UPSC Electrical Engineering 2026 — Paper I

(a) calculate: given > formula > substitution > result with units > interpretation | (b) analyse: intro > causes > effects > stakeholders/linkages > way forward | (c) calculate: given > formula > substitution > result with units > interpretation | (d) calculate: given > formula > substitution > result with units > interpretation | (e) calculate: given > formula > substitution > result with units > interpretation Full marks: Complete method with correct results and clear interpretation

• V_B = 20 V fixed by the ideal source; the 5 Ω resistor does not enter the nodal equation
• Power 16 W in 1 Ω gives |I| = 4 A; KCL at A: 10 = V_A/2 + I, so V_A = 12 V and i₀ = 6 A
• Dependent-source equation with the polarity as drawn: 20 = (V_A − I) + k i₀
• k = 2, with the check that the loss in 1 Ω is 16 W
• Both stages are linear and time-invariant, so the cascade is LTI
• h₁(t) = u(t) − u(t − 3) and h₂(t) = u(t) − u(t − 6)
• h = h₁ * h₂ is a trapezoid: t on 0 to 3, 3 on 3 to 6, 9 − t on 6 to 9, zero outside
• Plot with labelled breakpoints 0, 3, 6, 9 and peak 3
• B = μ₀I/(2πr) for each wire with r = √(x² + d²/4)
• Vector addition: components along x cancel by symmetry
• B = μ₀I x/(π(x² + d²/4)), directed along y (sign follows x)
• Checks: B = 0 at x = 0 and B → μ₀(2I)/(2πx) at large x
• First decoder: D₀ = X′Y′, D₃ = XY; only D₀ and D₃ are connected, D₁ and D₂ are open
• A₂ = D₀ ⊕ D₃ = X ⊙ Y (XNOR); B₂ = its complement = X ⊕ Y
• Second decoder: Y₁ = A₂′B₂ and Y₂ = A₂B₂′
• Y₁ = X ⊕ Y = XY′ + X′Y and Y₂ = X ⊙ Y = XY + X′Y′; Y₀ = Y₃ = 0
• Rated: I_L = 20.45 A, I_f = 0.2273 A, R_f = 968 Ω, I_a = 20.68 A, R_a = 0.4676 Ω, E = 229.67 V
• Flux and I_f fall by 1/1.2: I_f′ = 0.1894 A; total field resistance 1161.6 Ω, added resistance = 193.6 Ω
• Field-circuit loss = 220 × 0.1894 = 41.7 W (34.7 W in field, 6.9 W in the external resistor)
• Input = 4500 + 199.3 + 41.7 + 250 ≈ 4991 W; developed voltage ≈ 229.7 V

(a(i)) In a series circuit (6 ohm + 0.0255 H forming the Vb section, R + C forming the Va section) across 240 V, 50 Hz, find R and C so that |Vb| = 3|Va| AND Vb, Va are in quadrature; then state the phase relation between V and Vb, and between Va and I. (a(ii)) Find R in the two-source ladder so that the power supplied by the 100 V source equals the power supplied by the 10 A source. Requires expressing both supplied powers in terms of the node voltages and solving simultaneously. (b(i)) For the triangular pulse x(t) with vertices (-1,0), (1,2), (3,0), evaluate four Fourier-transform quantities WITHOUT computing X(omega) explicitly: X(0), the integral of X(omega) over all omega, the integral of |X(omega)|^2, and the phase angle of X(omega). Each of the four must be answered using a transform property. (b(ii)) Determine the two-sided Z-transform of the non-causal sequence x(n) = |n|(1/2)^|n| and state the corresponding ROC. Both the closed-form X(z) and the annular ROC must appear. (c) A plane wave in air is normally incident on a lossless dielectric SLAB of thickness lambda/8 with eta = 260 ohm (air on both sides). Compute the standing-wave ratio in the air region in front of the slab. Requires a quarter-wave-line-type input-impedance transformation followed by a reflection-coefficient/SWR calculation.

