GATE ECE 2024 question paper with solutions

GATE ECE 2024 answer key

General Aptitude Q.1 to Q.10

1. Q.1MCQ · 1 markOfficial answer: A

   General Aptitude › Reasoning

   An intensity analogy needs two things at once. The missing word must carry the same meaning as both ends of its chain, and it must be stronger than the first word and weaker than the last. 'Jaded' (dulled by too much of something) is the only option that names a degree of boredom, and it sits between 'bored' and 'weary'.

   WATCH OUT FOR

   Choosing 'dead' because of phrases like 'bored to death'. 'Dead' does not mean a degree of boredom, and an extreme word could not go in the middle of the chain. Another trap is 'worsted', which looks like 'worse' or 'worst' but means a woollen yarn or, in old usage, 'defeated'. 'Baffled' is a different feeling (confusion), not a stronger form of boredom.

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

   General Aptitude › Reasoning

   Use elimination. Single children cannot be in the sibling pair, and any two people who share a room in an accepted allocation cannot be the siblings. Remove every pair ruled out this way, and the one pair left is the answer.

   WATCH OUT FOR

   Reading the rule the wrong way round and picking a pair that shares a room (such as T and R, who share a room in the second allocation). Sharing a room rules a pair out; it does not make them siblings. Another slip is to forget that S and Q are single children and consider options (B) or (C).

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

   General Aptitude › Numerical

   Put one variable on the earlier ratio ($x$ and $4x$ five years ago), shift both ages by the same 10 years to reach the later ratio, and solve. The father's age when the son was born is just the age gap, which stays the same over time: $4x-x=3x$.

   WATCH OUT FOR

   Adding only 5 years (instead of 10) to the 'five years ago' ages when writing the 'five years from now' ratio. Another common slip is answering with the father's present age (45) or his age five years ago (40) instead of the age gap (30).

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

   General Aptitude › Numerical

   Swap the base with $\frac{1}{{\log }_{b}x}={\log }_{x}b$. All three terms then share the base $x$, so they add into one logarithm of the product: ${\log }_x(2\cdot 3\cdot 4)=1$, which gives $x=2\cdot 3\cdot 4$.

   WATCH OUT FOR

   Adding the bases ($2+3+4=9$, not an option) or taking their LCM (12, option B) when a sum of logarithms with the same base combines into the log of the PRODUCT, 24. Another slip is reading $\frac{1}{{\log }_{b}x}$ as ${\log }_b\frac{1}{x}=-{\log }_{b}x$. That identity is false: the reciprocal of a logarithm swaps the base and the argument, it does not invert the argument.

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5. Q.5MCQ · 1 markOfficial answer: A

   General Aptitude › Numerical

   Factor out the smaller power: $3^{199}-3^{196}=3^{196}(3^3-1)=3^{196}\times 26=2\times 3^{196}\times 13$. The huge power of 3 adds only the prime 3. The other primes all come from the small bracket $3^3-1$.

   WATCH OUT FOR

   Picking (C) 3 because the expression is 'made of powers of 3'. The subtraction brings in the new primes 2 and 13 through $3^3-1=26$. A second slip is to write $3^{199}-3^{196}=3^3$ by subtracting the exponents. That rule applies to division, $3^{199}÷3^{196}=3^3$, not to subtraction.

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6. Q.6MCQ · 2 marksOfficial answer: C

   General Aptitude › Reading

   Sentence ordering follows three anchors. Background comes before the incident: a timeless "once proclaimed" goes before "One day". Something is introduced before it is referred to: "a bull" goes before "the bull". A question comes before its answer.

   WATCH OUT FOR

   Starting with S because "One day" sounds like a story opening (options A and B). The student's objection in P only makes sense if we already know Shifu's claim that all life is illusion (R), and "the bull" in P and Q needs "a bull" in S to come first. So the belief comes first and the incident second.

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7. Q.7MCQ · 2 marksOfficial answer: D

   General Aptitude › Geometry

   A tight band around touching equal circles has two parts. The straight parts add up to the perimeter of the polygon joining the centres, here $4\times 2r=8r$. The curved parts always add up to exactly one full circle, $2\pi r$, because the band turns through $360^{\circ }$ in total. Unrolled, the tape is a rectangle: band length times stick length.

   WATCH OUT FOR

   Using the diameter in place of the radius in the arc term ($2\pi d$ or $\pi d$ counted per corner), or counting each straight side as $4r$ (two diameters) instead of the centre-to-centre distance $2r$. That gives $8+\pi$ and leads to option (C). Using $2l=20$ as the width gives options (A) and (B).

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8. Q.8MCQ · 2 marksOfficial answer: C

   General Aptitude › Data interpretation

   Read each bar off the 5% gridlines, then look at which way each ratio moves. When a ratio's numerator rises and its denominator falls, the ratio has to rise. Normal fell from 50% to 30% while Obese rose from 15% to 25%, so Obese/Normal went up. That makes (C) false without any division.

   WATCH OUT FOR

   One slip is swapping the legend colours (orange is 2023, blue is 2003). If you read the years the wrong way round, (A), (B) and (D) all become false and (C) becomes true, so you get three 'incorrect' options where the question allows one. That mismatch is itself a warning that the chart was misread. Another slip is forgetting the question asks for the INCORRECT statement and choosing one of the three true ones.

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9. Q.9MCQ · 2 marksOfficial answer: A

   General Aptitude › Spatial

   A mirror reflection followed by a water reflection is the same as rotating the object by 180 degrees in its own plane. A 180-degree rotation undoes itself, so the original is the image rotated by 180 degrees. Reverse the order of the characters and turn each one upside down in the rotational sense: in a seven-segment glyph, upper-left and lower-right swap, upper-right and lower-left swap, and top and bottom swap. Under this rotation 5 and 2 map onto themselves and 4 becomes h.

   WATCH OUT FOR

   Undoing only one of the two reflections. Undoing just the water reflection gives option (B), and undoing just the mirror reflection gives option (C). Both are traps. The other slip is to rotate each glyph but forget to reverse the order of the characters, or to reverse the characters but leave the dot where it was. That puts the dot between the h and the 2 and gives option (D).

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

    General Aptitude › Spatial

    Folding in half along an axis parallel to an edge keeps that edge's length and halves the side at right angles to the axis. The question defines both operations by the INITIAL edges, so each operation always halves the same original side: FO1 halves the 16 cm side and FO2 halves the 24 cm side.

