for some t 0 (D) A bounded signal is always finite A bounded signal always possesses some finite energy.

Transcription

1 GATE EE 6 Q.1 - Q. carry one mark each. MCQ 1.1 SOL 1.1 The following is true (A) A finite signal is always bounded (B) A bounded signal always possesses finite energy (C) A bounded signal is always zero outside the interval [ t, t] for some t (D) A bounded signal is always finite A bounded signal always possesses some finite energy. t E gt () dt< 3 Hence (B) is correct option. -# t MCQ 1. xt () is a real valued function of a real variable with period T. Its trigonometric Fourier Series expansion contains no terms of frequency ω π( k)/ T; k 1, g Also, no sine terms are present. Then xt () satisfies the equation (A) xt () xt ( T) (B) xt () xt ( t) x( t) (C) xt ( ) xt ( t) xt ( T / ) (D) xt ( ) xt ( T) xt ( T / ) SOL 1. Trigonometric Fourier series is given as 3 / n 1 xt () A+ an cos nωt+ bn sin nω t Since there are no sine terms, so bn T b n # xt ()sinnωtdt T T / T x() τ sin nωτ dτ+ x() t sin nωtdt T G Where τ T t & dτ dt # # T / T / T xt ( t) sin nω( T t)( dt) T ; # + xt ()sinnωtdte T # T /

2 Page GATE EE 6 T O T xt ( t)sinn π T t dt+ T ; # ` T j xt ()sinnωtdte T / O T # T / xt ( t) sin( nπ nω) dt T ; # + xt ()sinnωtdte T / T # T T / xt ( t) sin( nωtdt ) + T ; # xt ()sinnωtdte T / b n if xt () xt ( t) From half wave symmetry we know that if xt () xt! T ` j # T T / Then Fourier series of xt () contains only odd harmonics. Hence (C) is correct option. MCQ 1.3 In the figure the current source is 1+ A, R 1Ω, the impedances are ZC j Ω and Z j Ω. The Thevenin equivalent looking into the circuit across X-Y is L (A) + V,(1+ j) Ω (B) + 45 V,(1 j) Ω % (C) + 45 V,(1 + j) Ω % (D) + 45 V, (1 + j) Ω % SOL 1.3 Thevenin voltage: V th IR ( + ZL+ ZC) 1+ c[1+ j j] 1(1 +j) % + 45 V Thevenin impedance:

3 Page 3 GATE EE 6 Z th R+ ZL+ ZC 1+ j j (1 +j) Ω Hence (D) is correct option. MCQ 1.4 The three limbed non ideal core shown in the figure has three windings with nominal inductances L each when measured individually with a single phase AC source. The inductance of the windings as connected will be (A) very low (B) L/3 (C) 3L (D) very high SOL 1.4 MCQ 1.5 SOL 1.5 Hence ( ) is correct Option Which of the following statement holds for the divergence of electric and magnetic flux densities? (A) Both are zero (B) These are zero for static densities but non zero for time varying densities. (C) It is zero for the electric flux density (D) It is zero for the magnetic flux density From maxwell s first equation 4: D ρv ρ 4: E v ε (Divergence of electric field intensity is non-zero) Maxwell s fourth equation 4: B (Divergence of magnetic field intensity is zero) Hence (D) is correct option. MCQ 1.6 In transformers, which of the following statements is valid?

4 Page 4 GATE EE 6 (A) In an open circuit test, copper losses are obtained while in short circuit test, core losses are obtained (B) In an open circuit test, current is drawn at high power factor (C) In a short circuit test, current is drawn at zero power factor (D) In an open circuit test, current is drawn at low power factor SOL 1.6 MCQ 1.7 SOL 1.7 In transformer, in open circuit test, current is drawn at low power factor but in short circuit test current drawn at high power factor. Hence (D) is correct option. For a single phase capacitor start induction motor which of the following statements is valid? (A) The capacitor is used for power factor improvement (B) The direction of rotation can be changed by reversing the main winding terminals (C) The direction of rotation cannot be changed (D) The direction of rotation can be changed by interchanging the supply terminals A single-phase capacitor start induction motor. It has cage rotor and its stator has two windings. The two windings are displaced 9c in space. The direction of rotation can be changed by reversing the main winding terminals. Hence (B) is correct option. MCQ 1.8 In a DC machine, which of the following statements is true? (A) Compensating winding is used for neutralizing armature reaction while interpole winding is used for producing residual flux (B) Compensating winding is used for neutralizing armature reaction while interpole winding is used for improving commutation (C) Compensating winding is used for improving commutation while interpole winding is used for neutralizing armature reaction (D) Compensation winding is used for improving commutation while interpole winding is used for producing residual flux

