: A, B

    Engineering Mathematics › Complex analysis

    Cauchy's theorem runs one way only: if f is analytic in D, the integral is zero. The converse does not follow. A function with a singularity inside C can still have a zero integral whenever its residue there is zero. 1/z² is the standard example, because it has the single-valued antiderivative −1/z. The theorem is silent in that case, so you cannot use it, but you also cannot conclude the integral is non-zero.

    WATCH OUT FOR

    Reading Cauchy's theorem as "analytic if and only if the integral is zero". That reading makes (C) and (D) look true: (D) because "not analytic, therefore non-zero" and (C) because "must be analytic". The integral of 1/z² around the unit circle is 0 because the residue is 0. Some students also cite Morera's theorem to defend (C), but Morera needs f to be continuous throughout D, and 1/z² is not continuous at z = 0.

    Full solution →
40. Q.50MSQ · 2 marksOfficial answer: A, B, C

    Control Systems › Compensators & PID

    Form the closed-loop characteristic polynomial $D_c(s)(s^2-b^2)+a{N}_c(s)$ and apply the necessary condition for stability: every coefficient present and positive. The plant has no $s^1$ (damping) term. P cannot add one. An integrator raises the order to three but always leaves the $s^2$ coefficient at zero. Only derivative action ($K_{d}s$) supplies the missing term.

    WATCH OUT FOR

    Treating P control with $a{K}_p>b^2$ as stabilizing. It puts the poles at $\pm j\sqrt{a{K}_p-b^2}$, which is only marginally stable (sustained oscillation), not asymptotically stable. Another mistake is assuming PI must work because it has two gains to tune: the $s^2$ coefficient stays zero whatever $K_p$ and $K_i$ are. Also watch the word CANNOT, which makes some candidates mark only (D).

    Full solution →
41. Q.51MSQ · 2 marksOfficial answer: A, C

    Signals and Systems › LTI & convolution

    Every signal of the form $e^{st}$, for any complex $s$, passes through an LTI system as $H(s)e^{st}$. That includes real exponentials such as $2^t=e^{t\ln 2}$, not just $e^{j\omega t}$. The factor $e^{st}$ comes out of the convolution integral, so the only question is whether $H(s)$ converges. Stability puts the $j\omega$ axis inside the ROC, and causality plus stability puts the whole right half-plane inside it, including $s=\ln 2$. A real sinusoid is the sum of two exponentials whose eigenvalues $H(\pm j{\omega }_0)$ are generally different, so it comes back phase-shifted, not scaled.

    WATCH OUT FOR

    Choosing cos and sin because a sinusoid going in gives a sinusoid coming out. The output is a phase-shifted sinusoid, for example a delay turns $\cos$ into $\sin$. That is not a constant times the input, so cos and sin are not eigenfunctions. The opposite slip is dropping $2^t$ because it grows or is not an 'imaginary' exponential. In fact $2^t=e^{(\ln 2)t}$ is $e^{st}$ with $s=\ln 2$, and every causal stable system has $s=\ln 2$ in its ROC.

    The question and its official answer →
42. Q.52NAT · 2 marksOfficial answer: 0.72 to 0.76

    Engineering Mathematics › Probability & statistics

    Treat the 10 classes as independent trials with a fixed miss probability of 0.1. The number missed is then $\text{Binomial}(10,0.1)$. "Not more than one" means $X\le 1$, so add $P(X=0)$ and $P(X=1)$. Here $\left(\genfrac{}{}{0ex}{}{10}{1}\right)(0.1)=1$, so $P(X=1)=(0.9{)}^9$ and the whole sum factors to $1.9(0.9{)}^9$.

    WATCH OUT FOR

    Reading "not more than one" as "exactly one" gives only $P(X=1)=(0.9{)}^9\approx 0.39$. Reading it as "none" gives only $(0.9{)}^{10}\approx 0.35$. A third slip is swapping $p$ and $q$, using 0.9 as the miss probability, which gives a tiny wrong value. Another is using the Poisson approximation with $\lambda =np=1$: it gives $2e^{-1}\approx 0.7358$. That happens to round to 0.74 here, but it is an approximation, and the exact binomial is what the question asks for.

    Full solution →
43. Q.53NAT · 2 marksOfficial answer: 1.75 to 1.85
44. Q.54NAT · 2 marksOfficial answer: 5

    Signals and Systems › DTFT & DFT

    Taking $N$ samples of a DTFT and inverting them with an $N$-point IDFT gives the time-aliased sequence $y[n]={\sum }_{r}x[n+rN]$. Here the sequence has 4 samples and $N=3$, so $x[3]$ folds back onto $n=0$.

    WATCH OUT FOR

    Answering $y[0]=x[0]=1$ treats the 3-point DFT as though it simply cut off $x[3]$. That is wrong because $N=3$ is less than the sequence length 4, so the time-domain aliasing puts $x[3]$ on top of $x[0]$. A second slip is to take $y[0]$ as $Y[0]=9$, the sum of all the samples. $Y[0]$ is a frequency-domain value; $y[0]$ is the average of the three $Y[k]$.

