GATE IN 2022 question paper with solutions

GATE IN 2022 answer key

General Aptitude Q.1 to Q.10

1. Q.1MCQ · 1 markOfficial answer: C

   General Aptitude › Grammar & vocabulary

   Work out which part of speech each blank needs before you look at spellings. A noun follows the adjective "burning", and a base-form verb follows the modal "could". tyre is the Indian/British spelling of the wheel's rubber ring. tire is the verb meaning to exhaust. tier (pronounced like "tear" in "tears in her eyes") means a level or row.

   WATCH OUT FOR

   Picking (B) because tire is also the American spelling of the noun, without checking whether tyre can work as a verb after "could". It cannot. Another mistake is to mix up tier with tire because the spellings look alike. tier means a level, and it never means to exhaust someone.

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

   General Aptitude › Geometry

   A sphere inscribed in a cube touches all six faces, so the cube's side equals the sphere's diameter ($2r$), not its radius. Cube that side to get the volume.

   WATCH OUT FOR

   Taking the cube's side as the radius $r$ and choosing $r^3$ (B). Another slip is writing $2r^3$ (C) instead of $(2r{)}^3=8r^3$, which cubes only the $r$ and not the factor 2.

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

   General Aptitude › Numerical

   Turn each pipe into a rate in litres per minute. A pipe that fills a tank of $C$ litres in $t$ minutes pours in $\frac{C}{t}$ litres per minute. Rates add, drains count as negative, and the time to empty is capacity divided by the net outflow. Here that gives one linear equation in $C$.

   WATCH OUT FOR

   Setting the inflow equal to the outflow, $\frac{4C}{15}=34$, gives $C=127.5$, which is option (D). At that capacity the tank's level never changes, so it would never empty. The one-hour emptying time needs a net outflow of $\frac{C}{60}$ litres per minute, so the equation is $34-\frac{4C}{15}=\frac{C}{60}$. A second slip is adding the fill times ($10+6$) instead of the rates ($\frac{1}{10}+\frac{1}{6}$).

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

   General Aptitude › Reasoning

   Fix P in one seat. With six equally spaced seats, Q opposite P is three seats away. S and T take both of P's neighbouring seats, so the only seats left for R and U are the two next to Q. The distance statement only confirms the arrangement is possible. It is not needed to get the answer.

   WATCH OUT FOR

   Trying to use the S–R / T–U distance condition to pin down one exact arrangement, and then worrying that it allows two. Both arrangements give Q the same neighbours. Another slip is counting seats unevenly, so that Q is not placed exactly three seats from P.

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

   General Aptitude › Spatial

   Treat each enclosed space as a node and each door as a link between two spaces. A wall that stops short of another wall leaves an open gap, so the spaces on both sides of it are one room. The answer is the fewest links on any path from P's room to Q's room.

   WATCH OUT FOR

   Missing the gaps at the ends of the two short horizontal walls. If you miss the gap beside Q, the area around Q looks reachable only through the central room, and the route door 1, left wall of the central room, right wall of the central room gives 3, option (B). Answering 1, option (D), is also wrong: the bottom strip's only door opens into the left room, not into Q's room.

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

   General Aptitude › Reading

   A logical inference has to follow from the passage and go no further. GSR is still being developed and is meant to withstand the extreme weather that climate change brings, so the only safe conclusion is the hedged one: GSR may be used in future against climate change's adverse effects.

   WATCH OUT FOR

   Choosing (A) because GSR is linked with extreme weather. The passage says GSR is resilient under extreme weather, not that it grows only there. Look for absolute words like "only", "always" and "will continue" that the passage never supports. Also check (C) and (D) against what the passage states outright: "higher yields" and "a threat to sustained availability".

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

   General Aptitude › Statistics & probability

   Two independent spins of an 8-sector disk give $8\times 8=64$ equally likely ordered outcomes. Count the ordered pairs $(a,b)$ with $a+b=8$ and divide by 64. For a target sum $s$ with $2\le s\le 9$, there are $s-1$ such pairs, so here there are 7.

   WATCH OUT FOR

   Counting unordered pairs, so that $\{1,7\},\{2,6\},\{3,5\},\{4,4\}$ gives 4 outcomes, then dividing by 64 or by 36. The 64 outcomes are ordered pairs, so the favourable outcomes must be ordered too. A second slip is to include $(8,0)$ or $(0,8)$, but 0 is not a sector on the disk.

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

   General Aptitude › Numerical

   Inequalities that point the same way can be added. Adding (i) and (ii) cancels part of each variable: $(3p-p)+(3q-q)=2(p+q)$, which leaves $2(p+q)<16$.

   WATCH OUT FOR

   Trying to solve for $p$ and $q$ separately, as if the inequalities were equations, or subtracting one inequality from the other. Subtracting two inequalities that point the same way is not valid. Another mistake is to pick (C) or (D) by adding the right-hand sides $4+12=16$ and forgetting to divide by 2.

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

   General Aptitude › Reasoning

   Use the conversion rules. An I-type (\"Some X are Y\") and an E-type (\"No X is Y\") can both be reversed, so I comes from Statement 1 and III from Statement 2. An A-type (\"All X are Y\") cannot be reversed, so II fails. II is actually false: the engineers who are writers cannot be actors. IV contradicts III, which is true, so IV is false.

   WATCH OUT FOR

   Choosing (D) because III and IV look like a complementary pair. The \"either-or\" answer applies only when neither conclusion of the pair follows by itself. Here III follows directly from Statement 2, so you name it on its own, not as half of an either-or. A second trap is reversing \"All actors are engineers\" into \"All engineers are actors\" (Conclusion II). An A-type statement never converts to another A-type.

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

    General Aptitude › Spatial

    First use each piece's right-angled corners to decide where it goes. Then look only at the edges that surround the gap left inside the square. The hole is round only if every one of those edges is an arc of the same circle and the pieces fit with no overlap.

    WATCH OUT FOR

    Choosing (A) because it has the two biggest arcs, or (B) because its arcs look as if they could close up. In (A), once the pieces fit, the two arcs end up in separate places and each gap has straight sides. In (B) the right-hand piece has no V-cut, so it would overlap the large piece's peak. (D) is the version with no arcs, and its centre gap is a triangle.