5 Page 5 GATE EE 6 In DC motor, compensating winding is used for neutralizing armature reactance while interpole winding is used for improving commutation. Interpoles generate voltage necessary to neutralize the e.m.f of self induction in the armature coils undergoing commutation. Interpoles have a polarity opposite to that of main pole in the direction of rotation of armature. Hence (B) is correct option. MCQ 1.9 SOL 1.9 MCQ 1.1 SOL 1.1 MCQ 1.11 SOL 1.11 MCQ 1.1 The concept of an electrically short, medium and long line is primarily based on the (A) nominal voltage of the line (B) physical length of the line (C) wavelength of the line (D) power transmitted over the line With the help of physical length of line, we can recognize line as short, medium and long line. Hence (B) is correct option. Keeping in view the cost and overall effectiveness, the following circuit breaker is best suited for capacitor bank switching (A) vacuum (B) air blast (C) SF 6 (D) oil For capacitor bank switching vacuum circuit breaker is best suited in view of cost and effectiveness. Hence (A) is correct option. In a biased differential relay the bias is defined as a ratio of (A) number of turns of restraining and operating coil (B) operating coil current and restraining coil current (C) fault current and operating coil current (D) fault current and restraining coil current Ratio of operating coil current to restraining coil current is known as bias in biased differential relay. Hence (B) is correct option. An HVDC link consist of rectifier, inverter transmission line and other equipments. Which one of the following is true for this link? (A) The transmission line produces/ supplies reactive power (B) The rectifier consumes reactive power and the inverter supplies reactive power from/ to the respective connected AC systems (C) Rectifier supplies reactive power and the inverted consumers reactive power to/ from the respective connected AC systems

6 Page 6 GATE EE 6 (D) Both the converters (rectifier and inverter) consume reactive power from the respective connected AC systems SOL 1.1 MCQ 1.13 SOL 1.13 HVDC links consist of rectifier, inverter, transmission lines etc, where rectifier consumes reactive power from connected AC system and the inverter supplies power to connected AC system. Hence (B) is correct option. For a system with the transfer function 3( s ) Hs (), 4s s+ 1 the matrix A in the state space form Xo AX+ Bu is equal to R V R V S 1 W S 1 W (A) S 1 W (B) S 1 W S 1 4 W S 1 4 W T X T X R V R V S 1 W S 1 W (C) S3 1W (D) S 1 W S1 4 W S 1 4 W T X T X In standard form for a characterstic equation give as n n 1 s + an 1s a1s+ a in its state varibale representation matrix A is given as R V S 1 g W S 1 g W A S h h h h h W S W S a a1 a g an 1W T X characterstic equation of the system is 4s s + 1 so, a 4, a1, a 1 R 1 V R 1 V S W S W A S 1 W S 1 W S a a1 aw S 1 4W T X T X Hence (B) is correct option. MCQ 1.14 A discrete real all pass system has a pole at z + 3 % : it, therefore (A) also has a pole at % (B) has a constant phase response over the z -plane: arg Hz () constant constant (C) is stable only if it is anti-causal iω (D) has a constant phase response over the unit circle: arg He ( ) constant SOL 1.14 Z -transform of a discrete all pass system is given as 1 Hz () z z ) 1 zz 1

7 Page 7 GATE EE 6 it has a pole at z and a zero at 1/z ). given system has a pole at % z + 3 ( 3 + j) ( 3 +j) system is stable if z < 1 and for this it is anti-causal. Hence (C) is correct option. MCQ 1.15 The time/div and voltage/div axes of an oscilloscope have been erased. A student connects a 1 khz, 5 V p-p square wave calibration pulse to channel-1 of the scope and observes the screen to be as shown in the upper trace of the figure. An unknown signal is connected to channel-(lower trace) of the scope. It the time/div and V/ div on both channels are the same, the amplitude (p-p) and period of the unknown signal are respectively (A) 5 V, 1 ms (C) 7.5 V, ms (B) 5 V, ms (D) 1 V, 1 ms SOL 1.15 From the Calibration pulse we can obtain Voltage ( 3 V) 5 5. V Division Time ( 3 T) 1 ms 1 msec Division 4 4 So amplitude (p-p) of unknown signal is V P P 3 V# 5

8 Page 8 GATE EE 6 MCQ # V Time period T 3 T# # 8 ms Hence (C) is correct option. A sampling wattmeter (that computes power from simultaneously sampled values of voltage and current) is used to measure the average power of a load. The peak to peak voltage of the square wave is 1 V and the current is a triangular wave of 5 A p-p as shown in the figure. The period is ms. The reading in W will be (A) W (C) 5 W (B) 5 W (D) 1 W SOL 1.16 Reading of wattmeter (Power) in the circuit P av 1 T T # VIdt Common are between V I total common area (Positive and negative area are equal) So Pav Hence (A) is correct option. MCQ 1.17 What are the states of the three ideal diodes of the circuit shown in figure?