• X_L ≈ 8 Ω, Z_b = 6 + j8 = 10 Ω at 53.13°; |Z_a| = 10/3 Ω from V_b = 3V_a
• Quadrature: 53.13° + φ_a = 90°, so φ_a = 36.87°
• R = 2.67 Ω, X_C = 2 Ω, C ≈ 1592 μF
• V lags V_b by 18.43° (I = 22.77 A, V leads I by 34.7°); V_a lags I by 36.87°
• Node equations at the 10 Ω node and the R node
• P_v = 100(100 − V₁)/5 and P_i = 10 V₂, set equal
• V₁ = V₂ = 66.67 V, so no current in 20 Ω
• R = 20/3 ≈ 6.67 Ω, with both powers 666.7 W
• X(0) = area = 4
• ∫X(ω)dω = 2πx(0) = 2π
• ∫|X|²dω = 2π∫x²dt = 2π × 16/3 = 32π/3 by Parseval
• Phase = −ω: pulse is an even triangle shifted by 1, so X = G(ω)e^(−jω) with G real and non-negative
• Split into causal part n(1/2)ⁿ and anti-causal part with n < 0
• Right part 2z/(2z − 1)², |z| > 1/2; left part 2z/(z − 2)², |z| < 2
• X(z) = 2z/(2z−1)² + 2z/(z−2)² = 2z(5z² − 8z + 5)/[(2z − 1)²(z − 2)²]
• ROC is the annulus 1/2 < |z| < 2, including the unit circle
• Termination is η₀ = 377 Ω (air on both sides); electrical length π/4
• Z_in = η(η₀ + jη tan βd)/(η + jη₀ tan βd) ≈ 243 − j92.4 Ω
• |Γ| = |Z_in − η₀|/|Z_in + η₀| ≈ 0.26
• SWR = (1 + |Γ|)/(1 − |Γ|) ≈ 1.70

(a(i)) A first-order switching transient: with K open the circuit is in steady state; K closes at t = 0 connecting R_L = 20 ohm. Find i_RL(t) as an explicit time expression and state its initial (t = 0+) and final (t = infinity) values. (a(ii)) Find all four y-parameters (y11, y12, y21, y22) of the given two-port containing a 1 ohm shunt, a 2 ohm series arm, a 2 ohm shunt and a current-controlled current source 3*I1. The dependent source makes the network non-reciprocal, so y12 and y21 must differ. (b) A diode bridges the two mid-points of a resistive bridge fed by 20 V; the diode is described by a piecewise-linear forward characteristic. Find (i) the diode current, (ii) the diode power loss and (iii) the total power supplied by the 20 V source - all three parts must be answered. (c) Compute the a.c. resistance per metre of a coaxial copper line at 400 MHz given inner-conductor outer diameter 2 mm, outer-conductor inner diameter 7 mm, conductor thickness >> skin depth, resistivity 1.724e-8 ohm-m and mu_r = 1. Requires the skin-depth/surface-resistance route, not the d.c. cross-sectional formula.

• Inductor current before switching: 50/(5 + 20) = 2 A, continuous at t = 0
• Thevenin behind the inductor branch: 40 V and 4 Ω; τ = 1/24 s; i_L(∞) = 5/3 A
• i_RL(t) = 5/3 − (1/15)e^(−24t) A, from v = 20 i_L + L di_L/dt
• Initial value 1.6 A at 0⁺ (jump from 0) and final value 1.667 A
• Definition I₁ = y₁₁V₁ + y₁₂V₂, I₂ = y₂₁V₁ + y₂₂V₂
• Port 1: I₁ = 1.5V₁ − 0.5V₂, so y₁₁ = 1.5 S and y₁₂ = −0.5 S
• KCL at port-2 node with the source 3I₁ drawing current out of the node
• y₂₁ = 4 S and y₂₂ = −0.5 S
• Diode model from Fig. (2): v_d = 0.7 + 0.2 i_d for 0 ≤ i_d ≤ 1 A, r_d = 0.2 Ω
• Thevenin across the diode: V_th = 5 V, R_th = (2∥6) + (8∥8) = 5.5 Ω
• i_d = 4.3/5.7 = 0.754 A with v_d = 0.851 V, both in range
• Diode power 0.642 W; source power ≈ 78.8 W from I_s = 3.94 A
• Skin effect: R_s = √(πfμ₀ρ) = 5.22 × 10⁻³ Ω per square
• Skin depth 3.3 μm much less than the thickness, so surface resistance applies
• Radii a = 1 mm and b = 3.5 mm
• R = (R_s/2π)(1/a + 1/b) ≈ 1.07 Ω/m