    The question and its official answer →

Instrumentation Engineering Q.11 to Q.65

1. Q.11MCQ · 1 markOfficial answer: B

   Control Systems › Block diagrams & SFG

   A delay of $T$ seconds in time becomes a multiplying factor $e^{-sT}$ in the $s$-domain, because $\mathcal{L}\{x(t-T)u(t-T)\}=e^{-sT}X(s)$. The delay changes the exponent only. The gain $3$ and the pole at $s=-1$ come from the rest of the differential equation, so the answer is a gain of 3, a first-order lag $\frac{1}{s+1}$, and a delay factor $e^{-3s}$ multiplied together.

   WATCH OUT FOR

   Two traps match the wrong options. (1) Treating the time shift like time scaling and writing $e^{-s/3}$. That factor would come from $x(t/3)$-type manipulations, not from a shift, and it gives options (C) and (D). (2) Letting the input coefficient $3$ move into the denominator and writing $s+3$. The $3$ multiplies the input, so it sets the gain. The coefficient of $y(t)$ is $1$, so the pole stays at $s=-1$. That mistake gives options (A) and (D).

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

   Control Systems › Process control (IN)

   A nozzle-flapper is a pneumatic displacement-to-pressure transducer and amplifier. The fixed orifice and the variable nozzle gap form a pressure divider, so the back pressure is a steady-state function of the gap, $P_b=f(x)$. A travel of hundredths of a millimetre sweeps the full output pressure range, which gives a very large gain $K=-d{P}_b/dx$.

   WATCH OUT FOR

   Picking (C) by treating the flapper as a rate-sensitive element. The back pressure depends on the gap itself, not on how fast the gap changes: hold the flapper still and $P_b$ stays at a fixed value. Picking (A) or (B) is a separate slip. The medium is compressed air and the output is pressure, so temperature plays no part.

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

   Engineering Mathematics › Calculus

   Test for parity before you integrate anything. If $f(-x)=-f(x)$ over one period, then $a_0$ and every $a_n$ are zero and only sine terms are left. Here $f(-x)=-1+x=-(1-x)=-f(x)$, so the function is odd. $b_n=\frac{2}{n\pi }$ is non-zero for every $n$, so the series has a sine term at every harmonic, not only at the even or odd ones.

   WATCH OUT FOR

   Judging the function by its graph. The graph jumps at $x=0$ and each piece has a constant offset, so it can look as though it has a DC term or even symmetry, and students pick (A). The two constant offsets are $-1$ and $+1$, which is exactly what makes $f$ odd. A second trap is (D): the $2n\pi$ harmonics would appear only if the period were 1, but $f(x+1)\ne f(x)$ here (on $-1<x<0$, $f(x+1)=-x$ while $f(x)=-1-x$).

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

   Signals and Systems › LTI & convolution

   If $y(t)=g(t)x(t)$ with a fixed function $g(t)$, the system is always linear, because multiplying by any fixed function distributes over sums and scalings. It is time-invariant only when $g(t)$ is a constant. Any explicit $t$ outside the argument of $x$ (here $\sin (2\pi t)$) makes the system time-variant.

   WATCH OUT FOR

   There are two common mistakes. The first is to shift the $\sin (2\pi t)$ factor along with the input when checking time invariance. Only $x$ is delayed. The multiplier belongs to the system and stays at $\sin (2\pi t)$. The second is to test with an integer shift such as $t_0=1$. Because $\sin (2\pi (t-1))=\sin (2\pi t)$, the outputs agree for that shift, and the student wrongly concludes the system is time-invariant. Time invariance must hold for every shift, so one failing shift like $t_0=0.25$ settles the question. Seeing $\sin$ and calling the system non-linear is also wrong: the sine acts on $t$, not on $x$.

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

   Control Systems › Routh–Hurwitz

   Stability depends on the closed-loop poles, which are the roots of $1+L(s)=0$, not the open-loop poles. Here they are $s=2$ and $s=3$, so the system is unstable. Causality is a separate property. A feedback system built from causal physical parts is always causal, and its right-sided ROC ($\text{Re}(s)>3$) leaves out the $j\omega$ axis, which is why it is unstable.

   WATCH OUT FOR

   Treating instability as if it meant non-causality and choosing (D). Another mistake is judging stability from the open-loop pole at $s=5$ alone, or thinking that feedback with positive gain $6$ must stabilise it. The closed-loop polynomial $s^2-5s+6$ has mixed-sign coefficients, so the closed loop is still unstable.

   Full solution →
6. Q.16MCQ · 1 markOfficial answer: A
7. Q.17MCQ · 1 markOfficial answer: D
8. Q.18MCQ · 1 markOfficial answer: C

   Signals and Systems › Fourier transform

   Time scaling and the frequency axis move in opposite directions: $x(at)\leftrightarrow \frac{1}{|a|}X\left(\frac{f}{a}\right)$, so compressing by 2 doubles every band edge. A time shift only multiplies the spectrum by $e^{-j2\pi f{t}_0}$, whose magnitude is 1, so it never changes the band.

   WATCH OUT FOR

   Reading $X(f/2)$ as halving the frequencies and choosing (A), 50 to 100 Hz. $X(f/2)$ at $f=400$ is $X(200)$, so the band moves UP to 200 to 400 Hz. The other trap is thinking the shift by 5 changes the spectrum's support, or choosing (B) because the shift is mistaken for the only operation.

   Full solution →
9. Q.19MCQ · 1 markOfficial answer: D
10. Q.20MCQ · 1 markOfficial answer: C

    Electrical and Electronic Measurements › Bridges

    Work out the size of the resistance before picking a method. $R=\rho L/A\approx 2.2\text{m}\Omega$ is a low resistance (below $1\Omega$). At that size, lead and contact resistances are as large as the quantity being measured, so the method needs separate current and potential terminals: the four-terminal (Kelvin) connection.

    WATCH OUT FOR

    Picking the three wire method because it "compensates lead resistance". That scheme is meant for RTD bridges, depends on matched leads and does nothing about contact resistance at the specimen. Another mistake is skipping the estimate and treating this as an ordinary medium resistance, for which two-wire measurement would be good enough.

    Full solution →
11. Q.21MCQ · 1 markOfficial answer: D

    Digital Electronics › Boolean logic

    A NOR gate fed with complemented inputs acts as an AND gate: $\overline{\overline{P}+\overline{Q}}=PQ$. So the two first-level NOR gates produce the product terms $AB$ and $\overline{A}B$. The output NOR gives the complement of their sum. Since $AB+\overline{A}B=B$, the output does not depend on $A$ and $F=\overline{B}$.