9 Page 9 GATE EE 6 SOL 1.17 (A) D1ON, DOFF, D3 OFF (B) D1OFF, DON, D3OFF (C) D1ON, DOFF, D3 ON (D) D1OFF, DON, D3ON First we can check for diode D. Let diode D is OFF then the circuit is In the above circuit diode D 1 must be ON, as it is connected with 1 V battery now the circuit is Because we assumed diode D OFF so voltage across it VD # and it is possible only when D 3 is off. So, all assumptions are true. Hence (A) is correct option. MCQ 1.18 The speed of a 3-phase, 44 V, 5 Hz induction motor is to be controlled over a wide range from zero speed to 1.5 time the rated speed using a 3-phase voltage source inverter. It is desired to keep the flux in the machine constant in the constant torque region by controlling the terminal voltage as the frequency changes. The inverter output voltage vs frequency characteristic should be

10 Page 1 GATE EE 6 SOL 1.18 If we varying the frequency for speed control, Vf / should be kept as constant so that, minimum flux density (B m ) also remains constant so, V 4.44NBmAf Hence (D) is correct option. MCQ 1.19 For a given sinusoidal input voltage, the voltage waveform at point P of the clamper circuit shown in figure will be

11 Page 11 GATE EE 6 SOL 1.19 In the positive half cycle of input, Diode D 1 will be reverse biased and equivalent circuit is. Since there is no feed back to the op-amp and op-amp has a high open loop gain so it goes in saturation. Input is applied at inverting terminal so. V P V CC 1 V In negative half cycle of input, diode D 1 is in forward bias and equivalent circuit is shown below.

12 Page 1 GATE EE 6 Output VP V γ + V - Op-amp is at virtual ground so V+ V- and VP Vγ.7 V Voltage wave form at point P is Hence (D) is correct option. MCQ 1. A single-phase half wave uncontrolled converter circuit is shown in figure. A -winding transformer is used at the input for isolation. Assuming the load current to be constant and V V sin ωt, the current waveform through diode D will be m SOL 1. In first half cycle D 1 will conduct and D will not and at θ there is zero voltage. So current wave form is as following

13 Page 13 GATE EE 6 Hence (D) is correct option. Q.1 to Q.75 carry two marks each MCQ 1.1 xn [ ] ; n< 1, n>, x[ 1] 1, x[ ] is the input and yn [ ] ; n< 1, n>, y[ 1] 1 y[1], y[] 3, y[] is the output of a discrete-time LTI system. The system impulse response hn [ ] will be (A) h[ n] ; n <, n >, h[ ] 1, h[ 1] h[ ] 1 (B) hn [ ] ; n< 1, n> 1, h[ 1] 1, h[ ] h[ 1] (C) h[ n] ; n <, n > 3, h[] 1, h[1], h[] 1 (D) hn [ ] ; n<, n> 1, h[ ] h[ 1] h[ 1] h[ ] 3 SOL 1.1 According to given data input and output Sequences are xn [ ] { 1, }, 1# n # - yn [ ] { 1, 3, 1, }, 1 # n # - If impulse response of system is hn [ ] then output yn [ ] hn [ ]) xn [ ] Since length of convolution ( yn [ ]) is 1 to, xn [ ] is of length 1to so length of hn [ ] is to. Let hn [ ] { abc,, } - Convolution So, a 1 yn [ ] yn [ ] { a, a b, b c, c} - { 1,3, 1, } - a b 3 & b 1 a c 1 & c 1 Impulse response hn [ ] " 1, 1, 1, - Hence (A) is correct option. H MCQ 1. The expression V # πr ( 1 h/ H) dh for the volume of a cone is equal to

