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1 Index. Question Paper Analysis 2. Question Paper & Answer keys : , info@thegateacademy.com Copyright reserved. Web:

9% Signal & System Signal 8% & System 7% Digital Electronics Digital 6% Electronics 6% EMT 5% EMT 5% Measurement Measurement

2 ANALYSIS OF GATE 206 Electrical Engineering CN Mathematics % 4% Mathematics 5% Power Electronics Power 3% Electronics Power System 0% Power 9% System 8% GA 5% Electrical Machine 0% Network Theory Network 8% Theory 8% Control System 0% Control System 9% Signal & System Signal 8% & System 7% Digital Electronics Digital 6% Electronics 6% EMT 5% EMT 5% Measurement Measurement 3% 3% Analog Circuits Analog 5% Circuits 5% : , info@thegateacademy.com Copyright reserved. Web:

3 - EE 6 Feb Afternoon Session SUBJECT NO OF QUESTION Topics Asked in Paper Level of Toughness Total Marks Network Theory Control System Signal & System Digital Electronics Analog Circuits EMT Measurement Electrical Machines Power System Power Electronics Mathematics GA M: 4 2 M:2 M: 2 M: 4 M: 3 2 M: 2 M: 2 2 M: 2 M: 2 M:2 M: 2 M: 2 M: 2 M: M: 4 2 M: 3 M: 2 2 M: 3 M: 2 2 M: 4 M: 4 2 M: 5 M: 5 2 M: 5 Reduction of networks with controlled voltage and current sources; Transient; Resonance; AC circuit analysis Nyquist Plot; Routh Stability; Lag-lead system; Bode plot and transfer function LTI system; Fourier transform; Laplace transform ADC-Flash type; Multiplexer; Boolean algebra Transistor-DC biasing; Zener Diode with transistor Moderate 8 Tough 9 Tough 7 Easy 6 Easy 5 *Electrostatic field - Properties, Moderate 5 Moving coil and moving iron instrument Moderate 3 DC Machine-Shunt motor, efficiency; Transformer-Losses, Autotransformer; Induction machine, V/f control. Easy 0 Stability; LLG fault; Line parameters Tough 8 VSI; Rectifier - full wave - input power factor, efficiency; Chopper Complex integration; Differential Equation; infinite series; Probability-basic. Time & Work ;Paragraph; English fill in Blank;Number theory; Venn Diagram ; Mensuration& Area. Tough 0 Moderate 4 Easy 5 Total 65 Moderate 00 * Indicates Questions from New Syllabus Faculty Feedback: The question paper was bit tough and was little offbit than conventional GATE papers. Plenty of Numerical Answer Type (NAT) questions asked. Online Calculator was difficult to handle without Prectice. Numerical & Verbal ability was relatively easy. Practice previous Year Questions & Online Test Series will be beneficial. : , info@thegateacademy.com Copyright reserved. Web: 2

4 Question Paper & Answer Keys

5 GATE 206 Examination Electrical Engineering Test Date: 6/02/206 Test Time: 2:00 PM to 5:00 PM Subject Name: ELECTRICAL ENGINEERING Q NO. Section: General Aptitude [Ans. B] The man who is now municipal commissioner worked as a security guard at the university Q NO. 2 [Ans. A] Nobody knows how the Indian cricket team is going to put the difficult and seamer friendly wickets in Australia : , info@thegateacademy.com Copyright reserved. Web: 3

6 Q NO. 3 [Ans. C] Mock, Deride, Jeer all are synonym Q NO. 4 [Ans. D] In option A, B, C 2 nd letter and st letter is 2. In option D it is. Q NO. 5 [Ans. B] Let n =, then αn + β n α + βn α n = + β n α + = αβ = 4 β n n = 2, then α2 + β 2 α 2 + β 2 = α2 β 2 = 4 2 n = 3, then α3 + β 3 α 3 + β 3 = α3 β3 = 43 So, for n, solution is 4 n : , info@thegateacademy.com Copyright reserved. Web: 4

7 Q NO. 6 [Ans. A] Q NO. 7 [Ans. D] Q NO. 8 [Ans. D] : , info@thegateacademy.com Copyright reserved. Web: 5

8 Q NO. 9 [Ans. C] Q NO. 0 [Ans. C] 0 At x = 0 f(0) = Put x = 0 in all options. You will get f(x) = 2 x = x=0 Rest of the option do not match : , info@thegateacademy.com Copyright reserved. Web: 6