    WATCH OUT FOR

    Treating a crossing wire as a connection. The vertical wires from the $\overline{B}$ rail cross the $B$, $\overline{A}$ and $A$ rails with no dot, so they connect only to $\overline{B}$. If you read the right gate's inputs as $A$ and $B$, you get $F=A\oplus B$, which is not one of the options. The other slip is dropping the output bubble on the last gate. That gives $F=B$ instead of $\overline{B}$.

    Full solution →
12. Q.22MSQ · 1 markOfficial answer: A, C, D

    Electrical and Electronic Measurements › Bridges

    Write the balance condition $Z_1{Z}_4=Z_2{Z}_3$ and split it into real and imaginary parts. If $\omega$ cancels from both conditions (Maxwell, Schering) or never appears (Wheatstone), the balance is frequency-independent. If one condition fixes $\omega$ (Wien), the bridge balances at only one frequency.

    WATCH OUT FOR

    Marking the Schering bridge as frequency-dependent because its dissipation factor $D=\omega C_4{R}_4$ contains $\omega$. $D$ is calculated after balance, while the balance conditions themselves, $r_x=R_3{C}_4/C_2$ and $C_x=C_2{R}_4/R_3$, do not contain $\omega$. A related slip is to treat the Maxwell bridge like Hay's bridge. Hay's bridge puts $R$ and $C$ in series in arm 4, and that makes its $L_x$ depend on $\omega$. The Maxwell bridge uses $R_4\parallel C_4$, and $\omega$ cancels.

    Full solution →
13. Q.23MSQ · 1 markOfficial answer: A, B

    Digital Electronics › Combinational

    Work each 2×1 MUX with $Y=\overline{S}{I}_0+S{I}_1$. The first MUX, with inputs 1 and 0 and select B, is just an inverter: $Y_1=\overline{B}$. The second gives $F=\overline{A}+A\overline{B}=\overline{A}+\overline{B}$, which is NAND. This is an MSQ, so both equivalent forms of NAND (the De Morgan pair) must be marked.

    WATCH OUT FOR

    Marking only one of (A) and (B) and forgetting that De Morgan makes them the same function, or mixing up $\overline{A\cdot B}$ (NAND) with $\overline{A}\cdot \overline{B}$ (NOR) and picking (C) or (D). Another slip is wiring the first MUX output into $I_0$ of the second MUX instead of $I_1$, which gives a different function.

    Full solution →
14. Q.24MSQ · 1 markOfficial answer: A, D

    Engineering Mathematics › Linear algebra

    For a square matrix, a zero determinant, rank below full, and linearly dependent rows all say the same thing. Once you spot $R_3=2R_1$, (D) is true and (B) and (C) are false together. To fix the rank exactly you still need one non-zero $2\times 2$ minor.

    WATCH OUT FOR

    Ticking (D) and stopping there, or reasoning that because $\det M=0$ the rank must be 1. A zero determinant only shows the rank is below 3. You need a non-zero $2\times 2$ minor (here $-10$) to show the rank is exactly 2, which makes (A) correct as well.

    Full solution →
15. Q.25NAT · 1 markOfficial answer: 10

    Digital Electronics › ADC & DAC

    The resolution of an $n$-bit ADC is its full-scale range divided into about $2^n$ equal steps, so you need the smallest $n$ with $\frac{10\text{ V}}{2^n}\le 0.01\text{ V}$, which means $2^n\ge 1000$. Here, with $2^9=512<1000<1024=2^{10}$, it does not matter whether you divide by $2^n$ or $2^n-1$.

    WATCH OUT FOR

    Taking $n={\log }_{2}1000\approx 9.97$ and rounding it down to 9 bits. With 9 bits the step is about $0.0195\text{ V}$, nearly twice the required $0.01\text{ V}$. The number of bits has to be rounded up. A second slip is to divide 10 V by $0.01\text{ V}$, get 1000 levels, and write 1000 as the answer. The question asks for the number of bits, not the number of levels.

    Full solution →
16. Q.26NAT · 1 markOfficial answer: 0.42 to 0.44
17. Q.27NAT · 1 markOfficial answer: 870.0 to 880.0
18. Q.28NAT · 1 markOfficial answer: -3.0

    Engineering Mathematics › Calculus

    Remove the modulus by splitting at $x=0$. The function is non-negative for $x\ge 0$, so the minimum must be on $x<0$, where $f(x)=x^3{e}^x$. There $f^{\prime }(x)=x^2{e}^x(x+3)$ changes sign from negative to positive only at $x=-3$, so $f$ decreases down to $x=-3$ and rises after it. That monotonicity makes $x=-3$ the global minimum, not just a local one.

    WATCH OUT FOR

    Writing $f(x)=x^3{e}^{-x}$ on the negative side (forgetting that $|x|=-x$ there). That gives $f^{\prime }(x)=x^2{e}^{-x}(3-x)$ and the answer $x=3$, which is actually where the global maximum occurs. Another slip is to report $x=0$ from $f^{\prime }(x)=0$: $x^2=0$ there, but $f^{\prime }$ does not change sign at $0$, so $x=0$ is neither a maximum nor a minimum.

    Full solution →
19. Q.29NAT · 1 markOfficial answer: 1440.0

    Electrical Machines › Induction machines

    Rotor (mechanical) speed of an induction motor is related to synchronous speed and slip by $N=N_s(1-s)$, where $N_s=\frac{120f}{P}$ depends only on supply frequency and number of poles — the voltage, power rating and winding connection given in the stem are not needed for this calculation.

    WATCH OUT FOR

    Forgetting to convert slip into the (1-s) factor correctly, or mistakenly using the number of pole pairs instead of poles in the synchronous speed formula (using P=2 instead of P=4).

    Full solution →
20. Q.30NAT · 1 markOfficial answer: 3

    Control Systems › State space

    The minimum number of state variables is the order of the transfer function after all common pole-zero factors are cancelled. Here the zeros ($-1,-3$) and the poles ($-5,-7,-9$) share no factor, so the minimal order is the denominator degree, $3$.

    WATCH OUT FOR

    Answering 2 by counting zeros or taking the numerator degree, or answering 5 by adding the poles and the zeros. Another mistake is to cancel factors without checking that a zero and a pole actually coincide. Only an exact pole-zero cancellation lowers the minimal order.