14 Page 14 GATE EE 6 SOL 1. R R (A) # πr ( 1 h/ H) dr (B) # πr ( 1 h/ H) dh H R (C) # πrh( 1 r/ R) dh (D) # πrh`1 Volume of the cone H V R 1 H h # π b l dh By solving the above integral V 1 πrh 3 Solve all integrals given in options only for option (D) R rh 1 R r # π a k dr 1 πrh 3 Hence (D) is correct option. r R j dr MCQ The discrete-time signal xn [ ] Xz ( ) 3 n n z n /, where n + transform-pair relationship, is orthogonal to the signal denotes a (A) y [ n] Y( z) n n / z n `3j 3 n (B) y [ n] ) Y ( z) ( 5 n) z / (C) y [ n] ) Y ( z) z 3 3 / n 3 - n n -3 -n (D) y [ n] ) Y ( z) z + 3z SOL 1.3 Correct Option is ( ) ( n + 1) MCQ 1.4 A surface Sxy (, ) x+ 5y 3 is integrated once over a path consisting of the points that satisfy ( x+ 1 ) + ( y 1 ). The integral evaluates to (A) 17 (B) 17 (C) / 17 (D) SOL 1.4 Correct Option is ( ) MCQ 1.5 A continuous-time system is described by yt () e - xt (), where yt () is the output and xt () is the input. yt () is bounded (A) only when xt () is bounded (B) only when xt () is non-negative (C) only for t # if xt () is bounded for t $ (D) even when xt () is not bounded SOL 1.5 Hence (D) is correct option. Output yt () e - xt () If xt () is unbounded, xt ()" 3

15 Page 15 GATE EE 6 e - xt () " yt () ( bounded) So yt () is bounded even when xt () is not bounded. MCQ 1.6 The running integration, given by yt () xtdt () SOL 1.6 MCQ 1.7 (A) has no finite singularities in its double sided Laplace Transform Ys () (B) produces a bounded output for every causal bounded input (C) produces a bounded output for every anticausal bounded input (D) has no finite zeroes in its double sided Laplace Transform Ys () Hence (B) is correct option ' ' Given yt () xtdt () # t 3 Laplace transform of yt () Xs Ys () (), has a singularity at s s ' ' For a causal bounded input, yt () # xtdt () is always bounded. t 3 Two fair dice are rolled and the sum r of the numbers turned up is considered (A) Pr( r > 6) 1 6 (B) Pr ( r/ 3 is an integer) 5 6 (C) Pr ( r 8 ; r/ 4 is an integer) (D) Pr ( r 6 ; r/5 is an integer) SOL 1.7 Option (C) is correct. By throwing dice twice 6# 6 36 possibilities will occur. Out of these sample space consist of sum 4, 8 and 1 because r/ 4 is an integer. This can occur in following way : (1, 3), (, ), (, 6), (3, 1), (3, 5), (4, 4), (5, 3), (6, ) and (6, 6) Sample Space 9 Favourable space is coming out of 8 5 Probability of coming out MCQ 1.8 The parameters of the circuit shown in the figure are Ri 1 MΩ R 1 Ω, 6 A 1 V/V If vi 1 μv, the output voltage, input impedance and output impedance respectively are t -# 3 ' '

16 Page 16 GATE EE 6 (A) 1 V, 3,1 Ω (B) 1 V,,1 Ω (C) 1 V,, 3 (D) 1 V, 3, 1 Ω SOL 1.8 In the given circuit Output voltage v o Av i 1 6 # 1 μv 1 V Input impedance Z vi i i i v i 3 Output impedance Z vo o i o Avi R i 1 Ω Hence (A) is correct option. o o MCQ 1.9 In the circuit shown in the figure, the current source I 1 A, the voltage source V 5 V, R R R 1 Ω, L L L 1 H, C C 1F The currents (in A) through R 3 and through the voltage source V respectively will be

17 Page 17 GATE EE 6 (A) 1, 4 (B) 5, 1 (C) 5, (D) 5, 4 SOL 1.9 All sources present in the circuit are DC sources, so all inductors behaves as short circuit and all capacitors as open circuit Equivalent circuit is Voltage across R 3 is 5 IR I 1 (1) I 1 5 A (current through R 3 ) By applying KCL, current through voltage source 1 + I I 1 I A Hence (D) is correct option. MCQ 1.3 The parameter type and the matrix representation of the relevant two port parameters that describe the circuit shown are SOL 1.3 MCQ 1.31 (A) z parameters, G (B) h parameters, 1 1 G (C) h parameters, G (D) z parameters, 1 1 G Given Two port network can be described in terms of h-parametrs only. Hence (C) is correct option. δp Consider the following statements with reference to the equation δ t (1) This is a point form of the continuity equation. () Divergence of current density is equal to the decrease of charge per unit volume per unit at every point. (3) This is Max well s divergence equation

