GATE EC Electromagnetics Previous Year Questions with Solutions

Full question on page 24 of the official paper (PDF)

GATE 2026, Q.24

Electromagnetics1 markMCQ

A plane wave in a lossless medium has $\vec{E}(z, t) = E_0\left[\hat{x}\cos(\omega t - kz) + \hat{y}\sin(\omega t - kz)\right]$ . What is its polarization?

Option A: Right-handed circular
Option B: Left-handed circular
Option C: Linear
Option D: Linear, at $-45^\circ$ to $\hat{x}$

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Key idea: Equal amplitudes 90 degrees apart give circular polarization; the rotation sense fixes the hand.

Official answer: A

At $z = 0$ the field points along $\hat{x}$ at $t = 0$ and along $\hat{y}$ a quarter period later, with the same magnitude: it is circular, turning from $x$ towards $y$ .

The wave travels along $+z$ . With the right thumb along $+z$ , the fingers curl from $x$ to $y$ , so the wave is right-handed circularly polarized (IEEE convention). Option (A).

Full question on page 28 of the official paper (PDF)

GATE 2026, Q.28

Electromagnetics1 markMSQ

Which expressions describe the dielectric loss tangent of a substrate?

One or more options are correct.

Option A: The ratio of the real to the imaginary part of the total displacement current
Option B: $(\omega\epsilon'' + \sigma)/(\omega\epsilon')$
Option C: The ratio of electric susceptibility to permittivity
Option D: The ratio of the polarization $\vec{P}$ to the displacement $\vec{D}$

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Key idea: The loss tangent is the in-phase (lossy) current over the quadrature current.

Official answer: A and B

With $\epsilon = \epsilon' - j\epsilon''$ and conductivity $\sigma$ , Ampère's law gives the total current density

\nabla \times \vec{H} = (\sigma + \omega\epsilon'')\vec{E} + j\omega\epsilon'\vec{E}

\begin{aligned} \nabla \times \vec{H} &= (\sigma + \omega\epsilon'')\vec{E} \\ &\quad + j\omega\epsilon'\vec{E} \end{aligned}

The loss tangent is the ratio of the in-phase (lossy) part to the quadrature part:

\tan\delta = \frac{\omega\epsilon'' + \sigma}{\omega\epsilon'}

That is option (B), and option (A) describes the same ratio in words. Susceptibility over permittivity (C) and P over D (D) measure how polarizable the material is, not its loss.

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Full question on page 31 of the official paper (PDF)

GATE 2026, Q.33

Electromagnetics1 markNAT

Find the cutoff frequency of the TE10 mode in an air-filled rectangular waveguide of 0.28 inch by 0.14 inch.

Answer format: GHz, rounded off to two decimal places

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Key idea: $f_c = c/(2a)$ with $a$ the broad dimension.

Official answer: 20.50 to 21.50

a = 0.28 \times 2.54 = 0.7112\text{ cm}, \qquad f_c = \frac{c}{2a} = \frac{3 \times 10^{10}}{2 \times 0.7112}\text{ Hz} \approx 21.09\text{ GHz}

\begin{aligned} a &= 0.28 \times 2.54 \\ &= 0.7112\text{ cm}, \\ f_c &= \frac{c}{2a} \\ &= \frac{3 \times 10^{10}}{2 \times 0.7112}\text{ Hz} \\ &\approx 21.09\text{ GHz} \end{aligned}

Full question on page 31 of the official paper (PDF)

GATE 2026, Q.34

Electromagnetics1 markNAT

$|S_{11}|$ and $|S_{21}|$ cross at $-3$ dB for a lossless two-port and at $-4$ dB for a lossy one. What percentage of the incident power does the lossy network dissipate at the crossing?

Answer format: Percent, rounded off to two decimal places

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Key idea: Power lost inside is $1 - |S_{11}|^2 - |S_{21}|^2$ .