9 Q NO. Section: Technical [Ans. *]Range: 0 to 0 f(x) = x(x )(x 2) Put x = f(x) = 0 x = 2 f(x) = 0 x =.5 f(x) < 0 For all values of x between and 2 (Excluding and 2) f(x) is less than zero. Thus maximum value attained by f(x) is 0 Q NO. 2 [Ans. *]Range: 3 to 3 A = [ ] λ A λi = λ = 0 λ 0 λ λ = 0 [C 2 = C 2 C 3 ] λ λ ( λ)[ λ( λ) λ] + (λ + λ) = 0 ( λ)(λ 2 2λ) + 2λ = 0 λ[( λ)(λ 2) + 2] = 0 λ[ λ 2 + 3λ] = 0 λ 2 (λ 3) = 0 λ = 3 Q NO. 3 [Ans. A] f(t) = e 2t sin 5t u(t) We know L[sin at ] = a s 2 + a 2 L[e at f(t)] = f(s a) 5 Thus, L[sin 5(t)u(t)] = s L[e 2t sin 5t u(t)] = 5 s 2 4s + 29 : , info@thegateacademy.com Copyright reserved. Web: 7

10 Q NO. 4 [Ans. B] d 2 y dt dy dt + y = 0 (D 2 + 2D + ) = 0 D =, y = (C t + C 2 )e t Putting t = 0, y(0) =, C 2 = t =, y(s) = (C + C 2e t = 3e t C 2 =, C = 2 y(z) = (2.2 + )e 2 = 5e 2 Q NO. 5 [Ans. B] Singlarilies, Z = 2, 2 ± i Only Z = lies inside C 2 By residue theorem 2π(R /2 ) = 48πi c 3 Residue at 2 = R 2Z + 5 /2 = lim [(Z /2. z 2 (Z )] = 24 2 ) (Z2 + 4Z + 5) 3 : , info@thegateacademy.com Copyright reserved. Web: 8

11 Q NO. 6 [Ans. B] y(s) = s s + 2 (s/m) r(t) = cos t (i/p) Output y(t) = A cos(2t + Ψ) Ψ = 0 Phase angles ω = 2 ω y(t) = cos(2t + θ ± ϕ) ω ϕ = 90 tan ω 2 = 90 tan 2 2 ϕ = 45 Y(t) = 2 cos(2t + 45) 8 Ψ = 45 A = 2 8 = = 2 A = 2 ; Ψ = 45 Q NO. 7 [Ans. A] C(s) = 00 (s + ) 3 = 0 ( + jω) 3 G(jω) = 3 tan (ω) Phase cross over frequency (ω p ) 3 tan (ω p ) = 80 tan (ω p ) = 60 ω p = tan 60 = 3 : , info@thegateacademy.com Copyright reserved. Web: 9

12 Q NO. 8 [Ans. C] y(t) = x (t) cos(t) y 2 (t) = x 2 (t) cos(t) y 3 (t) = y (t) + y 2 (t) = x (t) cos t + x 2 (t) cos t = [x (t) + x 2 (t)] cos(t) Q NO. 9 [Ans. A] We know δ(at) = a δ(t) e t δ(2t 2)dt = 2 e t δ(t 2)dt = 2 e t = 2e Q NO. 0 [Ans. B] Temperature range pf 40 to 55 So total range in 95 Since resolution 0. So number of steps will be 950 To have 950 steps we need at least 0 bits : , info@thegateacademy.com Copyright reserved. Web: 0

13 Q NO. [Ans. D] 0 F A B B = ; F = A B = 0; F = A F = AB + A B = A B Q NO. 2 [Ans. *]Range: 8 to 20 : , info@thegateacademy.com Copyright reserved. Web:

0 V I 4.7 k 220 Ω I B 5.7 k 0.5 ma 470 Ω 0 5.7 I = = ma 4.7k I B = ( 0.5)mA = 0.5 ma 5.7 0.6 I E = = 0 ma 470 I E = I B + I C I C = 9.5 ma β = I C = 9.5 I B 0.5 = 9 Q NO. 3 [Ans.