    Full solution →
21. Q.31NAT · 1 markOfficial answer: 6.0

    Analog Electronics › Diode circuits

    In a shunt Zener regulator, the Zener fixes the output voltage, so the current through the series resistor is fixed at $(V_S-V_Z)/R_S$. The load and the Zener share that current. The load can take everything except the minimum current the Zener needs to stay in breakdown: $I_{L,\max }=I_S-I_{Z,\min }$.

    WATCH OUT FOR

    Computing $I_S=16/1200\approx 13.3\text{mA}$ from the full source voltage instead of the 12 V actually across the resistor. Another error is adding $I_{Z,\min }$ to the source current instead of subtracting it (14 mA). A third is answering the source current (10 mA) without taking out the Zener's minimum 4 mA.

    Full solution →
22. Q.32NAT · 1 markOfficial answer: 0.32 to 0.34

    Engineering Mathematics › Probability & statistics

    The first ball's colour is unknown, so condition on it with the total probability theorem. There is also a shortcut: without replacement, every draw position is symmetric, so the chance that draws 2 and 3 are red equals the chance that draws 1 and 2 are red, which is $\frac{6}{10}\cdot \frac{5}{9}=\frac{1}{3}$.

    WATCH OUT FOR

    Taking the first ball to be blue, or to be red, and computing only that case. This gives $\frac{6}{9}\cdot \frac{5}{8}\approx 0.42$ or $\frac{5}{9}\cdot \frac{4}{8}\approx 0.28$. Another mistake is computing the probability of the full sequence (a colour for draw 1, then red, red) and forgetting to add the two cases.

    Full solution →
23. Q.33NAT · 1 markOfficial answer: 6.22 to 6.42

    Networks (Electric Circuits) › AC, power & resonance

    With the output across $R_0$, the circuit is a series RLC with voltage transfer $R/Z$. That gives a bandpass response centred at ${\omega }_0=1/\sqrt{LC}$, with half-power bandwidth $B=R/L$, so $Q={\omega }_0/B=\frac{1}{R}\sqrt{\frac{L}{C}}$.

    WATCH OUT FOR

    Using the parallel-RLC formula $Q=R\sqrt{C/L}$ gives $50\sqrt{10^{-8}/10^{-3}}\approx 0.16$. The elements here are in series, so $R$ goes in the denominator. Another slip is mixing units: writing $C_0=10\text{nF}$ as $10^{-9}$ instead of $10^{-8}\text{F}$ gives $Q=20$.

    Full solution →
24. Q.34NAT · 1 markOfficial answer: 0.99 to 1.01

    Engineering Mathematics › Numerical methods (IN)

    One Newton-Raphson step replaces $f$ by its tangent line at $x_0$ and takes that line's zero: $x_1=x_0-f(x_0)/f^{\prime }(x_0)$. At $x_0=0$ we have $f(0)=-1$ and $f^{\prime }(0)=1$, so $x_1=1$.

    WATCH OUT FOR

    Answering with the converged root ($\approx 0.74$) when the question asks for only the first iterate $x_1$. Another slip is getting the sign of the update wrong ($x_0+f/f^{\prime }$), which gives $-1$. Note that the sign of $\frac{d}{dx}\cos x$ makes no difference here because $\sin 0=0$, but it does matter at every later step.

    Full solution →
25. Q.35NAT · 1 markOfficial answer: 9.9 to 10.0

    Analog Electronics › Op-amps

    When negative feedback samples the output voltage (shunt sampling, as in the series-shunt voltage buffer), it divides the open-loop output resistance by $(1+A\beta )$. A unity-gain buffer has $\beta =1$, so $R_{of}=R_o/(1+A)$. The input resistance $R_i$ ends up in parallel with this and has no practical effect.

    WATCH OUT FOR

    One mistake is multiplying $R_o$ by $(1+A\beta )$ instead of dividing, which gives about $1\text{M}\Omega$. That factor applies to the input impedance under series mixing, $R_{if}=R_i(1+A)\approx 10^{11}\Omega$. Another is entering the answer in ohms (0.01) when the question asks for milliohms (10.0). A third is claiming that $R_i$ carries no current in the output-impedance test. With the source zeroed, $R_i$ connects the output to ground and does carry current; it is left out only because it is negligibly large.

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

    Analog Electronics › Diode circuits

    A series ideal diode feeding a capacitor with no load is a peak detector. Its output is the largest of the initial capacitor voltage and every input value applied so far, $V_{out}(t)=\max \left(V_C(0),{\max }_{\tau \le t}{V}_{in}(\tau )\right)$. It is not the current input value. With no resistance in the loop, the size of $C$ does not matter.

    WATCH OUT FOR

    Choosing 0.6 V (B) by reading the input at t = 5 ms. The 0.6 V peak falls exactly at 5 ms, but the diode is reverse biased then because the capacitor is already at 1.0 V, and with an open-circuit output nothing discharges it. Other traps: keeping the initial 0.3 V (A) as if the input never exceeded it, and getting 0.7 V (C) by subtracting a silicon cut-in voltage or the initial 0.3 V from 1.0 V. The diode is ideal, so there is no drop, and the initial voltage only matters while the input stays below it.

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

    Control Systems › Block diagrams & SFG

    $G_2$ and $G_3$ are two parallel forward branches from the same node to the same node, and the $-1$ branch feeds back around both of them. That makes two forward paths and two loops that touch each other and both paths. Mason's formula, or block reduction (parallel, then unity negative feedback), gives $\frac{2G_1(G_2+G_3)}{1+G_2+G_3}$.

    WATCH OUT FOR

    Reading the $-1$ arc as a third forward branch from $A$ to $B$. It looks symmetric with the $G_3$ arc, but its arrowhead points into the left node, so it is feedback. Treating it as forward removes both loops and gives the wrong result $2G_1(G_2+G_3-1)$. A second slip is to count only the $G_2$ loop and forget that $G_3$ also closes a loop through the $-1$ branch, which gives a denominator of $1+G_2$.

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

    Signals and Systems › FIR & IIR filters

    Replace $s$ by $\frac{2}{T}\frac{1-z^{-1}}{1+z^{-1}}$ (here $\frac{2}{T}=20$) and write each factor $(s+a)$ over the common denominator $1+z^{-1}$, which gives $\frac{(20+a)-(20-a)z^{-1}}{1+z^{-1}}$. Each first-order pole factor puts one $(1+z^{-1})$ in the numerator. You can check the result two quick ways: the DC gain must survive ($z=1$ matches $s=0$), and a stable pole $s=p$ must land inside the unit circle at $z=\frac{20+p}{20-p}$.