    WATCH OUT FOR

    Two traps. (1) Mixing up which side gets halved: an axis parallel to the long edge halves the SHORT side, not the long side. Swapping them gives A = 6 x 16 and B = 24 x 4, the ratio 11:14, which is option (B). (2) Re-reading 'short edge' after the first fold: once B is folded to 12 x 16, the 16 cm side is now the longer one. If you then halve that side you get 12 x 8, perimeter 40, and a ratio of 7:5, which is not an option. The word 'initial' fixes the axis to the original 16 cm edge.

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Electronics and Communication Engineering Q.11 to Q.65

1. Q.11MCQ · 1 markOfficial answer: A

   Engineering Mathematics › Differential equations

   A factor of $t$ multiplying $e^{mt}$ in the complementary function means $m$ is a repeated root of the auxiliary equation. So $(At+B)e^{-2t}$ comes from $(m+2{)}^2=m^2+4m+4=0$. The forcing term $f(t)$ only changes the particular integral, never the CF.

   WATCH OUT FOR

   Choosing (D) or (C) because each has $-2$ as a root. They have two distinct roots, so their CF is a sum of two different exponentials with no $t{e}^{-2t}$ term. The other common slip is taking (B) because it shows a 4. But $m^2+4=0$ has roots $\pm 2j$, which give an oscillating CF, not a decaying one.

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

   Control Systems › Bode & margins

   Remember that $20\text{dB/decade}\approx 6\text{dB/octave}$, because $20{\log }_{10}2\approx 6.02$. Every slope is a whole multiple of that pair, so $40\text{dB/decade}$ (two poles or zeros) is $2\times 6=12\text{dB/octave}$.

   WATCH OUT FOR

   Some students divide by 10/2 = 5 and get 8 dB/octave. That treats the axis as linear, when it is logarithmic: a decade is ${\log }_{2}10\approx 3.32$ octaves, not 5. Others pick 20 dB/octave (C) because they remember \"20\" from the one-pole slope and forget that the unit is per decade there. A third error is halving the slope to 6 dB/octave (B), which is the value for 20 dB/decade, not 40.

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3. Q.13MCQ · 1 markOfficial answer: D

   Control Systems › Time response

   The final value theorem only applies when every pole of $sE(s)$ is in the open left half-plane. So check closed-loop stability first. Here $1+G(s)=(s+3)(s^2+2)$, which puts a pair of poles on the $j\omega$-axis at $\pm j\sqrt{2}$. The loop is marginally stable, so the error ends up as a steady sinusoid instead of settling to a limit.

   WATCH OUT FOR

   The common mistake is to apply the final value theorem without checking stability. Blindly, ${\lim }_{s\to 0}sE(s)=\frac{(1)(2)}{(3)(2)}=\frac{1}{3}$, which gives the trap option (B). A second trap: noticing $G(s)$ is type 1 and writing $e_{ss}=0$ for a step, which gives option (A). Both results assume a stable closed loop, and this loop is not asymptotically stable.

   Full solution →
4. Q.14MCQ · 1 markOfficial answer: Marks to all
5. Q.15MCQ · 1 markOfficial answer: A

   Electromagnetics › Transmission lines

   Looking into a short-circuited lossless line from a distance $d$, the normalized impedance is $z_{in}=j\tan \beta d$. It is purely reactive and repeats every $\lambda /2$. At $\lambda /8$ it is $+j1$, at $\lambda /4$ it is an open circuit, at $3\lambda /8$ it is $-j1$, and at $\lambda /2$ it is a short again.

   WATCH OUT FOR

   There are three common slips. The first is measuring $d$ from the generator end (point D), which reverses the sequence and leads to option (C). The second is using $\beta d=2\pi d/\lambda$ as if it were the Smith-chart rotation angle ($2\beta d$), which puts the open circuit at $\lambda /8$ instead of $\lambda /4$. The third is mixing up the sign: a short-circuited stub shorter than $\lambda /4$ is inductive ($+j$), not capacitive.

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6. Q.16MCQ · 1 markOfficial answer: B

   Electromagnetics › Electrostatics

   A vector field can be an electrostatic field exactly when it is conservative, meaning $\nabla \times \mathbf{E}=0$ (equivalently $\mathbf{E}=-\nabla V$). The divergence imposes no condition. It only fixes the charge density through $\nabla \cdot \mathbf{E}={\rho }_v/{ϵ}_0$.

   WATCH OUT FOR

   Testing the divergence instead of the curl. Both fields have zero divergence, so a divergence test would wrongly pick (A). The other common slip is a sign error in $\partial F_y/\partial x$ for ${\mathbf{F}}_2$: its $F_y=-Ax$, so the curl is $-2A$, not zero.

   Full solution →
7. Q.17MCQ · 1 markOfficial answer: A

   Analog Electronics › Amplifiers

   Drive the gate with a test source. With no gate current, $I_x$ flows entirely through $C_1$, so $v_{gs}=I_x/(j\omega C_1)$. The dependent current $g_m{v}_{gs}$ joins $I_x$ in charging $C_L$. The result is the two capacitors in series plus a term $g_m/((j\omega {)}^2{C}_1{C}_L)$, which is a negative real resistance. This is the standard negative-resistance cell at the heart of the Colpitts oscillator.

   WATCH OUT FOR

   There are three common slips. (1) Treating $(j\omega {)}^2$ as $+{\omega }^2$, which flips the sign and gives option (B). (2) Leaving out the $g_m{v}_{gs}$ current in the KCL at the source, which leaves only the series capacitors, option (C). (3) Combining $C_1$ and $C_L$ as if they were in parallel, which gives option (D). They are in series here: the same current $I_x$ flows through $C_1$, and the voltages across $C_1$ and $C_L$ add up to $V_x$.

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8. Q.18MCQ · 1 markOfficial answer: A

   Analog Electronics › Feedback & oscillators

   The output of the five-transistor OTA is tied straight to its inverting input, the gate of $M_2$. That makes it a unity-gain buffer with $\beta =1$. With a finite open-loop gain $A$, the closed-loop gain is $\frac{A}{1+A}$, not exactly 1.