(a(i)) x(t) = cos(20*pi*t + pi/4) is ideally sampled at 30 samples/s and the impulse train is filtered by h(t) = [sin(pi t)/(pi t)] cos(80*pi*t - pi/2). Find the closed-form output signal - this needs the sampled spectrum, identification of h(t) as a narrow band-pass filter, and the magnitude AND phase of H at the surviving component. (a(ii)) Given a 9-point sequence x(n) and Y(k) = e^(j 4 pi k/3) X(k), determine y(n) WITHOUT computing any DFT - i.e. recognise the circular time-shift property and apply it. (b(i)) For the Darlington emitter-follower with a 100 k / R2 divider bias, find R2 given V_o = +5.0 V across R_o = 1 kohm, V_CC = 12 V, beta1 = 35, beta2 = 30, V_BE = 0.7 V for both. The question explicitly asks the formula used to be DERIVED, so the Darlington current relation must be shown. (b(ii)) A 240 V, 50 Hz single-phase diode bridge feeds R_L = 100 ohm; diode D2 has failed to a 1 kohm resistance. SKETCH the load voltage, load current and source current waveforms under this fault, AND calculate the average load voltage and the r.m.s. source current. Both the sketches and the two numbers carry marks. (c) A 50 cm lossless line (L = 0.2 uH/m, C = 80 pF/m) at 100 MHz is short-circuited at z = 0, with Z_L = 50 + j20 ohm shunted ACROSS the line at z = -20 cm. With 100 V at the input (z = -50 cm), find the average power delivered to Z_L.

• f₀ = 10 Hz, f_s = 30 Hz; X_s(f) = f_s ΣX(f − kf_s), lines at ±10 + 30k Hz
• sinc ↔ rect; h(t) is a narrow band-pass filter, pass band 39.5–40.5 Hz
• H(40) = ½ e^(−jπ/2), gain ½ and phase −90°; only the 40 Hz line (k = 1) survives
• y(t) = 15 cos(80πt − π/4)
• N = 9 and e^(j4πk/3) = e^(j2πk·6/9)
• Circular time-shift property of the DFT
• y(n) = x((n − 3)) mod 9 (equivalently x((n + 6)) mod 9)
• y(n) = {−2, −2, 4, 3, 5, 1, 4, −3, 5}
• Derivation I_E2 = (1 + β₁)(1 + β₂)I_B1 and V_B1 = V_o + 2V_BE
• I_E2 = 5 mA, I_B1 = 4.48 μA, V_B1 = 6.4 V
• I_R1 = 56 μA and I_R2 = I_R1 − I_B1 = 51.5 μA
• R₂ = 6.4/51.5 μA ≈ 124 kΩ
• V_m = 339.4 V; positive half: 1100 Ω in the path, v_L = v_s/11 (peak 30.9 V)
• Negative half: v_L = |v_s| (peak 339.4 V) and source current |v_s|(1/100 + 1/1000), peak 3.73 A
• Three waveform sketches: unbalanced load voltage/current and asymmetric source current
• V_L(avg) ≈ 117.9 V and I_s(rms) ≈ 1.87 A
• Z₀ = 50 Ω, v = 2.5 × 10⁸ m/s, λ = 2.5 m (0.2 m = 0.08λ and 0.3 m = 0.12λ)
• Shorted stub Z = j50 tan 28.8° = j27.5 Ω in parallel with Z_L: Z_j = 7.94 + j19.9 Ω
• Z_in = 36.2 + j98.3 Ω by transforming through 0.12λ
• P = |V|²Re(Z_in)/|Z_in|² ≈ 32.9 W (100 V taken as r.m.s.)

(a) For the op-amp circuit (series C = 1 uF, shunt R1 = 10 k into the non-inverting input, feedback R3 = 10 k with R2 = 22 k to ground), derive the frequency-domain transfer function and give the output expression - magnitude AND phase - for each of the two input components 1.5 sin 62.8t and 0.5 sin 188.4t. (b) From DC, short-circuit and open-circuit tests on a 110/220 V, 1.5 kVA, 50 Hz transformer, predict its behaviour at 60 Hz with rated voltage and kVA: (i) the maximum efficiency at unity power factor and (ii) the no-load current, input power and power factor at the LV side. The core loss is purely eddy-current (no hysteresis), which is the key to the frequency scaling. (c) A thyristor is fired at t = 0 in an L-C circuit fed by V_s = 200 V, with an initial inductor current of 250 A and zero initial capacitor voltage. Determine (a) the conduction time of T (the time for the resonant current to reach zero) and (b) the capacitor voltage at that instant. Both parts must be answered. (d) X is Bernoulli with P(X=1) = p and P(X=0) = 1-p. Y is the sum of n i.i.d. copies. State and justify the probability distribution of Y - i.e. identify it as binomial, give the pmf/CDF and (desirably) the mean and variance. (e) A series R-L-C circuit on 300 V, 50 Hz gives a maximum current of 0.3 A as L is varied, and the capacitor voltage is then 900 V. Determine all three of R, L and C.