    WATCH OUT FOR

    Writing the factor as $(20-a)-(20+a)z^{-1}$, i.e. swapping the two coefficients. That gives $19-21z^{-1}$ or $17-23z^{-1}$, which put a pole outside the unit circle and flip the sign of the DC gain (the traps in options A and D). Another mistake is using $s=\frac{1}{T}\frac{1-z^{-1}}{1+z^{-1}}$ without the factor 2, or putting $(1-z^{-1})$ in the numerator (option B).

    Full solution →
29. Q.39MCQ · 2 marksOfficial answer: A
30. Q.40MCQ · 2 marksOfficial answer: B

    Signals and Systems › Laplace

    The signal is $t^{2}u(t)$ shifted right by 1 s, with the squared term and the step shifted together. So you can use the time-shift property: $X(s)=e^{-s}\cdot \frac{2}{s^3}$. Then check that $s=1$ is inside the ROC $\text{Re}(s)>0$ before you substitute.

    WATCH OUT FOR

    Two mistakes are common. (1) Using $\mathcal{L}\{t^{2}u(t)\}=1/s^3$ and dropping the $2!$, which gives $X(1)=1/e$, option (A). (2) Getting the sign of the shift exponent wrong ($e^{+s}$ in place of $e^{-s}$), which gives $2e$, option (C). The time-shift property applies directly only when the step is delayed by the same amount as the rest of the expression, as it is here.

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

    Control Systems › Time response

    With a PID controller the loop is type 1, and its velocity error constant is $K_v={\lim }_{s\to 0}sC(s)P(s)=K_i\cdot P(0)$. At $s=0$ the proportional and derivative terms drop out, so the steady-state ramp error $A/K_v$ is inversely proportional to $K_i$ alone.

    WATCH OUT FOR

    Thinking every gain increase reduces the error. Some students multiply the factors (2 x 3 = 6, not an option) or add them (2 + 3 = 5, which gives option D). Others take $K_p$ as the gain that sets the ramp error and pick option (B). In a PID loop, $K_p$ and $K_d$ vanish from ${\lim }_{s\to 0}sG(s)$, so only the factor of 3 on $K_i$ matters.

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

    Networks (Electric Circuits) › Theorems

    Reduce an R-2R ladder from the far end one node at a time. At every node you get two $2R$ branches in parallel, so the resistance collapses back to $R$ and the Thevenin voltage becomes the average of the two branch voltages (Millman's theorem). Each stage therefore halves whatever came before it, and that gives $V_{th}=V_{R}N/2^n$ with $R_{th}=R$ for any input code.

    WATCH OUT FOR

    One common slip is to read the switch blades the wrong way round, or to treat the bit next to the output as the LSB. A 4 V leg at node C would give 2 V, and one at node A would give 0.5 V, which is the trap in options (A) and (B). Another is to take $R_{th}$ as the $2\text{k}\Omega$ leg resistance (options B and D) and forget that the rest of the ladder, which also works out to $2\text{k}\Omega$, sits in parallel with it.

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

    Control Systems › Bode & margins

    Each margin is read at a different crossing of the Nyquist plot. Phase margin: at the unit-circle crossing, where |G| = 1, PM = 180° + ∠G. Gain margin: at the negative-real-axis crossing, where ∠G = −180°, GM = 1/|G|. Here PM = 180° − 140.3° = 39.7° (taken at P) and GM = 1/0.21 = 4.76 (taken at R). The sign of the phase at P comes from following the curve continuously from 0° at ω = 0.

    WATCH OUT FOR

    1. Taking GM = 1/|Q| = 1/14.40 ≈ 0.07. Q is the ω = 0 point on the positive real axis (phase 0°), not the −180° crossing. 2. Writing PM = −39.7° because P lies below the axis, or computing P's angle as +219.7° instead of following the curve from 0° to −140.3°. A gain-crossover point below the negative real axis means a positive PM. 3. Measuring positions off the figure: it is not to scale, and R is drawn near the circle's edge even though its value is −0.21.

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

    Analog Electronics › Amplifiers

    Looking into the source of a common-gate stage with $r_o\to \infty$ and no body effect (body tied to source), you see $1/g_m$, whatever resistor sits in the drain. That $1/g_m$ is in parallel with $R_S$, because both go from P to AC ground.

    WATCH OUT FOR

    Putting $R_L$ in parallel (option A) or in series with $R_S$ (option C). With $r_o$ infinite, $R_L$ is in series with an ideal current source and has no effect on the source-side impedance. $R_L$ would only matter through $r_o$: the general result is $\frac{r_o+R_L}{1+g_m{r}_o}$, which tends to $1/g_m$ as $r_o\to \infty$.

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

    Engineering Mathematics › Differential equations

    The equation is variables separable. Substituting $u=\ln y$ turns $\frac{dy}{y\ln y}$ into $\frac{du}{u}$, which gives $\ln y=A{e}^{-x}$: the logarithm of the solution decays exponentially. The initial condition $\ln y(0)=1$ fixes $A=1$, so $\ln y(1)=1/e$.

    WATCH OUT FOR

    Stopping at $\ln y=e^{-x}$ and reporting $y(1)=e^{-1}$, or mishandling the sign and getting $\ln y=e^x$ (which gives $e^e$, option A). A third slip is treating $\int \frac{dy}{y\ln y}$ as $\ln y$ rather than $\ln |\ln y|$, which leads to the wrong exponent.

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

    Digital Electronics › Sequential

    This is a 3-stage shift register with feedback $D_0=\overline{Q_1+Q_2}=\overline{Q_1}\overline{Q_2}$. A plain Johnson (twisted-ring) counter uses $D_0=\overline{Q_2}$ and has $2n=6$ states. Adding $Q_1$ to the NOR makes $110$ go to $011$ instead of $111$, so one state is skipped and the counter has $2n-1=5$ states. To find the modulus of any feedback shift register, trace the state sequence from $000$, then check every unused state for a separate loop.

    WATCH OUT FOR

    Treating the circuit as an ordinary Johnson counter (mod 6), or as a 3-bit binary counter (mod 8), without reading the second NOR input. Another mistake is to pick (D) after finding only the 5-state main cycle, without tracing the unused states $010$, $101$ and $111$. All three lead into the main cycle, so there is no disjoint loop.