18 Page 18 GATE EE 6 (4) This represents the conservation of charge Select the correct answer. (A) Only and 4 are true (B) 1, and 3 are true (C), 3 and 4 are true (D) 1, and 4 are true SOL 1.31 Hence Correct option is ( ) MCQ 1.3 The circuit shown in the figure is energized by a sinusoidal voltage source V 1 at a frequency which causes resonance with a current of I. The phasor diagram which is applicable to this circuit is SOL 1.3 At resonance reactance of the circuit would be zero and voltage across inductor and capacitor would be equal V L V C At resonance impedance of the circuit Z R R1+ R Current I V1 R + c R + R 1 Voltage V I R + j( V V ) R L C

19 Page 19 GATE EE 6 V V1 + c R R + R 1 Voltage across capacitor V C 1 I jωc # So phasor diagram is R 1 VR + c jωc # R + R 1 VR + 9c ωcr ( + R) 1 Hence (A) is correct option. MCQ 1.33 An ideal capacitor is charged to a voltage V and connected at t across an ideal inductor L. (The circuit now consists of a capacitor and inductor alone). If we let ω 1, the voltage across the capacitor at time t > is given by LC (A) V (B) V cos( ω t) (C) sin( ω t) (D) Ve cos( t) V -ω t ω SOL 1.33 This is a second order LC circuit shown below Capacitor current is given as dvc() t ic () t C dt Taking Laplace transform IC () s CsV() s V(), V () "initial voltage Current in inductor il () t 1 () L # v c t dt () IL () s 1 Vs L s for t >, applying KCL(in s-domain) I () s + I () s C L

20 Page GATE EE 6 () CsV() s V() 1 Vs + L s s 1 : + () LCsD Vs V o Vs () V s o s + ω, a ω 1 LC Taking inverse laplace transformation vt () Vocos ωot, t > Hence (B) is correct option MCQ 1.34 SOL 1.34 MCQ 1.35 SOL 1.35 A 4 V, 5 Hz, three phase balanced source supplies power to a star connected load whose rating is 1 3 kva,.8 pf (lag). The rating (in kvar) of the delta connected (capacitive) reactive power bank necessary to bring the pf to unity is (A) (B) 1.6 (C) 16.6 (D) 1.47 Rating of Δ-connected capacitor bank for unity p.f. real power P L S cos φ 1 3 # kw reactive power Q L S sin φ 1 3 # kw For setting of unity p.f. we have to set capacitor bank equal to reactive power kw Hence (D) is correct option. An energy meter connected to an immersion heater (resistive) operating on an AC 3 V, 5 Hz, AC single phase source reads.3 units (kwh) in 1 hour. The heater is removed from the supply and now connected to a 4 V peak square wave source of 15 Hz. The power in kw dissipated by the heater will be (A) (B) (C) 1.54 (D).87 Power dissipated in heater when AC source is connected P.3 kw V rms R # ( 3) R R 3 Ω (Resistance of heater) Now it is connected with a square wave source of 4 V peak to peak Power dissipated is So, P V rms V rms, Vp p 4 V & Vp V R V p (for square wave) ( ) P kw 3

21 Page 1 GATE EE 6 Hence (B) is correct option. MCQ 1.36 SOL 1.36 MCQ 1.37 SOL 1.37 MCQ 1.38 A V DC machine supplies A at V as a generator. The armature resistance is. ohm. If the machine is now operated as a motor at same terminal voltage and current but with the flux increased by 1%, the ratio of motor speed to generator speed is (A).87 (B).95 (C).96 (D) 1.6 Given: A 3 V, DC machine, A at V as a generator. R a Ω. The machine operated as a motor at same terminal voltage and current, flux increased by 1% So for generator E g V+ I a R a + #. E g 4 volt for motor E m V I a R a #. E m 196 volt Eg Ng φ So g E N # φ m N N m g m N N g # Hence (A) is correct option. m m # 1. 1 A synchronous generator is feeding a zero power factor (lagging) load at rated current. The armature reaction is (A) magnetizing (B) demagnetizing (C) cross-magnetizing (D) ineffective A synchronous generator is feeding a zero power factor(lagging) load at rated current then the armature reaction is demagnetizing. Hence (B) is correct option. Two transformers are to be operated in parallel such that they share load in proportion to their kva ratings. The rating of the first transformer is 5 kva ratings. The rating of the first transformer is 5 kva and its pu leakage impedance is.5 pu. If the rating of second transformer is 5 kva, its pu leakage impedance is (A). (B).1 (C).5 (D).5