   WATCH OUT FOR

   Answering 1.000 by treating the buffer as ideal, which assumes infinite open-loop gain. Other traps: halving the gain as if the single-ended output of a differential pair gave half, which leads to 0.488; getting the feedback sign wrong or using $1+1/A$, which leads to about 1.025. Check the polarity first. The gate of $M_2$ is the inverting input, so the feedback is negative and the gain is $A/(1+A)=40/41$.

   Full solution →
9. Q.19MCQ · 1 markOfficial answer: A

   Digital Electronics › Boolean logic

   A prime implicant is essential only if some minterm is covered by that prime implicant and by no other. Here $m_0,m_2$ can only be grouped into the corner quad $\overline{B}\overline{D}$, and $m_5,m_7$ only into $BD$. The remaining minterms $m_{12},m_{14}$ are covered equally well by $AB$ or by $A\overline{D}$, so neither of those is essential.

   WATCH OUT FOR

   Picking (D) because $AB$ appears in a minimum SOP. $AB$ is needed only to cover $m_{12}$ and $m_{14}$, and $A\overline{D}$ covers them just as well, so $AB$ is a selective prime implicant, not an essential one. A second mistake is to miss $A\overline{D}$ altogether: with only three prime implicants in view, $AB$ looks like the sole cover of $m_{12},m_{14}$ and wrongly seems essential.

   Full solution →
10. Q.20MCQ · 1 markOfficial answer: A
11. Q.21MCQ · 1 markOfficial answer: A

    Signals and Systems › LTI & convolution

    If you stretch both the input and the impulse response by a factor $a$ (replace $t$ with $t/a$), the convolution output is stretched by the same factor and also multiplied by $a$: $x(t/a)\ast h(t/a)=ay(t/a)$. The factor $a$ appears because the $d\tau$ in the convolution integral stretches too. Here $a=2$, so the output is $2y(0.5t)$.

    WATCH OUT FOR

    Many students say that scaling time in both signals just scales time in the output, and answer $y(0.5t)$ with no factor. That misses the Jacobian $d\tau =2du$. The other trap is option (B): each Fourier transform picks up a factor of 2, giving $4Y(2\omega )$, and a student who stops there forgets that $y(0.5t)$ already carries one factor of 2 of its own.

    Full solution →
12. Q.22MCQ · 1 markOfficial answer: A

    Electronic Devices › Semiconductors

    Carrier concentration in a non-degenerately doped semiconductor vs. $1/T$ shows three distinct regions on a $\log (n)$ vs $1/T$ plot: a steep intrinsic-region drop at high $T$ (low $1/T$), a flat extrinsic plateau at $n\approx N_D$ in the middle, and a steep freeze-out drop at low $T$ (high $1/T$).

    WATCH OUT FOR

    Confusing the direction of the horizontal axis — since it is $1/T$ (not $T$) increasing to the right, students sometimes place the intrinsic region on the wrong side, or forget that a non-degenerately doped sample must show a plateau (ruling out option D, which only applies to something like a fully degenerate/heavily doped sample with no distinct freeze-out/extrinsic split), or forget the freeze-out region altogether (ruling out option C).

    Full solution →
13. Q.23MCQ · 1 markOfficial answer: A

    Analog Electronics › Diode circuits

    In a centre-tapped full-wave rectifier each diode uses only half of the secondary. The output peak is therefore half the full-secondary peak, $V_m=\frac{V_{s,\text{peak}}}{2n}$, and the average load voltage is $\frac{2V_m}{\pi }$.

    WATCH OUT FOR

    Taking the full-secondary peak $10/n$ as $V_m$, as in a bridge rectifier, gives $\frac{20}{n\pi }=\frac{2.5}{\pi }$ and $n=8$, which is option (B). Reading $n:1$ as the ratio to each half-winding makes the same error. In a centre-tapped circuit only half of the secondary feeds the load at any instant.

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

    Signals and Systems › z-transform

    For a causal discrete-time system, poles decide stability (all inside $|z|=1$). Zeros matter too for minimum phase: poles and zeros must all be inside the unit circle. The initial value is $h[0]={\lim }_{z\to \infty }H(z)$. The final value is ${\lim }_{z\to 1}(z-1)H(z)$, and that formula is valid only when $(z-1)H(z)$ has no pole on or outside the unit circle.

    WATCH OUT FOR

    The common slip is to mark (B) true because both poles are well inside the unit circle, forgetting that minimum phase also requires the zeros inside. Here $|z|=\sqrt{1.5}>1$ for both zeros. A second slip is to apply the continuous-time rule (zeros in the left half-plane): these zeros sit on the imaginary axis of the z-plane, but what counts is their magnitude. A third is to take the initial value as $H(0)$ or $H(1)$ instead of ${\lim }_{z\to \infty }H(z)$.

    Full solution →
15. Q.25MSQ · 1 markOfficial answer: A, C

    Engineering Mathematics › Vector calculus

    When the integrand is continuous, a point-form equation holds everywhere exactly when its integral is zero over every volume. The Gauss divergence theorem then turns ${\int }_V\nabla \cdot (\rho u)dv$ into the outward flux $\iint \rho u\cdot \hat{n}ds$. So the integral forms (C) and (A) are equivalent to the point form, not merely consequences of it.

    WATCH OUT FOR

    The usual slip is dropping the minus sign when moving $\nabla \cdot (\rho u)$ to the right-hand side, which leads to (B) or (D). A second one is marking only (A) because (C) looks like 'just an intermediate step'. Since the equation must hold for every volume $V$, (C) is fully equivalent to the point form too. Remember that $\hat{n}$ points outward, so a positive flux means mass is leaving $V$.

    Full solution →
16. Q.26MSQ · 1 markOfficial answer: A, B, C

    Electronic Devices › Carrier transport

    At thermal equilibrium the total electron current is zero everywhere, so wherever n(x) has a non-zero slope, the built-in electric field must produce a drift current that exactly cancels the diffusion current. The sign of dn/dx on each side of the peak therefore fixes the diffusion-current direction, and forcing the drift term to be equal and opposite fixes both the drift-current direction and the electric-field direction on that segment.

    WATCH OUT FOR

    Reading only the diffusion current off the slope of n(x) and stopping there — e.g. concluding the field must point the same way as the diffusion current, or assuming the field is zero because the material is "at equilibrium." The whole point of the question is that equilibrium forces drift to exactly oppose diffusion, so the field's direction is opposite to what a naive glance at the concentration slope would suggest.