• High-pass RC at the input, R₁C = 0.01 s, ω_c = 100 rad/s
• Non-inverting gain 1 + R₃/R₂ = 1.4545; V_o/V_in = 1.4545 jωR₁C/(1 + jωR₁C)
• 1.16 sin(62.8t + 57.9°) for the first component
• 0.642 sin(188.4t + 28.0°) for the second component; phase lead in both
• R₁ = 0.667 Ω from the DC test; R_eq = 1.6 Ω and X_eq = 3.67 Ω from the short-circuit test
• I₀ = 2.273 A at 50 Hz, R₀ = 242 Ω, X₀ = 49.4 Ω
• Eddy-current core loss is independent of frequency at constant voltage: 50 W at 60 Hz
• η_max = 86.0% at x = 0.41; I₀(60 Hz) = 1.91 A, 50 W, power factor 0.238 lag
• ω₀ = 10⁵ rad/s and Z₀ = √(L/C) = 1 Ω
• i(t) = 250 cos ω₀t + 200 sin ω₀t A, peak about 320 A
• Conduction time until i = 0: 22.5 μs
• v_C = V_s(1 − cos ω₀t) + I₀Z₀ sin ω₀t = 520 V at turn-off
• Y is binomial B(n, p), justified from the product of moment generating functions or by counting successes
• P(Y = k) = C(n, k) pᵏ (1 − p)^(n−k), k = 0, 1, …, n
• Distribution function F_Y(y) as the sum of the pmf up to ⌊y⌋
• E[Y] = np, Var[Y] = np(1 − p)
• Maximum current means series resonance, X_L = X_C
• R = 300/0.3 = 1000 Ω
• X_C = 900/0.3 = 3000 Ω = X_L
• C = 1.06 μF and L = 9.55 H; Q = 3

(a) Full synchronous sequential design of a MOD-6 DOWN counter with J-K flip-flops: state sequence, state/excitation table, K-map simplification for all six J and K inputs, the logic diagram, and a check on the unused states. (b) For a 415 V, 50 Hz star-connected cage induction motor with starting current = 10 x no-load current, starting torque = (1/3) x pull-out torque, and no-load input 300 W at 0.2 pf lag, find the starting current AND the input power factor at 5 Hz with the rated V/f ratio maintained. Stator leakage impedance and no-load rotor current are neglected; core loss has no hysteresis component. (c) A three-stage cascade with F1 = 10, F2 = 20, F3 = 30 and G1 = 20, G2 = 40, G3 = 50 receives 1 x 10^-16 W of noise power. Compute the OUTPUT noise power, which needs Friis' formula for the overall noise figure and the definition of noise figure.

• Three flip-flops (2³ ≥ 6) with a common clock; down sequence 101-100-011-010-001-000-101
• J-K excitation table and the state/excitation table for all six transitions
• J₀ = K₀ = 1, J₁ = Q₂Q₀′, K₁ = Q₀′, J₂ = Q₁′Q₀′, K₂ = Q₀′ from K-maps with 110 and 111 as don't cares
• Logic diagram and check that 110 and 111 return to the sequence (self-correcting)
• V_ph = 239.6 V, I₀ = 2.087 A, R_c = 574 Ω, X_m = 117.2 Ω
• T_st/T_max = 2s_m/(1 + s_m²) = ⅓ gives s_m = 3 − 2√2 = 0.172 = R₂′/X₂′
• |I₀ + I₂′| = 10 I₀ gives X₂′ = 12.57 Ω and R₂′ = 2.16 Ω at 50 Hz
• At 5 Hz: V = 23.96 V, X₂′ = 1.257 Ω, X_m = 11.72 Ω, R₂′ and R_c unchanged; I_st = 10.8 A at pf 0.771 lag
• Friis' formula F = F₁ + (F₂ − 1)/G₁ + (F₃ − 1)/(G₁G₂)
• F ≈ 10.99 and G = 40 000
• Noise figure definition N_o = F G N_i
• P_no ≈ 4.39 × 10⁻¹¹ W

(a) Design a PLA that squares a 3-bit binary number: build the truth table, minimise all output functions, determine the MINIMUM number of product terms, and construct the PLA programming table. All three deliverables (minimum product-term count, simplified expressions, programming table) are demanded. (b(i)) For a single-phase half-wave (one-pulse) SCR converter with R-L-E load (Vs = 230 V, 50 Hz, R = 2 ohm, L = 1 mH, E = 120 V, alpha = 25 deg, beta = 220 deg), find three voltages across the SCR: (1) at the triggering instant, (2) when the current decays to zero, and (3) the peak inverse voltage. (b(ii)) Prove (not merely quote) that for a type-A chopper feeding an R-L-E load in continuous conduction, the maximum average freewheeling-diode current is I_fd(max) = (Vs/4R)(1 - E/Vs)^2. Requires the average-current expression and an optimisation over the duty ratio. (c) The given piecewise signal m(t) modulates a carrier once in FM and once in PM. Find the relation between k_p and k_f that makes the MAXIMUM phase deviation equal in the two cases. Requires max|m(t)| for PM and max|integral of m| for FM.