Official answer: 20.00 to 21.00

At the crossing both magnitudes are $-4$ dB:

|S_{11}|^2 = |S_{21}|^2 = 10^{-0.4} \approx 0.398, \qquad 1 - 2(0.398) \approx 0.204

\begin{aligned} |S_{11}|^2 &= |S_{21}|^2 \\ &= 10^{-0.4} \\ &\approx 0.398, \\ 1 - 2(0.398) &\approx 0.204 \end{aligned}

About 20.38% is dissipated. (For the lossless network, $2 \times 10^{-0.3} \approx 1$ , as it should be.)

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Full question on page 49 of the official paper (PDF)

GATE 2026, Q.52

Electromagnetics2 marksMCQ

A load has $\Gamma_L = 0.5\angle 30^\circ$ on a 50 Ω coaxial cable. Its input impedance moves to the diametrically opposite point on the same $|\Gamma|$ circle. Find the nearest input impedance.

Option A: $20.7 - j5.1$ Ω
Option B: $17.7 - j11.8$ Ω
Option C: $97.5 - j65.0$ Ω
Option D: $97.5 + j65.0$ Ω

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Key idea: The opposite point has $\Gamma_{in} = -\Gamma_L$ ; then $Z = 50\,\frac{1+\Gamma}{1-\Gamma}$ .

Official answer: B

Rotate $\Gamma_L$ by 180° (a quarter wavelength of line):

\Gamma_{in} = 0.5\angle 210^\circ = -0.433 - j0.25

\begin{aligned} \Gamma_{in} &= 0.5\angle 210^\circ \\ &= -0.433 - j0.25 \end{aligned}

Z_{in} = 50\,\frac{1 + \Gamma_{in}}{1 - \Gamma_{in}} = 50\,\frac{0.567 - j0.25}{1.433 + j0.25} \approx 17.7 - j11.8\ \Omega

\begin{aligned} Z_{in} &= 50\,\frac{1 + \Gamma_{in}}{1 - \Gamma_{in}} \\ &= 50\,\frac{0.567 - j0.25}{1.433 + j0.25} \\ &\approx 17.7 - j11.8\ \Omega \end{aligned}

Option (B). The load itself is $Z_L \approx 97.7 + j65.1$ Ω (close to option (D)), and the two satisfy $Z_{in}Z_L = 50^2$ , as a quarter-wave line requires.

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Full question on page 44 of the official paper (PDF)

GATE 2025, Q.49

Electromagnetics2 marksMCQ

A 4 m × 4 m metal sheet lies in the $x$ - $y$ plane with corners at $(\pm 2, \pm 2)$ , as in the figure. Its surface charge density is $\rho_s(x, y) = 4|y|$ µC/m². What is the total charge on the sheet, in µC?

x-y axes in metres with a square sheet whose corners are at (-2, 2), (2, 2), (-2, -2) and (2, -2), centred on the origin. — Figure from the official GATE 2025 EC paper, IIT Roorkee

Option A: 16
Option B: 85
Option C: 64
Option D: 256

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Key idea: Integrate the density over the area; the absolute value doubles the half where $y > 0$ .

Official answer: C

Q = \int_{-2}^{2}\int_{-2}^{2} 4|y|\,dy\,dx = 4 \times 4 \times 2\int_{0}^{2} y\,dy = 16 \times 2 \times 2 = 64\ \mu\text{C}

\begin{aligned} Q &= \int_{-2}^{2}\int_{-2}^{2} 4|y|\,dy\,dx \\ &= 4 \times 4 \times 2\int_{0}^{2} y\,dy \\ &= 16 \times 2 \times 2 \\ &= 64\ \mu\text{C} \end{aligned}

Option (C). Dropping the absolute value would make the two halves cancel to 0, and 16 is the total without the factor of 4.

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Full question on page 45 of the official paper (PDF)

GATE 2025, Q.50

Electromagnetics2 marksMCQ

A field of 0.01 V/m is applied along a copper wire of circular cross-section, 1 mm in diameter. Copper's conductivity is $5.8 \times 10^7$ S/m. What current flows, in amperes, to two decimal places?