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17. Q.27NAT · 1 markOfficial answer: 2047

    Digital Electronics › Processors & memory

    A field that must tell apart $N$ items needs $\lceil {\log }_{2}N\rceil$ bits. Give the opcode and every register field the smallest width that works, and whatever is left of the fixed 32-bit instruction goes to the immediate. An unsigned $n$-bit field can hold values up to $2^n-1$.

    WATCH OUT FOR

    Using ${\log }_2$ without rounding up: that gives a 5-bit opcode for 40 instructions, when 5 bits cover only 32, or a 4-bit register field for 24 registers, when 4 bits cover only 16. Other errors are counting only two register fields instead of three, and giving $2^{11}=2048$ as the answer instead of the largest value an 11-bit field can hold, which is $2^{11}-1=2047$.

    Full solution →
18. Q.28NAT · 1 markOfficial answer: 2.49 to 2.51
19. Q.29NAT · 1 markOfficial answer: 11

    Digital Electronics › Number systems

    Every multi-digit numeral in the equation is written in base $r$, so convert each one to decimal with positional weights: $(12{)}_r=r+2$ and $(37{)}_r=3r+7$. Then substitute the known root and solve the resulting linear equation in $r$. The digits that appear must all be smaller than $r$, so $r\ge 9$. You can confirm the answer with Vieta: sum of roots $=(12{)}_r$ and product $=(37{)}_r$.

    WATCH OUT FOR

    Reading $12$ and $37$ as decimal numbers. Then $x=8$ gives $64-96+37=5\ne 0$, and some students conclude the question has no answer or start guessing. Another slip is forgetting the digit constraint: the answer has to satisfy $r>8$ because the digits $8$ and $7$ appear.

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20. Q.30NAT · 1 markOfficial answer: 5

    Engineering Mathematics › Linear algebra

    Three vectors in ${\mathbb{R}}^3$ form a basis if and only if they are linearly independent, which happens if and only if the $3\times 3$ matrix built from them has a non-zero determinant. Set the determinant to zero and solve for $\alpha$.

    WATCH OUT FOR

    Getting the sign wrong on the middle cofactor. The first-row pattern is $+,-,+$ and the entry there is $-3$, so the term is $-(-3)\cdot 18=+54$, not $-54$. With the wrong sign you get $\det A=-38-14\alpha$ and a non-integer $\alpha$. Also, solving for the $\alpha$ that makes the set a basis gives every $\alpha \ne 5$, not one number: the question asks for the value where it is not a basis.

    Full solution →
21. Q.31NAT · 1 markOfficial answer: 2

    Networks (Electric Circuits) › Laws & node/mesh

    A floating dependent voltage source between two non-reference nodes calls for a supernode. Write KCL once around both nodes and use the source to link the node voltages ($V_B=V_A+1000I_0=2V_A$). A current source in series with a resistor fixes that branch current at 2 mA, whatever the resistor.

    WATCH OUT FOR

    (1) Reversing the dependent source's polarity, i.e. writing $V_B=V_A-1000I_0=0$. That gives $I_x=0$ and $V_A=3\text{V}$. (2) Putting $I_0$ into $1000I_0$ in mA instead of amperes, which inflates the source voltage by a factor of 1000. (3) Thinking the 1 kΩ in series with the 2 mA source changes that branch's current, or trying to write a KVL equation across a current source.

    Full solution →
22. Q.32NAT · 1 markOfficial answer: 0.75

    Networks (Electric Circuits) › Transients

    A single-inductor circuit has $\tau =L/R_{th}$, where $R_{th}$ is measured at the inductor terminals with independent sources switched off, and for the circuit as it is after switching. With the current source opened, the $2\Omega$ on one side is in series with the $4\Omega \parallel 4\Omega$ on the other.

    WATCH OUT FOR

    Replacing the current source with a short circuit (it must be opened), which removes the $2\Omega$ and gives $\tau =3/2=1.5\text{s}$. Another mistake is using the pre-switching circuit (S open), which gives $R_{th}=2+4=6\Omega$ and $\tau =0.5\text{s}$. Some students also treat the $2\Omega$ as parallel with the right-hand resistors when it is actually in series with them around the inductor loop.

    Full solution →
23. Q.33NAT · 1 markOfficial answer: 0.50

    Engineering Mathematics › Probability & statistics

    For independent uniform variables on $[0,1]$ the joint density is 1 on the unit square, so any probability equals the area of the event region. The region $x\ge y$ is half the square. Symmetry gives the same answer with no integration: for i.i.d. continuous variables, $P(X\ge Y)=P(Y\ge X)$ and $P(X=Y)=0$.

    WATCH OUT FOR

    Some students think the equality case $X=Y$ needs extra probability and give an answer above 0.5. It does not: for continuous variables $P(X=Y)=0$, so $P(X\ge Y)=P(X>Y)=0.5$. Another slip is setting the inner integral limits the wrong way round (for example ${\int }_x^{1}dy$), which describes the region $y\ge x$. Here that still gives 0.5 by symmetry, but the same slip gives a wrong answer when the variables are not identically distributed.

    Full solution →
24. Q.34NAT · 1 markOfficial answer: 4
25. Q.35NAT · 1 markOfficial answer: 0.250

    Analog Electronics › Diode circuits

    A reverse-biased ideal Zener behaves like a constant 5 V battery that absorbs energy. The capacitor discharges only until it reaches V_Z, so the charge moved is C(10 − 5) = 50 µC. The energy into any constant-voltage element is its voltage times the charge through it: W_Z = V_Z · Q. The energy balance (375 µJ released by the capacitor = 125 µJ in R + 250 µJ in the Zener) confirms it.

    WATCH OUT FOR

    (1) Letting the capacitor discharge fully to 0 V and giving the Zener a share of ½C·10² = 500 µJ. Once v_C falls to 5 V the current stops. (2) Giving the Zener all the capacitor's released energy, 375 µJ, and forgetting the 125 µJ lost in the resistor. (3) Reading the Zener as forward-biased: check which end the discharge current enters. It enters at the barred cathode, so the diode is in reverse breakdown.