• Squares 0, 1, 4, 9, 16, 25, 36, 49 need six output bits
• B₀ = A₀, B₁ = 0, B₂ = A₁A₀′, B₃ = A₂′A₁A₀ + A₂A₁′A₀, B₄ = A₂A₁′ + A₂A₀, B₅ = A₂A₁
• Minimum of 7 distinct product terms
• Programming table for the AND plane (3 inputs, 7 terms) and OR plane (6 outputs), a 3 × 7 × 6 PLA
• V_m = 325.3 V and v_SCR = v_s − E while the SCR is blocking
• +17.5 V at α = 25°
• −329 V at β = 220°
• PIV = V_m + E = 445 V
• V_o = δV_s and I_a = (δV_s − E)/R for continuous conduction
• Diode conducts for (1 − δ)T, so I_fd = (1 − δ)I_a
• Differentiation gives δ_opt = ½(1 + E/V_s), with a second-derivative check
• Substitution gives I_fd,max = (V_s/4R)(1 − E/V_s)²
• PM phase k_p m(t), maximum k_p × 1
• FM phase 2πk_f ∫m dt
• max|∫m| = 1.5 at t = 2 s, so maximum FM phase 3πk_f
• k_p = 3πk_f ≈ 9.42 k_f

(a) A 6.6 kV star-connected cylindrical-rotor alternator delivering MAXIMUM power to a 6.6 kV bus with I_f = 4.5 A. Using the OC (rated voltage at 5 A) and SC (100 A at 0.2 pf lag with 1.0 A) data, find (i) both the generated and the delivered power, (ii) the load angle and (iii) the power factor and reactive power. All three parts required. (b(i)) A 240 V, 1200 rpm, 15 A separately excited DC motor (R_a = 1.0 ohm) under chopper-controlled dynamic (rheostatic) braking with R_B = 20 ohm. Find (1) the duty ratio for 1000 rpm at 1.5 times rated braking torque and (2) the speed for duty ratio 0.5 at rated torque. (b(ii)) A 200 V, 875 rpm, 150 A separately excited DC motor (R_a = 0.06 ohm) fed by a single-phase fully controlled rectifier from 230 V, 50 Hz in continuous conduction. Compute (1) alpha for rated torque at 750 rpm, (2) alpha for rated torque at -500 rpm and (3) the speed at alpha = 160 degrees with rated torque. (c) An FM signal with 15 kHz deviation at a 5 kHz modulating frequency passes through cascaded frequency multipliers of 1.5 and 4. Find the output frequency deviation, the output modulation index and the approximate (Carson) bandwidth. All three quantities required.

• V = 3810.5 V/phase; E ∝ I_f so E = 3429.5 V at 4.5 A
• Z_s = 7.62 Ω at 78.46°, R_a = 1.524 Ω, X_s = 7.47 Ω
• Maximum power delivered when δ = θ = 78.5°; P_del = 4.00 MW
• P_gen = 5.66 MW (I = 602 A, copper loss 1.66 MW); pf 0.581 leading; Q ≈ 5.6 MVAr absorbed
• Kφ = 225/1200 = 0.1875 V/rpm
• Effective braking resistance (1 − δ)R_B and E = I_a[R_a + (1 − δ)R_B]
• 1.5 × rated torque gives I_a = 22.5 A; δ = 0.633
• δ = 0.5 at rated torque gives E = 165 V, N = 880 rpm
• V_dc = (2V_m/π) cos α = 207.07 cos α; Kφ = 0.2183 V/rpm
• 750 rpm: V_dc = 172.7 V, α = 33.5°
• −500 rpm: V_dc = −100.1 V, α = 118.9° (inverting mode)
• α = 160°: V_dc = −194.6 V, N ≈ −933 rpm
• Overall multiplication factor 6
• Deviation 6 × 15 = 90 kHz; f_m stays 5 kHz
• Modulation index β = 90/5 = 18
• Carson's rule BW = 2(Δf + f_m) = 190 kHz