22 Page GATE EE 6 SOL 1.38 MCQ 1.39 SOL 1.39 Given the rating of first transformer is 5 kva Per unit leakage impedance is.5 p.u. Rating of second transformer is 5 kva So, actual impedance Per unit impedance base impedance and, Per unit leakage impedance \ 1 kva Then 5 kva #.5 5 kva # x x # 5 1p.u. Hence (B) is correct option. The speed of a 4-pole induction motor is controlled by varying the supply frequency while maintaining the ratio of supply voltage to supply frequency ( Vf) / constant. At rated frequency of 5 Hz and rated voltage of 4 V its speed is 144 rpm. Find the speed at 3 Hz, if the load torque is constant (A) 88 rpm (B) 864 rpm (C) 84 rpm (D) 88 rpm Given speed of a 4-pole induction motor is controlled by varying the supply frequency when the ratio of supply voltage and frequency is constant. f 5 Hz, V 4 V, N 144 rpm So V \ f V V 1 f1 f So V 4 # V T \ V c S f m S S 1 bv V l f T # f # T 1 1 Given T 1 T Then S 4. 4 b4 l S. 66 N r N (1 S) s 1 # # 5 3

23 Page 3 GATE EE 6 a N r 1f P So N r 1 # ^ h N r 84.6 rpm Hence (C) is correct option. MCQ 1.4 SOL 1.4 A 3-phase, 4-pole, 4 V 5 Hz, star connected induction motor has following circuit parameters r 1 1. Ω, r'.5 Ω, X 1 X' 1. Ω, X m 35 Ω The starting torque when the motor is started direct-on-line is (use approximate equivalent circuit model) (A) 63.6 Nm (B) 74.3 Nm (C) 19.8 Nm (D).9 Nm Given a 3-φ induction motor P 4, V 4 V, f 5 Hz r 1 1. Ω, rl 5. Ω X 1 Xl 1. Ω, X m 35 Ω So Speed of motor is 1f N s 1 # 5 15 rpm P 4 Torque T st Nm Hence (A) is correct option. Vr 18 Ns ( r1 r ) l π # + l + X 4 c m # # # # ( 15. ) + ( 4. ) MCQ 1.41 A 3-phase, 1 kw, 4 V, 4-pole, 5Hz, star connected induction motor draws A on full load. Its no load and blocked rotor test data are given below. No Load Test : 4 V 6 A 1 W Blocked Rotor Test : 9 V 15 A 76 W Neglecting copper loss in no load test and core loss in blocked rotor test, estimate motor s full load efficiency (A) 76% (B) 81% (C) 8.4% (D) 85% SOL 1.41 Given that 3-φ induction motor star connected P 1 kw, V 4 V, Poles 4, f 5 Hz

24 Page 4 GATE EE 6 Full load current IFl A output Efficiency input So Cu losses at full load b15 l # ` Total losses MCQ 1.4 SOL 1.4 Efficiency # % Hence (B) is correct option. A 3-phase, 4 V, 5 kw, star connected synchronous motor having an internal reactance of 1 Ω is operating at 5% load, unity p.f. Now, the excitation is increased by 1%. What will be the new load in percent, if the power factor is to be kept same? Neglect all losses and consider linear magnetic circuit. (A) 67.9% (B) 56.9% (C) 51% (D) 5% The A, B, C, D constants of a kv line are : A D c, B c, C.1+ 9c Given 3-φ star connected synchronous motor internal reactance 1 Ω Operating at 5% load, unity power factor, 4 V, 5 kw Excitation increased 1% So, full load current 3 I Fl 5# # 4 # 1 E ( Vcos θ I R ) + ( Vsin θ I X ) a a a s So, E 4 c + ^1 # m h E Excitation will increase 1% then E E #. 1 E 36 IX a ( E ) V

25 Page 5 GATE EE 6 I a ( 36) 4 c 3 m I a Load(%) % 7. Hence (A) is correct option. MCQ 1.43 SOL 1.43 MCQ 1.44 If the sending end voltage of the line for a given load delivered at nominal voltage is 4 kv, the % voltage regulation of the line is (A) 5 (B) 9 (C) 16 (D) 1 Given ABCD constant of kv line A D.94+ 1c, B c, C. 1+ 9c, VS 4 kv % voltage regulation is being given as ( VR) No Load ( VR) Full %V.R. load 1 V ( Full load) # At no load I R ( V R ) NL VS / A, ( VR) Full load kv 4 %V.R # %V.R. 16 Hence (C) is correct option. R A single phase transmission line and a telephone line are both symmetrically strung one below the other, in horizontal configurations, on a common tower, The shortest and longest distances between the phase and telephone conductors are.5 m and 3 m respectively. The voltage (volt/km) induced in the telephone circuit, due to 5 Hz current of 1 amps in the power circuit is (A) 4.81 (B) 3.56 (C).9 (D) 1.7 SOL 1.44 Hence Correct Option is ( ) MCQ 1.45 Three identical star connected resistors of 1. pu are connected to an unbalanced 3-phase supply. The load neutral is isolated. The symmetrical components of the line voltages in pu. are: Vab X+θ 1 1, Vab Y+θ. If all the pu calculations are with the respective base values, the phase to neutral sequence voltages are