Option A: 0.46
Option B: 1.82
Option C: 0.58
Option D: 1.12

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Key idea: Ohm's law in point form, $J = \sigma E$ , times the cross-sectional area gives the current.

Official answer: A

J = \sigma E = 5.8 \times 10^7 \times 0.01 = 5.8 \times 10^5\ \text{A/m}^2

\begin{aligned} J &= \sigma E \\ &= 5.8 \times 10^7 \times 0.01 \\ &= 5.8 \times 10^5\ \text{A/m}^2 \end{aligned}

I = J \cdot \pi r^2 = 5.8 \times 10^5 \times \pi\,(0.5 \times 10^{-3})^2 \approx 0.4555\ \text{A}

\begin{aligned} I &= J \cdot \pi r^2 \\ &= 5.8 \\ &\quad \times 10^5 \\ &\quad \times \pi\,(0.5 \times 10^{-3})^2 \\ &\approx 0.4555\ \text{A} \end{aligned}

To two decimal places, 0.46 A. Option (A). Using the 1 mm diameter as the radius gives 1.82, option (B).

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Full question on page 52 of the official paper (PDF)

GATE 2025, Q.64

Electromagnetics2 marksNAT

A 50 Ω lossless transmission line is terminated in $Z_L = (50 - j75)$ Ω. If the average incident power is 10 mW, what average power reaches the load, in mW?

Answer format: mW, rounded off to one decimal place

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Key idea: On a lossless line the load gets the incident power minus the reflected power, $1 - |\Gamma|^2$ of it.

Official answer: 6.30 to 6.50

\Gamma = \frac{Z_L - Z_0}{Z_L + Z_0} = \frac{-j75}{100 - j75}, \qquad |\Gamma|^2 = \frac{75^2}{100^2 + 75^2} = \frac{5625}{15625} = 0.36

\begin{aligned} \Gamma &= \frac{Z_L - Z_0}{Z_L + Z_0} \\ &= \frac{-j75}{100 - j75}, \\ |\Gamma|^2 &= \frac{75^2}{100^2 + 75^2} \\ &= \frac{5625}{15625} \\ &= 0.36 \end{aligned}

P_L = (1 - 0.36) \times 10\ \text{mW} = 6.4\ \text{mW}

\begin{aligned} P_L &= (1 - 0.36) \times 10\ \text{mW} \\ &= 6.4\ \text{mW} \end{aligned}

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Full question on page 52 of the official paper (PDF)

GATE 2025, Q.65

Electromagnetics2 marksNAT

A loop of area 5 m² in the $x$ - $y$ plane contains two 2 Ω resistors in series, as in the figure. A flux density $B(t) = 0.5t$ T is applied along the positive $z$ -axis. What is the magnitude of the current $I$ in the loop, in amperes?

A rectangular loop in the x-y plane with a 2 ohm resistor on each of its left and right sides and a current I marked on the top side. The loop area is labelled. Axes show x to the right, y up and z out of the page. — Figure from the official GATE 2025 EC paper, IIT Roorkee

Answer format: Amperes, rounded off to two decimal places

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Key idea: The induced EMF is the rate of change of flux through the loop, and it drives current through the loop's total resistance.

Official answer: 0.62 to 0.63

|\mathcal{E}| = \frac{d\Phi}{dt} = A\,\frac{dB}{dt} = 5 \times 0.5 = 2.5\ \text{V}

\begin{aligned} |\mathcal{E}| &= \frac{d\Phi}{dt} \\ &= A\,\frac{dB}{dt} \\ &= 5 \times 0.5 \\ &= 2.5\ \text{V} \end{aligned}

|I| = \frac{2.5}{2 + 2} = 0.625\ \text{A}

\begin{aligned} |I| &= \frac{2.5}{2 + 2} \\ &= 0.625\ \text{A} \end{aligned}

The key accepts 0.62 to 0.63, either rounding of 0.625. Using one resistor instead of both gives 1.25 A.