14 0 V I 4.7 k 220 Ω I B 5.7 k 0.5 ma 470 Ω I = = ma 4.7k I B = ( 0.5)mA = 0.5 ma I E = = 0 ma 470 I E = I B + I C I C = 9.5 ma β = I C = 9.5 I B 0.5 = 9 Q NO. 3 [Ans. B] V is given as static field in time invariant Hence E = 0 Q NO. 4 [Ans. *]Range: 99 to 0 Toroid has field B μ As μ = 00 (inside field) Magnetic field density B at any point at a distance at r is B = μl 2πr Now, B at r = μ 0μ r I (Just inside toriod) 2πr and B at r + = μ 0I 2πr + (Just outside toriod) B at r = μ B r = 00 at r + : , info@thegateacademy.com Copyright reserved. Web: 2

15 Q NO. 5 [Ans. D] 2 Ω 2 Ω R A Ω Ω 2 Ω 2 Ω R B Ω Ω Ω R B Ω R A = R A + R A Q NO. 6 [Ans. C] : , info@thegateacademy.com Copyright reserved. Web: 3

16 f 0 = Nominal frequency ω s = ( f ) ω f s0. (i) 0 S max,t = ( f 0 f ) ( R 2 I ). (ii) X 20 From (i) and (ii) S max,t = ( ω s 0 ) ( R 2 ω s I ) X 20 S max,t = ( ω s 0 ) ( R 2 ω s I ) X 20 S max,t = ω s Q NO. 7 [Ans. *]Range:.9 to 2. 4 Ω 6 Ω l 5 Ω 4 Ω 6 Ω 2l 5 Ω And (2l) 2 5 = 0 I 2 = = = 0.5 So, I 2 4 = = 2 cal/sec : , info@thegateacademy.com Copyright reserved. Web: 4

17 Q NO. 8 [Ans. *] Range: 0.5 to0.5 I Ω I 2 Ω V I 2 Ω Here, I = 2I 2 = A I 2 = 2 A = 0.5 Q NO. 9 [Ans. B] Q NO. 20 [Ans. *] Range: 0.83 to 0.85 : , info@thegateacademy.com Copyright reserved. Web: 5

18 Q NO. 2 [Ans. A] Synchronous generator working at a lagging power factor, will supply active power and lagging reactive power, so this is source Synchronous motor working at an overexcited condition i.e., leading p.f. operation so it will take active supplying lagging reactive power. Induction generator generates active power but as there is no dc excitation in rotor so it takes lagging reactive power. Q NO. 22 [Ans. *]Range: 0.39 to 0.4 Average voltage across inductor is zero. V L(Avg) = 0 30(T ON ) 20(T OFF ) = 0 : , info@thegateacademy.com Copyright reserved. Web: 6

19 30(αT) = 20( α)t 30α + 2α = 20 50α = 20 α = 2 5 = 0.4 Q NO. 23 [Ans. *]Range: 69 to 7 I D = 0.0 V D V D = 0.0 I D =.7 V (y = mx + c) I D = V D 0.0 = P = V D I D =.7 ( ) P = 70 W : , info@thegateacademy.com Copyright reserved. Web: 7

20 Q NO. 24 [Ans. C] 4 pole, lap-connected separately excited dc motor, N = 6000 rpm Parallel path = 4 = Number of poles Current in each parallel path = A Current in aramture conductor = 0A 40 A 0 A 0 A 0 A 0 A Speed = 600 rpm N = 20 f = f = = 20 Hz 20 Time period, T = f = 20 = 50 msec So, 0 A 0 A I 50 msec : , info@thegateacademy.com Copyright reserved. Web: 8

21 Q NO. 25 [Ans. B] n: L Port Port 2 At port i.e, high voltage side impedance will be high and current will be low, So n 2 L. Q NO. 26 [Ans. *] Range: 0.2 to 0.2 Q NO. 27 [Ans. *] Range: 0.28 to 0.3 S = n n=0 n S = S = : , info@thegateacademy.com Copyright reserved. Web: 9

22 S ( ) = S ( ) = S = ( ) 2 According to question S = 2 Thus, 2 = = ± 2 0 < < So, = 2 = 0.29 ( ) 2 ( )2 = 2 Q NO. 28 [Ans. A] Eigen values of 2 2 matrix A be, 2 Eigen values of A 2 matrix A be, 4 Eigen values of 3A matrix A be 3, 6 Eigen values of 4I matrix A be 4 Eigen values of A 2 3A + 4I be 2, 4; Eigen vectors of A 2 3A + 4I be same as A Q NO. 29 [Ans. B] Result Rank (A T A) = Rank (A) Q NO. 30 : , info@thegateacademy.com Copyright reserved. Web: 20