26 Page 6 GATE EE 6 SOL 1.45 (A) V X+ ( θ + 3 c), V Y( θ 3 c) an 1 an 1 (B) Van X+ ( θ1 3 c), Van Y+ ( θ+ 3 c) 1 (C) V 1 X ( 3 ), V 1 an + θ1 c an Y+ ( θ 3 c) (D) V 1 X ( 6 ), V 1 an + θ1 c an Y+ ( θ 6 c) Given that, Vab1 X+θ1, Vab Y+θ, Phase to neutral sequence volt? First we draw phasor of positive sequence and negative sequence. MCQ 1.46 From figure we conclude that postive sequence line voltage leads phase voltage by 3c V AN1 X+θ 1 3c V AN 4+θ + 3c Hence (B) is correct option. A generator is connected through a MVA, 13.8/138 kv step down transformer, to a transmission line. At the receiving end of the line a load is supplied through a step down transformer of 1 MVA, 138/69 kv rating. A.7 pu. load, evaluated on load side transformer ratings as base values, is supplied from the above system. For system base values of 1 MVA and 69 kv in load circuit, the value of the load (in per unit) in generator will be (A) 36 (B) 1.44 (C).7 (D).18 SOL 1.46 For system base value 1 MVA, 69 kv, Load in pu(z new )? (MVA) old Z new Z kvnew old # (MVA) # b new kv l old Z new # # b l MCQ pu Hence (A) is correct option. The Gauss Seidel load flow method has following disadvantages. Tick the incorrect statement. (A) Unreliable convergence

27 Page 7 GATE EE 6 (B) Slow convergence (C) Choice of slack bus affects convergence (D) A good initial guess for voltages is essential for convergence SOL 1.47 Unreliable convergence is the main disadvantage of gauss seidel load flow method. Hence (A) is correct option. 1 ( 1 + jω) MCQ 1.48 The Bode magnitude plot Hj ( ω) ( 1 + jω)( 1 + jω) 4 is SOL 1.48 Given function is, 4 1 ( 1 + jω) Hjω ( ) ( 1 + jω)( 1 + jω) Function can be rewritten as, 4 1 ( 1 + jω) Hjω ( ) 1 1 j ω j ω 9 + C 9 + 1C 1. ( 1+ jω) Hjω ( ) 1 j ω j 1 1 ω a + ka + 1k The system is type, So, initial slope of the bode plot is db/decade. Corner frequencies are

28 Page 8 GATE EE 6 ω 1 1 rad/sec ω 1 rad/sec ω 3 1 rad/sec As the initial slope of bode plot is db/decade and corner frequency ω 1 1 rad/ sec, the Slope after ω 1 rad/sec or logω is( + ) + db/dec. After corner frequency ω 1 rad/sec or log ω 1, the Slope is ( + ) db/ dec. Similarly after ω 3 1 rad/sec or log ω, the slope of plot is ( # ) 4 db/dec. Hence (A) is correct option. MCQ 1.49 A closed-loop system has the characteristic function ( s 4)( s+ 1) + K( s 1). Its root locus plot against K is SOL 1.49 Closed loop transfer function of the given system is, Ts () s + 4 ( s + 1)( s + 4) Tjω ( ) ( jω) + 4 ( jω+ 1)( jω+ 4)

29 Page 9 GATE EE 6 If system output is zero 4 ω Tjω ( ) ^jω+ 1( h jω+ 4) 4 ω ω 4 & ω rad/sec Hence (C) is correct option. MCQ 1.5 yn [ ] denotes the output and xn [ ] denotes the input of a discrete-time system given by the difference equation yn [ ]. 8yn [ 1] xn [ ] xn [ + 1]. Its right-sided impulse response is (A) causal (B) unbounded (C) periodic (D) non-negative SOL 1.5 Hence Correct Option is ( ) MCQ 1.51 The algebraic equation Fs () s 3s + 5s 7s + 4s+ is given. Fs () has (A) a single complex root with the remaining roots being real (B) one positive real root and four complex roots, all with positive real parts (C) one negative real root, two imaginary roots, and two roots with positive real parts (D) one positive real root, two imaginary roots, and two roots with negative real parts SOL 1.51 Hence Correct Option is ( ) MCQ 1.5 Consider the following Nyquist plots of loop transfer functions over ω to ω 3. Which of these plots represent a stable closed loop system?