    Full solution →
45. Q.55NAT · 2 marksOfficial answer: 24.5 to 25.5
46. Q.56NAT · 2 marksOfficial answer: 1.55 to 1.65

    Electrical and Electronic Measurements › Bridges

    With a series RL arm $Z_1$ and a parallel RL arm $Z_2$ made of the same R and L, $Z_1/Z_2=(R+j\omega L{)}^2/(j\omega LR)$ always has real part 2. The resistor ratio $R_1/R_2=2$ therefore holds at every frequency, and balance only needs the imaginary part to vanish, which gives $\omega L=R$. The bridge balances at only one frequency, so it can be used to measure frequency.

    WATCH OUT FOR

    Reading the top-right arm as a second series RL branch: then $Z_1=Z_2$, the ratio is 1 rather than 2, and the bridge can never balance. Other slips: pairing adjacent arms ($Z_1{R}_1=Z_2{R}_2$), which gives a ratio of 1/2 that the real part 2 can never match; stopping at $\omega =10^4\text{rad/s}$ without dividing by $2\pi$; and entering 1591.55 (in Hz) when the question asks for kHz.

    Full solution →
47. Q.57NAT · 2 marksOfficial answer: 4.00 to 4.40

    Electromagnetics › Electrostatics

    A current source fixes the charge, not the voltage. With charge fixed, the pull between the plates is $\frac{Q^2}{2{\epsilon }_{0}A}$ whatever the gap. The spring's pull-back $ky$ grows as the gap closes and is largest at contact, $k{d}_0$. So the gap closes when $\frac{Q^2}{2{\epsilon }_{0}A}=k{d}_0$.

    WATCH OUT FOR

    Using the voltage-drive result. At fixed voltage the force $\frac{{\epsilon }_{0}A{V}^2}{2x^2}$ grows as the gap shrinks, which gives the familiar pull-in at one third of the gap. Setting the balance at $y=d_0/3$ gives $Q\approx 2.43\times 10^{-14}\text{C}$, which is wrong here: at fixed charge the force is constant and there is no pull-in. A second trap is treating the charge as switched on suddenly. A constant force applied in one step makes the plate overshoot to twice its equilibrium displacement, which gives $Q=\sqrt{{\epsilon }_{0}Ak{d}_0}\approx 2.97\times 10^{-14}\text{C}$. A current source builds the charge up gradually, so there is no overshoot.

    Full solution →
48. Q.58NAT · 2 marksOfficial answer: 2160 to 2210
49. Q.59NAT · 2 marksOfficial answer: Marks to all

    Control Systems › Time response

    Rejecting a step disturbance needs the disturbance-to-output transfer function to have at least one zero at the origin. Here the loop's feedback $1/s^2$ puts $s^2$ in the numerator of $Y/D$, and the disturbance path's $1/s^n$ removes $n$ of those zeros. That leaves $s^{2-n}$, and a step (one more $1/s$) is rejected only if $2-n\ge 1$. A second rule matters just as much: the final value theorem only holds for a stable closed loop, so do a Routh check first. Here it fails, because $s^3+s^2+1$ has no $s$ term.

    WATCH OUT FOR

    (1) Using the final value theorem without checking stability. The characteristic polynomial $s^3+s^2+1$ has a missing coefficient, so the loop is unstable and the theorem's result is only formal. (2) Answering $n=2$ by asking only for a finite steady state ($2-n\ge 0$) rather than a zero one ($2-n>0$). (3) Reading the $-1/s^2$ block going into a $+$ junction as positive feedback. That gives $s^3+s^2-1$, which is also unstable, so it changes the algebra but not the formal answer.

    Full solution →
50. Q.60NAT · 2 marksOfficial answer: 31.00 to 31.50

    Networks (Electric Circuits) › Coupled circuits

    Reflect each resistance back toward the source by dividing by n^2 at every ideal 1:n step-up transformer. That turns the whole network into a single 10 Ω load on a 25 V rms source. Ideal transformers are lossless, so the power that reaches the reflected 500 Ω is exactly the power in the 50 kΩ resistor.

    WATCH OUT FOR

    Three mistakes are common. (1) Reading 25√2 V as the rms value, which doubles the answer to 62.50 W. The √2 means it is the peak value, so V_rms = 25 V. (2) Reflecting the wrong way, multiplying by n^2 instead of dividing, when moving an impedance from the high-turns side to the low-turns side. (3) Forgetting the 400 Ω resistor in the middle loop, which takes 400/900 of the power that gets past the 1 Ω resistor.

    Full solution →
51. Q.61NAT · 2 marksOfficial answer: 9

    Networks (Electric Circuits) › Theorems

    A slightly unbalanced Wheatstone bridge with a galvanometer of finite resistance is solved with Thevenin's theorem. Remove G. The open-circuit voltage between the two mid-nodes is the difference between the two voltage dividers. With the source shorted, the Thevenin resistance is (upper-left ∥ lower-left) + (upper-right ∥ lower-right). Then $I_G=V_{th}/(R_{th}+R_G)$.