23 [Ans. A] Magnitude (db) 20 db/dec 2 db 40 db/dec ω ω ω For measuring ω, 0 2 = 20[log 0.5 log ω ] = log ; ω 20 ω = 2 For measuring ω 2, 2 0 = 40(log ω 2 log 8) 2 40 = log ω 2 8 ; ω 2 = 4 Initial slope is 20 db/dec, thus in numerator jω term will be there. From point ω, slope decreased to 0, i.e., 20 db/dec applied. So, in denominator ( + term will be there 2 s) From point ω 2, slope is 40 db/dec. Thus in denominator ( + 2 term will be there. 4 s) Thus, transfer function looks like Q NO. 3 s ( + 0.5s)( s) 2 : , info@thegateacademy.com Copyright reserved. Web: 2

24 [Ans. *]Range: 5.9 to 6. The time response is given by x(t) = ϕ(t)x(0); Where, ϕ(t) = L (si A) (si A) = [ s 0 0 s ] [ ] = [s 0 0 s 2 ] ϕ(s) = (si A) = (s )(s 2) [s s ] s ϕ(s) = [ 0 0 ] s 2 ϕ(t) = [ et 0 0 e 2t] x(t) = [ et 0 0 e 2t] [ 2 2 ] = [ et 2 2 e 2t] y(t) = [ Q NO. 32 ] [ et e 2t] = et + e 2t ; y(t) = e log 2 + e 2 log e 2 t =log2 e = 6 [Ans. A] Nyquist plot of G(s)H(s) = s + 3 s 2 is as shown below (s 3) w = w = w = 0 w From the Nyquist plot G(s) H(s) encircle + j0 once in clockwise direction. Q NO. 33 : , info@thegateacademy.com Copyright reserved. Web: 22

25 [Ans. *]Range: 2 to 2 s s s + 3 = 0 s 3 + 3s s s + s + 3 = 0 (s + 3)(s s + ) = ±.5 = 3, 2 = 3, 2, 2 So, roots strictly less than is 2 and 3. Q NO. 34 [Ans. C] By observing X (jω)and X 2 (jω) we can say that they are not conjugate symmetric. Sin ce the Fourier transform is not conjugate symmetric the signal will not be real So, x (t). x 2 (t)are not real. Now the Fourier transform of x (t) x 2 (t) will be 2π X (jω) X 2(jω) and by looking at X (jω) and X 2 (jω) X (jω) X 2 (jω) will be conjugate symmetric and thus, x (t) x 2 (t) will be real Q NO. 35 : , info@thegateacademy.com Copyright reserved. Web: 23

[Ans. D] Given that impulse response is real and even, Thus H(jω) will also be real and even LTI e jω 0t System H(jω 0 )e jω 0t LTI e jω 0t System H( jω 0 )e jω 0t Since H(jω) is real and even thus

26 [Ans. D] Given that impulse response is real and even, Thus H(jω) will also be real and even LTI e jω 0t System H(jω 0 )e jω 0t LTI e jω 0t System H( jω 0 )e jω 0t Since H(jω) is real and even thus H(jω 0 ) = H( jω 0 ) Now cos(t)is input i. e., ejt + e jt is input 2 Output will be H(j)ejt + H( jω)e jt = H(j) [ ejt + e jt ] = H(j) cos(t) 2 2 If sin(t)is input i. e., ejt e jt is input 2 Output will be H(j)ejt H( jω)e jt = H(j) [ ejt e jt ] = H(j) sin(t) 2 2 So sin(t)and cos(t)are Eigen signals with same Eigenvalues. Q NO. 36 [Ans. C] : , info@thegateacademy.com Copyright reserved. Web: 24

27 5V J A K Q A J B K Clk Next when clock applied Q A toggles as J, K input of A connects to 5 V. Also, Q B toggles as previous state Q A values is J which is the input of J, K of B. Thus, output of next state is Q B Q NO. 37 [Ans. A] X 0 X X 2 B B X 2 X X X 0 [x 2 x + x 2 x ] = x 0 [x x 2 ] : , info@thegateacademy.com Copyright reserved. Web: 25