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

    Engineering Mathematics › Calculus

    A latitude circle cuts a spherical cap off the sphere. Convert latitude to polar angle (${\theta }_0=90^{\circ }-\text{latitude}$) and integrate $dA=R^2\sin \theta d\theta d\varphi$ to get $A=2\pi R^2(1-\cos {\theta }_0)$. This is the same as Archimedes' $A=2\pi Rh$, where $h$ is the cap height.

    WATCH OUT FOR

    Using $60^{\circ }$ as the polar angle, i.e. mixing up latitude (measured from the equator) with the angle from the pole. That gives $2\pi R^2(1-\cos 60^{\circ })=\pi R^2$, which is not among the options. A second slip is taking the cap height as $R\cos 60^{\circ }$ or $R-R\cos 60^{\circ }$ instead of $R-R\sin 60^{\circ }$.

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

    Control Systems › Routh–Hurwitz

    "Decays faster than $e^{-\sigma t}$" is a relative-stability requirement: every closed-loop pole must lie to the left of $s=-\sigma$. Substitute $s=z-\sigma$ to move that line onto the imaginary axis, then apply the ordinary Routh test to the polynomial in $z$.

    WATCH OUT FOR

    Running the Routh test on the original polynomial $s^3+6s^2+11s+6+K$. That gives absolute stability, $-6<K<60$, which only puts the poles in the left half plane. With that range, options (A), (B) and (C) all look acceptable. Another mistake is to include the boundary values $K=0$ or $K=6$: at those values a pole sits exactly on $\text{Re}(s)=-1$, so the response decays at the same rate as $e^{-t}$, not faster.

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

    Control Systems › Compensators & PID

    The loop $\frac{K(s+\alpha )}{s^2(s+4)}$ gives a third-order characteristic equation $s^3+4s^2+Ks+K\alpha =0$, so there are three closed-loop poles. Two of them are specified. Add an unknown third pole $(s+p)$ and match coefficients. The $s^2$ coefficient contains no $K$ or $\alpha$, so it fixes the third pole first: the poles must sum to $-4$, which puts it at $s=-2$. The other two coefficients then give $K=8$ and $K\alpha =8$.

    WATCH OUT FOR

    Treating the closed loop as second order, so that $s^2+2s+4$ is matched straight against a truncated characteristic equation and the third pole is dropped. A second trap is assuming the third closed-loop pole stays at the compensator pole $s=-4$. Closing the loop moves it: the pole sum forces it to $s=-2$. Assuming $-4$ gives $K\alpha =16$ and an $s^2$ coefficient of $6\ne 4$, which is inconsistent. Some also try pole-zero cancellation ($\alpha =4$), which is not an option and does not place the poles at $-1\pm j\sqrt{3}$.

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

    Electromagnetics › Plane waves

    When a wave hits a perfect conductor at normal incidence it forms a standing wave. $\mathbf{E}$ goes as $\sin (\beta x)$ and has a null at the wall. $\mathbf{H}$ goes as $\cos (\beta x)$ and has a maximum at the wall. So the $\mathbf{H}$ nulls sit at odd multiples of $\lambda /4$ from the wall, and here $\lambda =2\pi /\beta =3\text{m}$.

    WATCH OUT FOR

    The usual slip is to use the electric-field pattern for the magnetic field. $\mathbf{E}$ is zero at the wall and its next null is at $x=-\lambda /2=-\frac{3}{2}\text{m}$ (option B). $\mathbf{H}$ is at its maximum at the wall, so its first null is at $-\lambda /4$. Choosing $-3\text{m}$ (D) means the null spacing was taken as $\lambda$ instead of $\lambda /2$. Choosing $-6\text{m}$ (C) usually means $\beta =\pi /3$ was read off the exponent instead of $2\pi /3$, which doubles $\lambda$. Some candidates also forget that the air is the region $x<0$ and give a positive answer.

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

    Digital Electronics › Combinational

    Each input of a priority encoder is gated by the complements of every higher-priority input. Write $B$ as the sum of the conditions whose code has LSB 1, then set the all-zero input (no valid input) to 1 as a don't-care and simplify with $X+\overline{X}Y=X+Y$.

    WATCH OUT FOR

    Leaving out the don't-care gives $B=\overline{D_3}{D}_2+\overline{D_3}\overline{D_1}{D}_0$, which is not among the options. Other common slips: dropping the priority gating (writing just $D_2+D_0$), or choosing (A), which is the expression for the other output bit $A$.

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

    Digital Electronics › Combinational

    With S = 1 the MUX connects Y to its own complement, delayed by the MUX propagation delay: $Y(t)=\overline{Y}(t-t_{pd})$. One inversion in a feedback loop gives an oscillator (a ring oscillator). The output toggles once every $t_{pd}$, so a full cycle takes $2t_{pd}$ and $f=\frac{1}{2t_{pd}}$.

    WATCH OUT FOR

    Taking the period as one propagation delay ($T=10\text{ns}$) and choosing 100 MHz (A). One delay is only half a cycle: the output needs one 10 ns delay to go from 0 to 1 and another to go back to 0, so $T=20\text{ns}$. Another mistake is to treat the circuit as static and pick a constant output (C or D), which ignores the inverting feedback. The input-0 path does not matter when S = 1.

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

    Digital Electronics › Sequential

    Trace the wiring to get the excitation equations $T_1=Q_1+\overline{Q_0}$ and $T_0=Q_1+Q_0$. Then use $Q^{+}=T\oplus Q$ to find the next state for all four present states. Because $Q_1$ feeds both OR gates, both flip-flops toggle whenever $Q_1=1$. That fact alone gives $10\to 01$ and $11\to 00$.

    WATCH OUT FOR

    Treating the T input as if it were a D input ($Q^{+}=T$) instead of a toggle ($Q^{+}=T\oplus Q$). Doing that gives $10\to 11$, the wrong transition in options (A) and (C). A second common slip is misreading the long feedback wires: taking $Q_0$ into gate 1 instead of $\overline{Q_0}$, or missing the $Q_1$ feedback into gate 1. Students also often skip state $11$. It is outside the main cycle, but option (B) starts from it, so you have to check it.

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

    Engineering Mathematics › Complex analysis

    Check what the numerator does at each root of the denominator before you pick a pole order. Here $\sin (\pi z)$ vanishes at every integer. That drops $z=0$ from a double pole to a simple pole and makes $z=2$ removable. Only the residue $-\pi /2$ at $z=0$ contributes, so the integral is $2\pi j\cdot (-\pi /2)=-{\pi }^{2}j$.