Full question on page 13 of the official paper (PDF)

GATE 2024, Q.15

Electromagnetics1 markMCQ

A lossless line is short-circuited at $z = 0$ , as in the figure. Moving towards the generator, the points A, B, C and D are $\lambda/8$ , $\lambda/4$ , $3\lambda/8$ and $\lambda/2$ from the load. What are the normalised input impedances $z_{inA}$ , $z_{inB}$ , $z_{inC}$ and $z_{inD}$ ?

A lossless transmission line shorted at z = 0 on the right, extending to z = -l on the left. Four reference points spaced lambda/8 apart, from the load towards the generator, are marked zinA, zinB, zinC and zinD. — Figure from the official GATE 2024 EC paper, IISc Bengaluru

Option A: $z_{inA} = +j1$ , $z_{inB} = \infty$ , $z_{inC} = -j1$ , $z_{inD} = 0$
Option B: $z_{inA} = \infty$ , $z_{inB} = +j0.4$ , $z_{inC} = 0$ , $z_{inD} = +j0.4$
Option C: $z_{inA} = -j1$ , $z_{inB} = 0$ , $z_{inC} = +j1$ , $z_{inD} = \infty$
Option D: $z_{inA} = +j0.4$ , $z_{inB} = \infty$ , $z_{inC} = -j0.4$ , $z_{inD} = 0$

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Key idea: A shorted lossless line of length $l$ has normalised input impedance $j\tan\beta l$ .

Official answer: A

With $\beta = 2\pi/\lambda$ , each $\lambda/8$ step adds $45°$ to $\beta l$ :

A, $\lambda/8$ : $j\tan 45° = +j1$
B, $\lambda/4$ : $j\tan 90° = \infty$ (an open circuit)
C, $3\lambda/8$ : $j\tan 135° = -j1$
D, $\lambda/2$ : $j\tan 180° = 0$ (a short again)

Option (A).

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Full question on page 14 of the official paper (PDF)

GATE 2024, Q.16

Electromagnetics1 markMCQ

With $\hat{i}$ and $\hat{j}$ the unit vectors along $x$ and $y$ and $A$ a positive constant, which statement is true of $\mathbf{F}_1 = A(y\,\hat{i} + x\,\hat{j})$ and $\mathbf{F}_2 = A(y\,\hat{i} - x\,\hat{j})$ ?

Option A: Both are electrostatic fields.
Option B: Only $\mathbf{F}_1$ is an electrostatic field.
Option C: Only $\mathbf{F}_2$ is an electrostatic field.
Option D: Neither is an electrostatic field.

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Key idea: An electrostatic field is the gradient of a potential, so its curl must be zero.

Official answer: B

(\nabla \times \mathbf{F}_1)_z = \frac{\partial (Ax)}{\partial x} - \frac{\partial (Ay)}{\partial y} = A - A = 0

\begin{aligned} (\nabla \times \mathbf{F}_1)_z &= \frac{\partial (Ax)}{\partial x} \\ &\quad - \frac{\partial (Ay)}{\partial y} \\ &= A - A \\ &= 0 \end{aligned}

(\nabla \times \mathbf{F}_2)_z = \frac{\partial (-Ax)}{\partial x} - \frac{\partial (Ay)}{\partial y} = -2A \neq 0

\begin{aligned} (\nabla \times \mathbf{F}_2)_z &= \frac{\partial (-Ax)}{\partial x} \\ &\quad - \frac{\partial (Ay)}{\partial y} \\ &= -2A \\ &\neq 0 \end{aligned}

Only $\mathbf{F}_1$ is curl-free; it is $-\nabla V$ for $V = -Axy$ . Option (B).

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Full question on page 30 of the official paper (PDF)

GATE 2024, Q.39

Electromagnetics2 marksMCQ

A uniform plane wave $\mathbf{E}(x) = A_y\,\hat{a}_y\,e^{-j\frac{2\pi}{3}x}$ V/m travels in air in the $+x$ direction and is incident normally on a perfect electric conductor at $x = 0$ . Where is the first null of the total magnetic field in the air, measured from $x = 0$ , in metres?