28 Q NO. 38 [Ans. D] Charge, Q is located at (0, 0) and 2θ is located at (6, 0) To find V at any point (x, y) Q V Q = 4π x 2 + y 2 2Q V 2Q = 4π ( (x 6) 2 + y 2 ) Q V total = 0 = 4π ( x 2 + y 2 ) + 2Q 4π ( (x 6) 2 + y 2 ) (x 6) 2 + y 2 = 2 ( x 2 + y 2 ) x x + y 2 = 4x 2 + 4y 2 3x 2 + 3y 2 + 2x = 36 x 2 + y 2 + 4x = 2 (x + 2) 2 + y 2 = 6 Q NO. 39 [Ans. *]Range:.9 to 2. S Ω S 2 2 Ω 3 V H 3 V At t = 0 : , info@thegateacademy.com Copyright reserved. Web: 26

2 Ω H 3 V At t = 0 +, Ω 3 V H 2 Ω 3 V KCL at node A, V A 3 + 3 2 + V A 3 = 0 2 2(V A 3) + 3 + (V A 3) = 0 3V A = 6, V A = 2 V A = L di(0+ ) = 2 dt di(0 + ) = 2 dt L = 2 = 2 A/sec Q NO. 40 [Ans.

29 2 Ω H 3 V At t = 0 +, Ω 3 V H 2 Ω 3 V KCL at node A, V A V A 3 = 0 2 2(V A 3) (V A 3) = 0 3V A = 6, V A = 2 V A = L di(0+ ) = 2 dt di(0 + ) = 2 dt L = 2 = 2 A/sec Q NO. 40 [Ans. *] Range: 47 to 49 KVA = = 000 KVA Without excessive heat dissipation means current should be constant (i.e.) KVA rating must be constant. In second case Active power, P = = 900 KW Reactive power in second case Q 2 = = KVAR Reactive power supplied by the three phase bank = = 64. KVAR Q bank ph = 64. = 54.7 KVAR 3 V ph = 3.3 =.9052 KV 3 (V ph Qc ph = )2 X c : , info@thegateacademy.com Copyright reserved. Web: 27

X c = (.9052 03 ) 2 54.7 0 3 = 66.36 Ω C = = 2πfX c 2π 50 66.36 = 47.96 μ Q NO. 4 [Ans. *] Range:.05 to.5 Base current I B = I f = 4270 A I p.u. = 4270 255.09 30 03 3 3.8 = 255.09 A = 3.402 p. u.

30 X c = ( ) = Ω C = = 2πfX c 2π = μ Q NO. 4 [Ans. *] Range:.05 to.5 Base current I B = I f = 4270 A I p.u. = = A = p. u. E a I g = X + (X 2 X 0 ) Where, X 0 = X 0 + 3(Z n + Z f ) X 0 = 3Z n = (3Z n ) Z n By solving the equation Z n =.07 p. u. Q NO. 42 [Ans. D] : , info@thegateacademy.com Copyright reserved. Web: 28

31 V B V A (V line in delta Δ) V C V AB lags V ab by 30 V B According to negative sequence phasors V A V B = V AB (V line in Star Y) Q NO. 43 [Ans. C] Input power factor = Power factor at ac mains = C.D.F. D.F. = 2 2 π cos α = 2 2 cos 30 = 0.78 π : , info@thegateacademy.com Copyright reserved. Web: 29

32 Q NO. 44 [Ans. *] Range: 9.9 to 0. m a = 0.8 V d (V 0 ) peak = m a [m 2 a ] = = 200 V (I 0 ) peak = (V 0 ) peak Z = 200 R 2 + (ωl) 2 = = 0 A : , info@thegateacademy.com Copyright reserved. Web: 30

33 Q NO. 45 [Ans. *] Range: 74 to 76 T Energy = V. i dt + V. i dt 0 T 2 = V [ 2 IT ] + I [ 2 VT 2] 0 = 600 [ ] + 00 [ ] Energy = 75 mj Q NO. 46 [Ans. *] Range:.4 to V, 50 Hz transformer, P i = 5000 Watt When, 200 V, 25 Hz Pi = 2000 Watt 46 V, 52 Hz P h P e =? P i = P h + P e P h fb m x : , info@thegateacademy.com Copyright reserved. Web: 3