    WATCH OUT FOR

    Treating $z=0$ as a double pole and using the derivative formula carelessly. Done correctly, $\frac{d}{dz}{\left[\frac{\sin \pi z}{z-2}\right]}_{z=0}=-\frac{\pi }{2}$ gives the same residue, but students often slip on the quotient rule. Some then get $+\pi /2$ and pick (A). Others take the residue at $z=2$ to be nonzero, or leave $z=2$ out of the count altogether. That costs nothing here, since its residue is $0$, but it is a habit that fails on other questions.

    Full solution →
34. Q.44MCQ · 2 marksOfficial answer: Marks to all

    Signals and Systems › Fourier transform

    Write $y$ as a sign flip, a time scaling and a time shift of $x$. Then apply the scaling and shift properties in the same order you built the argument. A 4-times stretch in time compresses the spectrum ($X(4f)$) and multiplies it by 4. Shifting the stretched signal by 2 multiplies the transform by $e^{-j4\pi f}$. As a quick check, $y(t)$ is real and even, so $Y(f)$ must be real. Only option (B) gives that.

    WATCH OUT FOR

    The most common slip is to apply the shift of $\frac{1}{2}$ after the scaling and so write $e^{-j\pi f}$ (option A). Once $x$ has been stretched by 4, the shift that lands its centre at the origin is 2, not $\frac{1}{2}$. If you shift first, the scaling turns $e^{-j\pi f}$ into $e^{-j4\pi f}$. The second slip is to mix up stretching and compression and write $\frac{1}{4}X(f/4)$ (options C and D). A signal that is wider in time has a narrower spectrum, which is $X(4f)$ with a factor of 4 in front.

    Full solution →
35. Q.45MCQ · 2 marksOfficial answer: B
36. Q.46MCQ · 2 marksOfficial answer: B

    Digital Electronics › ADC & DAC

    ENOB is found by putting the measured SINAD into the ideal-quantiser formula $\text{SQNR}=6.02N+1.76\text{dB}$ and solving for $N$. The nominal bit count does not enter. Both quantities given are powers, so use $10{\log }_{10}$.

    WATCH OUT FOR

    Three traps. First, answering 10 (option D) because the ADC is 10-bit: ENOB depends on the measured SINAD, not on the nominal resolution. Second, misreading $10\mu \text{W}$ as $10^{-6}\text{W}$. That gives 60 dB and ENOB $\approx 9.67$, which rounds to 10, also option D. The correct value is $10\times 10^{-6}=10^{-5}\text{W}$. Third, using $20{\log }_{10}$ on a power ratio, which gives 100 dB and an ENOB of about 16, not one of the options.

    Full solution →
37. Q.47MCQ · 2 marksOfficial answer: A
38. Q.48MCQ · 2 marksOfficial answer: Marks to all

    Signals and Systems › Sampling

    Ideal impulse-train sampling copies the spectrum to every multiple of $f_s$ and scales each copy by $f_s$. The impulse response is a sinc low-pass multiplied by a 19 Hz cosine, so it is a narrow band-pass filter centred at 19 Hz. That filter picks out only the image at $f_s+f_0=15+4=19$ Hz. The factor $f_s=15$ in every option confirms the ideal impulse-sampling model, and the $-\pi /2$ inside the modulating cosine shows up as a $-\pi /2$ phase shift on the passed tone.

    WATCH OUT FOR

    Students often drop the $\frac{1}{2}$ from modulation and get amplitude 30, or forget the $f_s$ scaling of the images and get amplitude 1. Some take the image at 19 Hz to be $-4+15k$ and so give it phase $-\pi /3$ instead of $+\pi /3$ (19 is $4+15$). Others ignore the $-\pi /2$ phase of the band-pass filter. Finally, many do not notice that $\sin (\theta +\pi /3)=\cos (\theta -\pi /6)$, which makes (A) and (C) the same function.

    Full solution →
39. Q.49MCQ · 2 marksOfficial answer: D

    Analog Electronics › Filters & waveforms

    OPA1 is a non-inverting Schmitt trigger with thresholds $\pm \frac{R_1}{R_2}{V}_{sat}=\pm 1\text{V}$, so its square-wave output lags the triangle by the time the triangle takes to climb from $0$ to $1\text{V}$, which is $1\mu \text{s}$. OPA2 (series $L$, feedback $R$) integrates with gain $R/L=10^6{\text{s}}^{-1}$. It would need to swing $40\text{V}$ per half-period but can only swing $20\text{V}$, so it saturates and the triangle becomes a trapezoid.

    WATCH OUT FOR

    Two traps. Reading OPA1 as a plain zero-crossing comparator (no delay, options B or C), which ignores the positive feedback through the $100\text{k}\Omega$ resistor. And calling any integrator's output 'triangular' (A or C) without checking whether its ramp fits between the rails. Here $10\text{V}/\mu \text{s}$ for $4\mu \text{s}$ is far more than $\pm 10\text{V}$ allows, so the output clips.

    Full solution →
40. Q.50MCQ · 2 marksOfficial answer: B

    Analog Electronics › Feedback & oscillators

    The amplifier is non-inverting, so it adds $0^{\circ }$ of phase. The feedback path is a CR high-pass section followed by an unloaded RC low-pass section, which together act as a band-pass network with $\beta (s)=\frac{sRC}{(sRC{)}^2+3sRC+1}$. That network has zero phase at ${\omega }_0=1/(RC)$, where $\beta =1/3$. Sustained oscillation therefore needs a gain of exactly 3, which means $R_1=2R$. The numbers match the Wien-bridge oscillator.

    WATCH OUT FOR

    The usual mistake is to see alternating R and C sections and assume a phase-shift oscillator. That leads to a gain of 29 and $f=1/(2\pi \sqrt{6}RC)$, which are options (A) and (C). Those results belong to an inverting amplifier with three identical ladder sections giving $180^{\circ }$. Here there are only two sections, one high-pass and one low-pass, and the amplifier is non-inverting, so the network must supply $0^{\circ }$, not $180^{\circ }$. A second mistake is to multiply the two sections' unloaded transfer functions together. That ignores how the second section loads node A: it gives $\beta =x/(1+x{)}^2$ and $\beta ({\omega }_0)=1/2$ instead of the correct $1/3$.