Option A: $-\tfrac{3}{4}$
Option B: $-\tfrac{3}{2}$
Option C: $-6$
Option D: $-3$

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Key idea: At a perfect conductor the electric field has a null and the magnetic field a maximum, so the magnetic nulls are a quarter wavelength away.

Official answer: A

From the exponent, $\beta = 2\pi/3$ rad/m, so $\lambda = 2\pi/\beta = 3$ m. The reflected wave makes a standing wave with

E \propto \sin\beta x, \qquad H \propto \cos\beta x

$H$ is largest at the conductor and first vanishes where $\beta x = -\pi/2$ , at $x = -\lambda/4 = -0.75$ m. Option (A). The electric field's first null in the air is at $-\lambda/2 = -1.5$ m, option (B).

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Full question on page 44 of the official paper (PDF)

GATE 2024, Q.58

Electromagnetics2 marksNAT

A lossless line with $Z_0 = 50$ Ω has an unknown load with $|\Gamma| = 0.6$ . Moving towards the generator, what is the maximum magnitude of the input impedance looking towards the load, in ohms?

Answer format: Ohms

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Key idea: At a voltage maximum the input impedance is purely real and equal to $Z_0$ times the VSWR.

Official answer: 200

\text{VSWR} = \frac{1 + 0.6}{1 - 0.6} = 4, \qquad |Z_{in}|_{max} = 4 \times 50 = 200\ \Omega

\begin{aligned} \text{VSWR} &= \frac{1 + 0.6}{1 - 0.6} \\ &= 4, \\ |Z_{in}|_{max} &= 4 \times 50 \\ &= 200\ \Omega \end{aligned}

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Full question on page 14 of the official paper (PDF)

GATE 2023, Q.20

Electromagnetics1 markMCQ

A narrow-band signal propagates in a lossless dielectric medium ( $\varepsilon_r = 4$ , $\mu_r = 1$ ) with phase velocity $v_p$ and group velocity $v_g$ . Which statement is true? ( $c$ is the velocity of light in vacuum.)

Option A: $v_p > c$ , $v_g > c$
Option B: $v_p < c$ , $v_g > c$
Option C: $v_p > c$ , $v_g < c$
Option D: $v_p < c$ , $v_g < c$

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Key idea: A lossless dielectric with constant $\varepsilon_r$ is not dispersive, so $v_g = v_p$ .

Official answer: D

v_p = \frac{\omega}{\beta} = \frac{c}{\sqrt{\mu_r \varepsilon_r}} = \frac{c}{2}

$\beta = \omega\sqrt{\mu\varepsilon}$ is proportional to $\omega$ , so $v_g = d\omega/d\beta$ is also $c/2$ . Both are less than $c$ . Option (D).

Full question on page 37 of the official paper (PDF)

GATE 2023, Q.51

Electromagnetics2 marksMCQ

The S-parameters of a two-port network are $[S] = \begin{bmatrix} S_{11} & S_{12} \\ S_{21} & S_{22} \end{bmatrix}$ , with reference to $Z_0$ . Two lossless transmission line sections of electrical lengths $\theta_1 = \beta l_1$ and $\theta_2 = \beta l_2$ are added to the input and output ports, respectively. What are the S-parameters $[S']$ of the resulting two-port?

Option A: $\begin{bmatrix} S_{11}e^{-j2\theta_1} & S_{12}e^{-j(\theta_1 + \theta_2)} \\ S_{21}e^{-j(\theta_1 + \theta_2)} & S_{22}e^{-j2\theta_2} \end{bmatrix}$
Option B: $\begin{bmatrix} S_{11}e^{j2\theta_1} & S_{12}e^{-j(\theta_1 + \theta_2)} \\ S_{21}e^{-j(\theta_1 + \theta_2)} & S_{22}e^{j2\theta_2} \end{bmatrix}$
Option C: $\begin{bmatrix} S_{11}e^{j2\theta_1} & S_{12}e^{j(\theta_1 + \theta_2)} \\ S_{21}e^{j(\theta_1 + \theta_2)} & S_{22}e^{j2\theta_2} \end{bmatrix}$
Option D: $\begin{bmatrix} S_{11}e^{-j2\theta_1} & S_{12}e^{j(\theta_1 + \theta_2)} \\ S_{21}e^{j(\theta_1 + \theta_2)} & S_{22}e^{-j2\theta_2} \end{bmatrix}$

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Key idea: Each pass along a lossless line delays a wave by its electrical length, a factor $e^{-j\theta}$ .