34 P e f 2 2 B m As in the problem V f = = = = 8 = constant P h = Af And P e = Bf 2 From given data, 2000 = (P 25Hz ) i = A(25) + B(25) = (P 50 Hz ) i = A(50) + B(50) 2. 2 Solving and 2 A = 60, B = 0.8 (P h ) 52 Hz = Af = = 320 Watt (P e ) 52 Hz = Bf 2 = 0.8 (52) 2 = Watt (P h ) 52 Hz = 320 (P e ) 52 Hz =.4423 Q NO. 47 [Ans. *] Range: 86 to A 0.0 Ω I a I sh 44 Ω 220 V Stay losses = 375 Watt Total copper losses = I 2 a R a + I 2 sh R sh = 50 2 (0.0) + (5) 2 44 = 25 Watt O p η = O p + losses = (25) = ro 86.84% : , info@thegateacademy.com Copyright reserved. Web: 32

Q NO. 48 [Ans. *] Range:.58 to.62 Synchronous motor at leading p.f. X d = 0.8, ϕ = 36.86 X q = 0.6, R q = 0 tan ψ = V sin ϕ + I a. X a V cos ϕ ψ = 56.30 For synchronous motor at leading p.f. ψ = ϕ + δ δ = 9.

35 Q NO. 48 [Ans. *] Range:.58 to.62 Synchronous motor at leading p.f. X d = 0.8, ϕ = X q = 0.6, R q = 0 tan ψ = V sin ϕ + I a. X a V cos ϕ ψ = For synchronous motor at leading p.f. ψ = ϕ + δ δ = 9.70 Now, e = V cos δ + I d X d I d = I a sin ϕ = 0.83 E = () cos(9.7) + (0.83)(0.8) =.606 Q NO. 49 [Ans. C] 22 KVA, 2200 V/220 V, 50 Hz Distribution transformer is to be connected as on auto transformer to get an output voltage of 2420 V (kva) maximum as an auto transformer =? As voltage rating = is 2420 i.e. ( ) V Additive polarity, (kva) auto = (a 2winding + ) kva 2winding Where, a 2winding = = 0 (kva)auto = (0 + ) 22 = 242 Q NO. 50 : , info@thegateacademy.com Copyright reserved. Web: 33

36 [Ans. A] V 0 (rms) = 2 2 π V s sin d Pulse width where 2d = 20 d = 60 V 0 (rms) = 2 2 π V s sin 60 = 2 2 π V 3 s 2 = 234 V Q NO. 5 [Ans. *] Range:.42 to.45 In first case m 0 mm A L = ln D r = ln ( ) L = 0.97 μh m L 2 = =.085 μh m 3 L 2 = ln ( D ) m B 0 mm D D 3 ln ( D ) = = : , info@thegateacademy.com Copyright reserved. Web: 34

37 3 e D2 = D = 42.7 cm =.427 m Q NO. 52 [Ans. *] Range: to If we observe the parallel LC combination we get that at ω = 000 rad sec the parallel LC is at resonance thus it is open circuited. The circuit given in question can be redrawn as 4 Ω Ω 5 Ω So, I = ~ 0 sin(000 t) 00 sin 000t 0 = sin 00t ;So, peak value is Amp Q NO. 53 [Ans. *] Range: 0.30 to 0.33 Moving coil, V(t) = sin(ωt) 5 sin(3ωt) volt V = V avg = 00 V Moving iron, V 2 = V rms = ( ) : , info@thegateacademy.com Copyright reserved. Web: 35

38 = V 2 V = 0.32 Q NO. 54 [Ans. *] Range: 4.0 to 4.2 The resonant frequency for the circuit is ω 0 = 2 LC R L L C R 2 C L C Since, (R L = R C = R) So, the circuit will have zero real part of admittance when, R = L C So, R = μf = 4.4 Ω Q NO. 55 [Ans. *] Range:.25 to.50 : , info@thegateacademy.com Copyright reserved. Web: 36

39 A I 5 Ω 5 Ω 5 A 5 Ω 0I Ω Applying KCL at node A, we get V A 5 + V A 0 + V A + 0 I = So, 2V A + V A 0 + 2V A + 20 I = 5 5V A + 20 I = 60 Since, I = V A 0 0 So, 5V A + 2V A 20 = 60 7V A = 80 V A = 80 7 =.42 0 V : , info@thegateacademy.com Copyright reserved. Web: 37

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GATE : , Copyright reserved. Web:www.thegateacademy.com GATE-2016 Index 1. Question Paper Analysis 2. Question Paper & Answer keys : 080-617 66 222, info@thegateacademy.com Copyright reserved. Web:www.thegateacademy.com ANALYSIS OF GATE 2016 Electrical Engineering