    Full solution →
41. Q.51MCQ · 2 marksOfficial answer: B

    Analog Electronics › Amplifiers

    A diode-connected MOSFET looks like a resistance of $\frac{1}{g_m}$ to ground, and an ideal bias current source is an open circuit in AC. So the bias network $R_B+\frac{1}{g_{m1}}$ forms a voltage divider with $R_s$ at the gate of $M_2$, and that reduced input feeds a common-source stage with gain $g_{m2}{R}_D$.

    WATCH OUT FOR

    Choosing (A) because "the gate draws no current, so the whole of $v_{in}$ reaches the gate". The gate draws nothing, but the bias path $R_B+\frac{1}{g_{m1}}$ does carry AC current, so there is a divider loss across $R_s$. A second mistake is choosing (D), which drops $R_B$ from the path to ground. A third is treating the diode-connected $M_1$ as an AC ground (it is $\frac{1}{g_{m1}}$, not zero), or treating $I_{ref}$ as a short, which grounds the top of $R_B$ and gives an answer that is not among the options.

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

    Electronic Devices › BJT

    The $\beta$ vs $I_C$ curve of a BJT rises then falls: it is low at very small currents (base current dominated by $n=2$ recombination, so $\beta$ actually increases with $I_C$ there), peaks at moderate currents, and falls at high currents due to high-level injection (Webster effect). Separately, $\beta$ drops in saturation (extra reverse base-current injection) and a larger $\beta$ forces a lower $B{V}_{CEO}$ through $B{V}_{CEO}=B{V}_{CBO}/{\beta }^{1/n}$.

    WATCH OUT FOR

    Assuming $\beta$ always decreases with increasing $I_C$ everywhere on the curve (over-generalizing the high-injection roll-off in option A to the low-injection, recombination-dominated regime in option B), which leads students to wrongly mark B as true as well.

    Full solution →
43. Q.53MSQ · 2 marksOfficial answer: A, C

    Control Systems › State space

    A state-space realisation is not unique. Every option here has the same characteristic polynomial $s^2+3s+2$, so the poles cannot tell them apart and only the numerator can. For a controllable canonical form, read the numerator straight from $C$ (when $B=[01{]}^T$). For a diagonal form, compare the products $c_i{b}_i$ with the partial-fraction residues of $G_S$.

    WATCH OUT FOR

    Deciding on eigenvalues alone. All four options have poles at $-1$ and $-2$, so that test passes every one of them. A second trap is option (B): it has the companion $A$ matrix, but $B=[10{]}^T$ is not the controllable-canonical input vector, so you cannot read the numerator off $C$. Its actual transfer function is $-4/(s^2+3s+2)$. In the diagonal forms, only the products $c_i{b}_i$ matter, not $b_i$ or $c_i$ on their own. That is why (C) matches even though its $C=[11]$ looks nothing like the original.

    Full solution →
44. Q.54MSQ · 2 marksOfficial answer: A, B, D

    Engineering Mathematics › Vector calculus

    For a smooth field, these three statements say the same thing: the line integral is path independent, the integral around every closed path is zero, and $\vec{F}=\nabla f$ for some scalar potential $f$. Each of them gives $\nabla \times \vec{F}=\vec{0}$, whose components are the cross-derivative equalities in (D). Path independence says nothing about the divergence, $\nabla \cdot \vec{F}={\nabla }^{2}f$.

    WATCH OUT FOR

    Ticking (C) by mixing up "conservative" (curl-free) with "solenoidal" (divergence-free). A conservative field has zero divergence only if its potential also satisfies Laplace's equation, which is an extra condition. For example, $\vec{F}=2x\hat{i}$ is conservative but has divergence 2. A second slip is to pair the wrong derivatives in (D). Check each equation against a component of $\nabla \times \vec{F}$, e.g. the $\hat{i}$-component gives $\partial F_3/\partial y=\partial F_2/\partial z$.

    Full solution →
45. Q.55MSQ · 2 marksOfficial answer: A, B

    Engineering Mathematics › Linear algebra

    For a 2×2 matrix, one row of $(M-\lambda I)\mathbf{x}=0$ fixes the ratio $y/x$ of the eigenvector. Here $y/x=(\lambda -1)/k=\pm \sqrt{2k}/k=\pm \sqrt{2/k}$. A 2×2 matrix with two distinct eigenvalues has exactly two eigenvector directions. The four options point in four different directions, so at most two of them can be correct. Match each option by its ratio $y/x$.

    WATCH OUT FOR

    Stopping at $y/x=\pm \sqrt{2k}$ without dividing by $k$, or writing the eigenvector as $\left[\begin{array}{c}\lambda -1\\ 1\end{array}\right]$. Both lead to the look-alike options (C)/(D), which have ratio $\pm 1/\sqrt{2k}$. Note that $\sqrt{2k}/k=\sqrt{2/k}$, not $\sqrt{2k}$. A second trap is choosing only (B) because it is the one with the positive sign. Scaling an eigenvector by $-1$ keeps it an eigenvector, but (A) is not just $-1$ times (B). It belongs to the other eigenvalue, and it is also correct.

    Full solution →
46. Q.56MSQ · 2 marksOfficial answer: A, D or A, C, D

    Signals and Systems › Signals basics

    A discrete-time sinusoid $\cos (\Omega n+\varphi )$ is periodic only when $\Omega /2\pi$ is rational. Its fundamental period is the denominator $N$ of $\Omega /2\pi =k/N$ once the fraction is reduced to lowest terms. The phase does not affect the period.

    WATCH OUT FOR

    One mistake is using the continuous-time formula $N=2\pi /\Omega$. For (A) that gives $13.33$, and for (D) it gives $4.44$, so both correct options get rejected as \"not 40\" or even as aperiodic. The other mistake is leaving $\Omega /2\pi$ unreduced: $0.3\pi$ gives $6/40$, which looks like period 40, but it reduces to $3/20$, so the fundamental period is 20.

    Full solution →
47. Q.57NAT · 2 marksOfficial answer: 45.00 to 45.30
48. Q.58NAT · 2 marksOfficial answer: 200

    Electromagnetics › Transmission lines

    On a lossless line, $|\Gamma |$ stays the same everywhere and only its phase rotates. So $|Z_{in}|=Z_0|1+\Gamma |/|1-\Gamma |$ is largest where $\Gamma (d)$ is real and positive, which is a voltage maximum. There $Z_{in}=Z_0\cdot \text{VSWR}$, which is real. You do not need the load's phase: it only moves the maximum along the line.