Official answer: A

A wave reflected at port 1 crosses the input line twice, in and back out: $S'_{11} = S_{11}e^{-j2\theta_1}$ . Likewise $S'_{22} = S_{22}e^{-j2\theta_2}$ .

A wave passing through crosses both lines once each: $S'_{21} = S_{21}e^{-j(\theta_1 + \theta_2)}$ , and the same for $S'_{12}$ . Every factor is a delay, so every exponent is negative. Option (A).

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Full question on page 38 of the official paper (PDF)

GATE 2023, Q.52

Electromagnetics2 marksMSQ

The standing wave ratio on a 50 Ω lossless line terminated in an unknown load is 2.0. Successive voltage minima are 30 cm apart, and the first minimum is 10 cm from the load. $Z_L$ can be replaced by an equivalent length $l_m$ of the same line terminated in a resistance $R_m$ . What are $R_m$ and $l_m$ ?

One or more options are correct.

Option A: $R_m = 100\ \Omega$ , $l_m = 20$ cm
Option B: $R_m = 25\ \Omega$ , $l_m = 20$ cm
Option C: $R_m = 100\ \Omega$ , $l_m = 5$ cm
Option D: $R_m = 25\ \Omega$ , $l_m = 5$ cm

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Key idea: The impedance is purely resistive at every voltage minimum ( $Z_0/S$ ) and maximum ( $Z_0 S$ ), and these repeat every $\lambda/2$ , so the line can be extended past the load to one of them.

Official answer: B and C

Minima are $\lambda/2$ apart, so $\lambda = 60$ cm. At a voltage minimum the line looks like $Z_0/S = 25$ Ω, and at a maximum like $Z_0 S = 100$ Ω. The load is then the input impedance of a length $l_m$ of line ending in that resistance, measured from the load away from the generator.

Minima sit at 10 cm, 40 cm, ... towards the generator, so also at $10 - 30 = -20$ cm, 20 cm beyond the load: $R_m = 25$ Ω with $l_m = 20$ cm.
Maxima sit $\lambda/4 = 15$ cm from each minimum, at 25 cm and at $-5$ cm, 5 cm beyond the load: $R_m = 100$ Ω with $l_m = 5$ cm.

Options (B) and (C).

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Full question on page 39 of the official paper (PDF)

GATE 2023, Q.53

Electromagnetics2 marksMSQ

The electric field of a plane wave is $\mathbf{E} = \mathbf{a}_x C_{1x}\cos(\omega t - \beta z) + \mathbf{a}_y C_{1y}\cos(\omega t - \beta z + \theta)$ V/m. Which combination(s) give a left-handed elliptically polarised (LHEP) wave?

One or more options are correct.

Option A: $C_{1x} = 1$ , $C_{1y} = 1$ , $\theta = \pi/4$
Option B: $C_{1x} = 2$ , $C_{1y} = 1$ , $\theta = \pi/2$
Option C: $C_{1x} = 1$ , $C_{1y} = 2$ , $\theta = 3\pi/2$
Option D: $C_{1x} = 2$ , $C_{1y} = 1$ , $\theta = 3\pi/4$

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Key idea: The wave travels along $+z$ ; when $E_y$ leads $E_x$ ( $0 < \theta < \pi$ ), the field turns from $+x$ towards $-y$ , left-handed about $+z$ .