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

    Analog Electronics › Amplifiers

    The gate draws no current, so the two lower $R_0$ resistors form a divider and $v_{gs}=v_o/2$. The MOSFET therefore loads the output like a resistance $2/g_m$ to ground, in parallel with $2R_0$. The supply fluctuation reaches the output through a divider of $R_0$ against $2R_0\Vert (2/g_m)$.

    WATCH OUT FOR

    There are two common mistakes. The first is to take $v_{gs}=v_o$, forgetting the gate divider, and also leave out the current in the $2R_0$ feedback path. KCL then reduces to $(v_d-v_o)/R_0=g_m{v}_o$, which gives $(1+g_m{R}_0{)}^{-1}$, the trap option (B). The second is to treat the circuit as a common-source amplifier driven at the gate, which gives an inverting gain like (A) or (C). Here the signal enters at the supply, not the gate, and the gain is positive and below 1.

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

    Digital Electronics › Boolean logic

    A comparator reads MSB first. $A>B$ when $a_1{\overline{b}}_1=1$, or when the MSBs are equal and $a_0{\overline{b}}_0=1$. On the K-map, the quad $a_1{\overline{b}}_1$ covers four of the six 1-cells. The other two, $(3,2)$ and $(1,0)$, each need their own essential prime implicant: $a_1{a}_0{\overline{b}}_0$ and $a_0{\overline{b}}_1{\overline{b}}_0$.

    WATCH OUT FOR

    Two mistakes are common. One is stopping at (A): its terms handle $a_1>b_1$ and $A=3$, but not $A=1,B=0$, which needs $a_0{\overline{b}}_1{\overline{b}}_0$. The other is choosing (D): writing $b_1$ instead of ${\overline{b}}_1$ in the third term makes $F=1$ at $A=1,B=2$, which is wrong because $1<2$.

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

    Engineering Mathematics › Linear algebra

    A nonzero vector $x$ is an eigenvector of $A$ exactly when $Ax$ is a scalar multiple of $x$. For a $2\times 2$ matrix with known eigenvalues, one row of $(A-\lambda I)x=0$ gives the direction for each eigenvalue: here $x_2=x_1$ for $\lambda =7$ and $x_2=-3x_1$ for $\lambda =-5$. Any nonzero scalar multiple of an eigenvector is also an eigenvector, so $\left[\begin{array}{c}2\\ -6\end{array}\right]$ counts even though it is not the simplest form $\left[\begin{array}{c}1\\ -3\end{array}\right]$.

    WATCH OUT FOR

    Rejecting (C) because it is not written in its simplest form $\left[\begin{array}{c}1\\ -3\end{array}\right]$. Scaling an eigenvector by a nonzero number gives another eigenvector. The other common slip is a sign error in $(A+5I)$, which gives $x_2=3x_1$ instead of $x_2=-3x_1$ and can steer you towards a vector like (B).

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

    Engineering Mathematics › Complex analysis

    $a-jb$ is the conjugate of $a+jb$. The two have the same modulus and opposite angles. Dividing a complex number by its conjugate therefore gives a number of magnitude $1$ whose angle is twice the original angle: $\frac{r{e}^{j\theta }}{r{e}^{-j\theta }}=e^{j2\theta }$.

    WATCH OUT FOR

    Doubling inside the inverse tangent, writing ${\tan }^{-1}\frac{2b}{a}$ instead of $2{\tan }^{-1}\frac{b}{a}$ (option B). The inverse tangent is not linear, and $\tan 2\theta =\frac{2\tan \theta }{1-{\tan }^2\theta }$, not $2\tan \theta$. A second trap: rationalising to $z=\frac{(a^2-b^2)+j2ab}{a^2+b^2}$ and then taking the phase as ${\tan }^{-1}\frac{2ab}{a^2-b^2}$ gives the wrong quadrant when $b>a$, because the real part is then negative. A third: taking $|a-jb|=\sqrt{a^2-b^2}$ (option D). The minus sign is inside the complex number, and the modulus is still $\sqrt{a^2+b^2}$.

    Full solution →
41. Q.51NAT · 2 marksOfficial answer: 0.75 to 0.85
42. Q.52NAT · 2 marksOfficial answer: 0.8

    Networks (Electric Circuits) › AC, power & resonance

    Power factor is $\cos$ of the load impedance angle, i.e. $\text{Re}(Z)/|Z|$. The unusual values ($C_1=1000/\pi \mu \text{F}$, $L_1=20/\pi \text{mH}$) are chosen so the $\pi$ cancels against $\omega =100\pi$. That gives clean reactances ($X_C=10\Omega$, $X_L=2\Omega$). Reduce $R_1\parallel C_1$ to one series impedance, then add the series elements.

    WATCH OUT FOR

    Adding $R_1$ and $X_C$ as if they were in series, which gives $Z=24-j8$ and a power factor of about $0.95$. The same error happens when $\text{Re}(Z)$ is taken as $R_1+R_2=24\Omega$ instead of using the real part of the parallel combination ($4\Omega$) plus $R_2$. A second slip is forgetting the $10^{-6}$ or $10^{-3}$ unit factors on $C_1$ and $L_1$.

    Full solution →
43. Q.53NAT · 2 marksOfficial answer: -60

    Control Systems › Bode & margins

    An asymptotic Bode slope at a given frequency depends only on how many poles and zeros have corner frequencies below it: start from $-20N$ for $N$ poles at the origin, then add $-20$ for each pole corner and $+20$ for each zero corner already passed. Here, by $\omega =3$, the integrator and the poles at $0.1$ and $0.5$ have all acted, and the zero at $10$ has not yet, so $-20-20-20=-60\text{dB/decade}$.

    WATCH OUT FOR

    Reading the compensator corners wrongly: in $\frac{s+10}{s+0.1}$ the pole is at $0.1\text{rad/s}$ and the zero is at $10\text{rad/s}$ (a lag network puts its pole first), not the other way round. Swapping them would place the zero below $\omega =3$ and give $-20\text{dB/decade}$. Another slip is taking the $(2s+1)$ corner to be $2\text{rad/s}$ instead of $0.5\text{rad/s}$. That does not change this answer, because both values are below 3, but it gives wrong slopes elsewhere. Also, the constant $0.001$ has no effect on slope.