    WATCH OUT FOR

    Using $I_G=V_{th}/R_G$ gives about 99.5 μA because it leaves out the bridge's own Thevenin resistance of about 1002.5 Ω. Another common error is to put the arm resistors in series (1000 + 1010) instead of in parallel when finding $R_{th}$: once the source is shorted, the top and bottom vertices are the same node. Rounding $V_A$ to 2.01 V too early gives $V_{th}$ = 10 mV and about 9.07 μA. That still rounds to 9 here, but on a tighter question the lost precision would matter.

    Full solution →
52. Q.62NAT · 2 marksOfficial answer: 4.5 to 5.5

    Electrical and Electronic Measurements › Errors & standards

    For a product or quotient of independent measured quantities, you find the uncertainty of the result by adding the relative uncertainties in quadrature (root-sum-square), each one weighted by its power. With $Q={\omega }_{0}L/R$ and ${\omega }_0$ given as fixed, both $L$ and $R$ enter with power $\pm 1$, so $\delta Q/Q=\sqrt{4^2+3^2}\%=5\%$.

    WATCH OUT FOR

    Adding the relative errors linearly, $4\%+3\%=7\%$. That is the worst-case (limiting) error. An 'uncertainty' in independent measurements is combined by root-sum-square, which gives $5\%$. Another slip is to treat the capacitor as the fixed element and use $Q=\frac{1}{R}\sqrt{L/C}$, so that $L$ gets weight $\frac{1}{2}$. The question gives ${\omega }_0$ as a fixed number, so $Q={\omega }_{0}L/R$ is the relation to use.

    Full solution →
53. Q.63NAT · 2 marksOfficial answer: 4.60 to 5.00

    Networks (Electric Circuits) › Transients

    When the switch opens, the ideal 10 mA source is forced to drive a single RL network. The 50 Ω and 100 Ω resistors in series with the source affect neither how the current splits nor the time constant. The current splits between 120 Ω and the path 40 Ω + L + 40 Ω: all of it goes through 120 Ω at $t=0^{+}$, because the inductor starts with zero current, and it settles to the resistive current-division value. The inductor sees $40+120+40=200\Omega$, so $\tau =L/R=1\text{ms}$.

    WATCH OUT FOR

    Counting the 50 Ω and 100 Ω resistors in the time constant (e.g. $R=40+40+120+50+100$), or shorting the current source instead of opening it when finding $R_{eq}$. Opening the source leaves those resistors on an open branch, so they drop out. Another slip is taking $i(0^{+})=0$ by treating the 120 Ω resistor's current as continuous. Only the inductor current is continuous; the resistor current jumps from 0 to 10 mA at $t=0^{+}$.

    Full solution →
54. Q.64NAT · 2 marksOfficial answer: 1.50 to 1.54

    Networks (Electric Circuits) › Transients

    While $S$ is closed, the ideal source holds the node at $V_M(t)$. When $S$ opens at $t=0.5$ s, the node starts at $5\sin (\pi /6)=2.5\text{V}$ and discharges through the ADC's $10\text{M}\Omega$. The ADC's own $0.1\mu \text{F}$ is in parallel with $C$, so $\tau =R(C+C_{ADC})=10\text{M}\Omega \times 0.2\mu \text{F}=2\text{s}$.

    WATCH OUT FOR

    The most common slip is leaving out the ADC's input capacitance. That gives $\tau =1\text{s}$ and $V_x=2.5e^{-1}\approx 0.92\text{V}$. Other slips:<br>(a) starting the decay from the peak value 5 V instead of the instantaneous value $5\sin (\pi /6)=2.5\text{V}$;<br>(b) evaluating $\sin (0.5\pi /3)$ as if the angle were in degrees;<br>(c) measuring the decay from $t=0$ instead of from the switching instant, which gives $2.5e^{-1.5/2}\approx 1.18\text{V}$.

    Full solution →
55. Q.65NAT · 2 marksOfficial answer: 590 to 610

    Networks (Electric Circuits) › AC, power & resonance

    Reduce the network to one impedance seen by the source, $Z=15-j5\Omega$. The active power is $\text{Re}(V{I}^{\ast })$, which equals $|I{|}^2\text{Re}(Z)$. Only resistors absorb active power, so adding $|I{|}^{2}R$ over the two resistors checks the answer independently.

    WATCH OUT FOR

    One slip is to report the apparent power $|V||I|=100\sqrt{40}\approx 632.5\text{VA}$, or to use $VI$ without conjugating the current. Another is to treat $100\angle 0^{\circ }$ as a peak value and halve the result to $300\text{W}$. GATE phasors are rms unless the question says otherwise. A third is to use the source current for the $5\Omega$ resistor. That resistor carries only its branch current, $|I_5{|}^2=40$, which here happens to have the same magnitude as the source current.

    Full solution →