Official answer: A, B and D

At $z = 0$ , take $\theta = \pi/2$ : $E_x = C_{1x}\cos\omega t$ and $E_y = -C_{1y}\sin\omega t$ . As $t$ grows from 0, $\mathbf{E}$ moves from $+x$ towards $-y$ : clockwise as seen looking along $+z$ , opposite to the right-hand rule about the direction of travel. That is left-handed, and the same holds for any $0 < \theta < \pi$ , since the sign of $\sin\theta$ sets the sense. $\pi < \theta < 2\pi$ gives right-handed.

(A) $\theta = \pi/4$ : left-handed, and elliptical (equal amplitudes, but $\theta \neq \pi/2$ ). Valid.
(B) $\theta = \pi/2$ with unequal amplitudes: left-handed elliptical. Valid.
(C) $\theta = 3\pi/2$ : right-handed. Not valid.
(D) $\theta = 3\pi/4$ : left-handed elliptical. Valid.

Options (A), (B) and (D). This uses the IEEE convention, in which handedness is judged looking along the direction of travel.

Full question on page 40 of the official paper (PDF)

GATE 2023, Q.54

Electromagnetics2 marksMSQ

Which of the circuits in the figure represent a lumped-element equivalent of an infinitesimal section $\Delta z$ of a transmission line?

Four lumped circuits, as described in the options: (A) an L section with C delta z over 2 at the input and G delta z over 2 at the output; (B) an L section with G delta z and C delta z at the input; (C) a pi section with G delta z over 2 and C delta z over 2 at each end; (D) a T section with R delta z over 2 and L delta z over 2 on each side of a shunt G delta z and C delta z. — Figure from the official GATE 2023 EC paper, IIT Kanpur

One or more options are correct.

Option A: Circuit (A): series $L\Delta z$ and $R\Delta z$ , with $C\Delta z/2$ across the input and $G\Delta z/2$ across the output
Option B: Circuit (B): $G\Delta z$ and $C\Delta z$ across the input, then series $L\Delta z$ and $R\Delta z$
Option C: Circuit (C): series $L\Delta z$ and $R\Delta z$ , with $G\Delta z/2$ and $C\Delta z/2$ across each end
Option D: Circuit (D): series $R\Delta z/2$ and $L\Delta z/2$ on each side of a shunt $G\Delta z$ and $C\Delta z$

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Key idea: A section of length $\Delta z$ must contain the full series $R\Delta z$ , $L\Delta z$ and the full shunt $G\Delta z$ , $C\Delta z$ , however they are split.

Official answer: B, C and D

As $\Delta z \to 0$ , it doesn't matter where the shunt elements sit (an L, $\pi$ or T section all reduce to the telegrapher's equations), but the totals must be $R\Delta z$ , $L\Delta z$ , $G\Delta z$ and $C\Delta z$ .

(A) has only $C\Delta z/2$ and $G\Delta z/2$ in total: half the shunt elements. Not valid.
(B) is an L section with the full shunt elements. Valid.
(C) is a $\pi$ section: two halves of each shunt element. Valid.
(D) is a T section: two halves of each series element. Valid.

Options (B), (C) and (D).

Full question on page 42 of the official paper (PDF)

GATE 2023, Q.60

Electromagnetics2 marksNAT

A transparent dielectric coating is applied to glass ( $\varepsilon_r = 4$ , $\mu_r = 1$ ) to eliminate the reflection of red light ( $\lambda_0 = 0.75$ µm). What is the minimum thickness of the coating, in µm?

Answer format: µm, rounded off to two decimal places

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Key idea: A quarter-wave layer cancels reflection when its index is the geometric mean of the media on either side.

Official answer: 0.12 to 0.14

Glass has $n = \sqrt{4} = 2$ and air $n = 1$ , so the coating needs $n_c = \sqrt{1 \times 2} = \sqrt{2}$ and a thickness of a quarter wavelength in the coating:

d = \frac{\lambda_0}{4 n_c} = \frac{0.75}{4\sqrt{2}} = 0.1326\ \mu\text{m}

\begin{aligned} d &= \frac{\lambda_0}{4 n_c} \\ &= \frac{0.75}{4\sqrt{2}} \\ &= 0.1326\ \mu\text{m} \end{aligned}

Answer: 0.13 µm.