    Full solution →
44. Q.54NAT · 2 marksOfficial answer: 0.0

    Networks (Electric Circuits) › Theorems

    Circuit B has the same resistor network as Circuit A, so you don't need $R_0$. Use superposition. With only the left source acting, Circuit A is scaled by homogeneity: 8 V instead of 16 V gives half of $I_1$. With only the right source acting, apply reciprocity to Circuit A: 16 V in the left branch gives $\frac{1}{3}$ A in the right branch, so 16 V in the right branch gives $\frac{1}{3}$ A in the left branch. Then scale by 48/16 and fix the sign from the source's polarity.

    WATCH OUT FOR

    The usual mistake is the sign of the 48 V contribution. Reciprocity needs the moved source to push current in the same sense as the reference direction of the branch whose current is measured, here downward through the right-hand $R_0$. The 48 V source has + on top and pushes current upward, so its contribution is $-1$ A, not $+1$ A. Getting this wrong gives 2.0 A. Two other slips: forgetting that the 8 V source gives only half of $I_1$ (1 A, not 2 A), and making algebra errors while solving for $R_0$ when it isn't needed.

    Full solution →
45. Q.55NAT · 2 marksOfficial answer: 141.80 to 146.20

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

    A wattmeter reads the product of the voltage across its pressure coil, the current in its current coil and the cosine of the angle between them. Here that is $V_{RY}$ and $I_R$. In a balanced star load that angle is $30^{\circ }+\varphi$, with $\varphi =-90^{\circ }$ for pure capacitance, so the angle is $60^{\circ }$ and the reading is $V_L{I}_L\cos 60^{\circ }=V_L^2\omega C_0/(2\sqrt{3})$. The load takes no real power, but the meter still reads a non-zero value.

    WATCH OUT FOR

    Answering 0 W because a purely capacitive load takes no real power. This meter does not measure the load's real power: it pairs $I_R$ with the line voltage $V_{RY}$, not with $V_{RN}$. Other slips: using the phase angle $90^{\circ }$ instead of $30^{\circ }+\varphi$; using the line voltage 440 V in place of the phase voltage when working out $I_R$; taking the RBY sequence, which flips the angle to $120^{\circ }$ and gives $-143.97$ W.

    Full solution →
46. Q.56NAT · 2 marksOfficial answer: 15.80 to 16.00

    Analog Electronics › Filters & waveforms

    With $R_f\parallel C$ in the feedback path, the inverting amplifier becomes a first-order low-pass with gain $-\frac{R_f/R_1}{1+j\omega R_{f}C}$. Its DC gain is $10$ (20 dB). The 0 dB frequency is where $\sqrt{1+(\omega R_{f}C{)}^2}$ equals the DC gain, so $\omega R_{f}C=\sqrt{(R_f/R_1{)}^2-1}=\sqrt{99}$.

    WATCH OUT FOR

    Using the ideal-integrator shortcut $\omega =1/(R_{1}C)$, which drops the 1 under the square root. That gives $\omega =10^5\text{rad/s}$ and $f\approx 15.92\text{kHz}$. It happens to sit inside the key range, but the reasoning is wrong. A second mistake is to give the 3 dB corner, $f=1/(2\pi R_{f}C)\approx 1.59\text{kHz}$, or to forget to divide $\omega$ by $2\pi$ and answer in rad/s.

    Full solution →
47. Q.57NAT · 2 marksOfficial answer: 200.0

    Electromagnetics › Maxwell's equations

    The half-disc and the field region $x\ge 0$ are both sectors of angle $\pi$ centred on the pivot, so their overlap is always a sector. Its area $\frac{1}{2}{R}^2\alpha$ rises and falls linearly at the rate $\omega$. That makes the flux a triangular wave, so the emf has a constant magnitude $\frac{1}{2}B{R}^2\omega$ and flips sign every half-turn. The current is a $\pm 100\text{mA}$ square wave, and its peak-to-peak value is twice the magnitude.

    WATCH OUT FOR

    There are three common mistakes. (1) Treating this like a coil turning in a uniform field and assuming a sinusoidal emf. Here the flux is a triangular wave and the emf is a square wave. (2) Reporting the magnitude, 100 mA, and forgetting that the current reverses when the overlap starts to shrink. The peak-to-peak value is 200 mA. (3) Deciding the current is DC because the emf magnitude never changes, and answering 0. It is also easy to use the whole semicircle's area $\pi R^2/2$ in place of the rate of change of the overlap sector, $\frac{1}{2}{R}^2\omega$.

    Full solution →
48. Q.58NAT · 2 marksOfficial answer: 148.00 to 152.00

    Analog Electronics › Biasing

    This is collector-feedback bias. Because the emitter is grounded, $V_{CE}$ gives the collector voltage directly. $R_1$ runs from collector to base, so the voltage across it fixes $I_B=(V_C-V_{BE})/R_1$. $R_2$ carries $I_C+I_B$, which is the same as $I_E$. Subtract $I_B$ from the current in $R_2$ to get $I_C$, then take the ratio.

    WATCH OUT FOR

    The usual slip is to treat the whole 6.8 mA in $R_2$ as $I_C$, which gives $\beta =6.8/0.045=151.11$. That lands inside the answer range by luck, but it ignores that $R_2$ also supplies the base current through $R_1$. The other common error is to use $V_{CE}=5.2\text{V}$ as the drop across $R_1$ and forget that the base end of $R_1$ sits at $0.7\text{V}$.

    Full solution →
49. Q.59NAT · 2 marksOfficial answer: 32.8 to 33.9

    Electromagnetics › Electrostatics

    Between two concentric spheres the field is that of a point charge, $E=Q/(4\pi {\epsilon }_0{r}^2)$. Integrating it from $b$ to $a$ gives $V=\frac{Q}{4\pi {\epsilon }_0}\left(\frac{1}{a}-\frac{1}{b}\right)$, so $C=\frac{4\pi {\epsilon }_{0}ab}{b-a}$. Only the geometry factor $ab/(b-a)$, measured in metres, depends on the radii.

    WATCH OUT FOR

    Substituting the radii in centimetres. That gives $ab/(b-a)=30$ instead of $0.3\text{m}$ and makes the answer 100 times too large (about 3338 pF). The other common slip is using the isolated-sphere formula $4\pi {\epsilon }_{0}a$ (about 11.13 pF), which ignores the outer sphere.