    WATCH OUT FOR

    The usual slips are these. Some students give $Z_0/\text{VSWR}=12.5\Omega$, which is the minimum, not the maximum. Some stop at VSWR $=4$ and forget to multiply by $Z_0$. Some decide the problem cannot be solved because the load is unknown, but the load's phase only shifts where the maximum occurs, never its value.

    Full solution →
49. Q.59NAT · 2 marksOfficial answer: 112

    Signals and Systems › DTFT & DFT

    Applying the N-point DFT twice gives $N$ times the circularly time-reversed sequence, $Nx[(-n{)}_N]$. Applying it four times gives back the original sequence scaled by $N^2$. So $y[n]=N^{2}x[n]$, and then $Y[0]=N^2{\sum }_{n}x[n]$.

    WATCH OUT FOR

    Assuming ${\mathcal{F}}^4$ is the identity (true only for the unitary DFT with a $1/\sqrt{N}$ factor), which gives $Y[0]=7$. Another slip is reading $y[n]$ itself as the answer and forgetting that $Y[0]$ needs one more DFT, or treating $Y[0]$ as $y[0]=16$.

    Full solution →
50. Q.60NAT · 2 marksOfficial answer: 1.5

    Networks (Electric Circuits) › Two-port

    For $Y_{21}$, short port 2 ($V_2=0$). Here terminal 2 is wired straight to the middle node, so that node is at 0 V. The two branch currents then follow from Ohm's law. The $8V_1$ source stays active because $V_1\ne 0$. KCL at the middle node, with both port currents taken as entering, gives $I_2$.

    WATCH OUT FOR

    Misreading the polarity of the dependent source. Its '-' terminal is at the top, so the node above it is at $-8V_1$. Taking '+' at the top puts that node at $+8V_1$, which gives $I_{4\Omega }=-2V_1$ and the wrong answer $Y_{21}=-2.5\text{S}$. The other common slips are taking $I_2$ as leaving terminal 2, which gives $-1.5$, and switching off the $V_1$-controlled source when port 2 is shorted, which gives $-0.5$.

    Full solution →
51. Q.61NAT · 2 marksOfficial answer: 4.10 to 4.50

    Electronic Devices › MOS devices

    The flatband voltage relates the metal-semiconductor work function difference to the effect of oxide charge: $V_{FB}={\varphi }_{ms}-\frac{Q_{ox}}{C_{ox}}$, with ${\varphi }_{ms}={\varphi }_m-{\varphi }_s$. Setting $V_{FB}=0$ gives ${\varphi }_s={\varphi }_m-Q_{ox}/C_{ox}$ directly, once $C_{ox}$ is found from the oxide thickness and permittivity.

    WATCH OUT FOR

    Forgetting to convert the oxide thickness from nanometres to centimetres (100 nm = $1\times 10^{-5}$ cm) before computing $C_{ox}$, or mixing up the sign in $V_{FB}={\varphi }_{ms}-Q_{ox}/C_{ox}$ and adding the charge term instead of subtracting it, which shifts the answer outside the accepted key range.

    Full solution →
52. Q.62NAT · 2 marksOfficial answer: 2.5

    Networks (Electric Circuits) › Theorems

    For maximum power transfer, $R_L=R_{th}$. When the network has a dependent source, find $R_{th}$ from the terminal V–I relation, as $V_{oc}/I_{sc}$ or with a test source. Here both branches from A to B carry $V/5$, so the network behaves like $2.5\Omega$ in series with $50\text{V}$.

    WATCH OUT FOR

    One mistake is to kill the dependent source as if it were independent, which gives $5\parallel 2=\frac{10}{7}\approx 1.43\Omega$. Another is a sign error on $V_x$: taking $V_x=+3I_1$ with current flowing towards the $+$ end makes the middle branch carry $\frac{4V}{5}$, which gives $R_{th}=1\Omega$. $V_x$ is measured with $+$ at node B, so it is negative when current flows from A to B.

    Full solution →
53. Q.63NAT · 2 marksOfficial answer: 0.17 to 0.19

    Electronic Devices › Semiconductors

    Use the degeneracy-corrected donor occupation (neutral-donor) probability, $f_D=\frac{1}{1+\frac{1}{g}\exp \left(\frac{E_D-E_F}{kT}\right)}$, set it to the given neutral fraction, solve for $E_D-E_F$, then combine with $E_C-E_F$ to get $E_C-E_D$.

    WATCH OUT FOR

    Using the plain Fermi–Dirac form without the donor degeneracy factor $g=2$ (dropping the $1/g$ term), or confusing whether "5% neutral" refers to $f_D$ (occupied/neutral) versus $1-f_D$ (ionized), which flips the sign of the final answer.

    Full solution →
54. Q.64NAT · 2 marksOfficial answer: 52.40 to 52.60

    Electronic Devices › MOS devices

    Even though the transistor starts in the triode (linear) region, raising $V_{DS}$ at fixed $V_{GS}$ can push it into saturation once $V_{DS}$ exceeds the overdrive voltage $V_{OV}=V_{GS}-V_T$; $g_m$ must then be computed from the saturation current equation, not the triode one.

    WATCH OUT FOR

    Assuming the device stays in the linear/triode region for the whole problem and differentiating the triode $I_{DS}$ expression (or scaling the triode current) at $V_{DS}=1.5$ V without first checking the saturation boundary $V_{DS}\ge V_{OV}$.

    Full solution →
55. Q.65NAT · 2 marksOfficial answer: 4.00 to 4.26

    Electronic Devices › Optoelectronics

    At the maximum power point of a solar cell's I–V curve, $dP/dV=0$, which gives the condition $I+V\frac{dI}{dV}=0$; substituting the diode equation $I=I_L-I_0\left(e^{V/V_T}-1\right)$ and solving for $I_0$ links the given photocurrent, MPP voltage and thermal voltage directly to the reverse saturation current.

    WATCH OUT FOR

    Forgetting the product-rule term $VdI/dV$ and instead setting $I=0$ (an open-circuit-style condition) at the "maximum power point," which drops the $\left(1+\frac{V}{V_T}\right)$ factor and gives a badly wrong (much larger) value of $I_0$.

    Full solution →