    Full solution →
50. Q.60NAT · 2 marksOfficial answer: 2.35 to 2.43

    Networks (Electric Circuits) › AC, power & resonance

    The internal resistance and the capacitor form a series R–C divider. V_R and V_C are in quadrature, so the voltages obey a right-angle (Pythagoras) relation: V_R = sqrt(10² − 8²) = 6 V p-p. Since the same current flows through both, X_C/R = 8/6. That gives X_C = 66.67 Ω, and C = 1/(2πf X_C).

    WATCH OUT FOR

    Treating the divider as purely resistive, i.e. 8/10 = X_C/(R + X_C). That gives X_C = 200 Ω and C ≈ 0.80 µF, because it ignores the 90° phase difference between V_R and V_C. A second mistake is taking the "missing" 2 V p-p as the resistor drop. The drops add as phasors, not as numbers, so the resistor drop is 6 V p-p.

    Full solution →
51. Q.61NAT · 2 marksOfficial answer: 0.49 to 0.51

    Engineering Mathematics › Complex analysis

    At a simple pole $z_0$, the residue is ${\lim }_{z\to z_0}(z-z_0)f(z)$. For a rational function with distinct linear factors, this means covering up the factor $(z-z_0)$ and evaluating the rest at $z_0$. The result is the same as the partial-fraction coefficient of $\frac{1}{z-z_0}$.

    WATCH OUT FOR

    Substituting $z=-1$ as $(z+2)(z+3)=(1)(2)$ is easy to get right. The usual slips are sign errors, such as writing $(-1+3)$ as $-2$ and getting $-\frac{1}{2}$. Another is reporting the residue at a different pole, such as $-1$ at $z=-2$. A third is using the higher-order-pole derivative formula on what is only a simple pole.

    Full solution →
52. Q.62NAT · 2 marksOfficial answer: -10.1 to -9.9

    Networks (Electric Circuits) › Transients

    A diode in series with a lossless LC loop lets current flow for only one half-cycle. The current $i=C_0{V}_0{\omega }_0\sin {\omega }_{0}t$ returns to zero at $t=\pi /{\omega }_0$. At that instant the capacitor voltage has swung from $+V_0$ to $-V_0$. The diode then blocks the reverse current and the loop is open, so the capacitor holds $-V_0$ for ever after. Any time later than $\pi /{\omega }_0$ (here 0.314 ms, far shorter than 0.5 s) gives $V_c=-V_0$.

    WATCH OUT FOR

    One common mistake is to ignore the diode and evaluate the free LC oscillation at 0.5 s: $10\cos (10^4\times 0.5)=10\cos (5000)\approx 1.5\text{V}$. Another is to assume the capacitor fully discharges and answer 0 V. A third is to forget that the charge reverses and answer $+10$ V. The diode stops the current at its first zero crossing, which comes when the capacitor is at $-10$ V, so the sign of the answer matters.

    Full solution →
53. Q.63NAT · 2 marksOfficial answer: 14.40 to 14.60

    Electrical and Electronic Measurements › Bridges

    This is a Maxwell inductance-capacitance bridge. The unknown $R_3+j\omega L_3$ arm is opposite the parallel $R_2\parallel C_2$ arm. Balance gives $R_1{R}_4=Z_2{Z}_3$, so $Z_3=R_1{R}_4\left(\frac{1}{R_2}+j\omega C_2\right)$. That yields $L_3=R_1{R}_4{C}_2$ and $R_3=R_1{R}_4/R_2$. Both are independent of frequency, so the 2 kHz and $10{\text{V}}_{\text{p-p}}$ data are distractors.

    WATCH OUT FOR

    Pairing the wrong arms. Using adjacent arms, for example $L_3=R_1{R}_2{C}_2$, gives $60.9\text{mH}$. The capacitance multiplies the two resistors that form the other pair of opposite arms, here $R_1$ and $R_4$. A second trap is treating $C_2$ as in series with $R_2$ (Hay's bridge). That brings in $\omega =2\pi \times 2000\text{rad/s}$ and gives about $0.24\text{mH}$. A third is quoting $0.0145$, the value in henry, when the answer is asked for in millihenry.

    Full solution →
54. Q.64NAT · 2 marksOfficial answer: 0.365 to 0.369

    Networks (Electric Circuits) › Transients

    The comparator switches when the capacitor voltage crosses the fixed 2 V reference. Before the switch opens, the 1 kΩ / 5 kΩ divider charges the capacitor to 25/6 V. After it opens, the capacitor discharges through 5 kΩ alone with $\tau =0.5\text{ms}$. Setting $\frac{25}{6}{e}^{-t/\tau }=2$ gives the crossing time $t=\tau \ln \frac{25}{12}$.

    WATCH OUT FOR

    The usual slips are: taking the initial capacitor voltage as the full 5 V and forgetting the 1 kΩ / 5 kΩ divider (that gives $0.5\ln 2.5\approx 0.458$ ms); using $(1+5)\text{k}\Omega$ or $1\text{k}\Omega \parallel 5\text{k}\Omega$ as the discharge resistance, when the open switch leaves only 5 kΩ; and taking the threshold as 3 V by reading the divider upside down (600 kΩ is on top, so the reference is 2 V).

    Full solution →
55. Q.65NAT · 2 marksOfficial answer: 1.62 to 1.72

    Electrical and Electronic Measurements › Bridges

    Closing the switch puts the 30 kΩ branch in parallel with $R_1$, which lowers arm AD by a finite $\delta R_1=-\frac{1}{14}\text{k}\Omega$. Only the divider potential at D changes; $V_C$ stays at 5 V. Work out $V_{CD}$ before and after, and divide the finite change in output by the finite change in resistance, exactly as the question defines it.

    WATCH OUT FOR

    Using the small-signal derivative $\left|\frac{d{V}_{CD}}{d{R}_1}\right|=\frac{10R_3}{(R_1+R_3{)}^2}=\frac{10}{2.5^2}=1.60\text{V/k}\Omega$ at $R_1=1.5\text{k}\Omega$. That value falls outside the key, because the change of $-0.0714\text{k}\Omega$ is too large to treat as infinitesimal. The question asks for the ratio of the actual changes. Another slip is to add 30 kΩ in series with $R_1$. In the figure, the switch and the 30 kΩ resistor join both ends of arm AD, so they form a parallel path.

    Full solution →