30 Page 3 GATE EE 6 (A) (1) only (B) all, except (1) (C) all, except (3) (D) (1) and () only SOL 1.5 In the given options only in option (A) the nyquist plot does not enclose the unit circle ( 1, j), So this is stable. Hence (A) is correct option. MCQ 1.53 A current of 8+ 6 ( sin ωt + 3 ) A is passed through three meters. They are a centre zero PMMC meter, a true rms meter and a moving iron instrument. The respective reading (in A) will be (A) 8, 6, 1 (B) 8, 6, 8 (C) 8,1,1 (D) 8,, SOL 1.53 PMMC instrument reads only dc value so I PMMC 8 A rms meter reads rms value so I rms ( 8) + ( 6 ) A Moving iron instrument also reads rms value of current So I MI 1 ma Reading are ( IPMMC, Irms, IMI) ( 8A,1A,1A) Hence (C) is correct option. MCQ 1.54 A variable w is related to three other variables x,y,z as w xy/ z. The variables are measured with meters of accuracy! 5. % reading,! 1% of full scale value and! 15. % reading. The actual readings of the three meters are 8, and 5 with 1 being the full scale value for all three. The maximum uncertainty in the measurement of w will be (A)! 5. % rdg (B)! 55. % rdg (C)! 67. rdg (D)! 7. rdg SOL 1.54 Hence (D) is correct option.

31 Page 31 GATE EE 6 Given that ω xy z log ω log x+ log y log z Maximum error in ω % d ω ω dx dy!!! dz x y z dx x dy y dy y dz z! 5. % reading! 1% full scale! 1 1 # 1! 1! % 1 # 1! 5 reading 15. % reading So % d ω ω! 5. %! 5%! 15. %! 7% MCQ 1.55 SOL 1.55 MCQ 1.56 SOL 1.56 A /1 Current transformer (CT) is wound with turns on the secondary on a toroidal core. When it carries a current of 16 A on the primary, the ratio and phase errors of the CT are found to be 5. % and 3 minutes respectively. If the number of secondary turns is reduced by 1 new ratio-error(%) and phase-error(min) will be respectively (A) 3., (B).5,35 (C) 13., (D) 1.,5 Hence ( ) is Correct Option R 1 and R 4 are the opposite arms of a Wheatstone bridge as are R 3 and R. The source voltage is applied across R 1 and R 3. Under balanced conditions which one of the following is true (A) R R R / R (B) R R R / R (C) R R R / R (D) R R + R + R Hence (B) is correct option

32 Page 3 GATE EE 6 In balanced condition there is no current in CD arm so V Writing node equation at C and D VC V + VC & V V R3 C R R b R + R l V V + VD & V V R4 D R R b R + R l 4 C V D MCQ 1.57 So V R3 b R + R l V R4 b R + R l RR RR RR + RR R 1 RR/ R Assuming the diodes D 1 and D of the circuit shown in figure to be ideal ones, the transfer characteristics of the circuit will be SOL 1.57 In the circuit when V i < 1 V, both D 1 and D are off. So equivalent circuit is,

33 Page 33 GATE EE 6 Output, V o 1 volt When V i > 1 V (D 1 is in forward bias and D is off So the equivalent circuit is, Output, Vo Vi Transfer characteristic of the circuit is Hence (A) is correct option. MCQ 1.58 Consider the circuit shown in figure. If the β of the transistor is 3 and I CBO is ma and the input voltage is + 5V, the transistor would be operating in (A) saturation region (C) breakdown region (B) active region (D) cut-off region SOL 1.58 Assume that BJT is in active region, thevenin equivalent of input circuit is obtained as

34 Page 34 GATE EE 6 Vth Vi Vth ( 1) Vth Vi + 3Vth V th # 5 36, Vi 5 V V th 78. V Thevenin resistance R th 15 KΩ 1 KΩ 13.4 KΩ So the circuit is Writing KVL for input loop 78. RthIB 7. I B. 157 ma Current in saturation is given as, So, I B(sat) I C(sat) I B(sat) I C(sat) β ma 545mA ma 3 MCQ 1.59 Since IB(sat) > IB, therefore assumption is true. Hence (B) is correct option. A relaxation oscillator is made using OPAMP as shown in figure. The supply voltages of the OPAMP are! 1 V. The voltage waveform at point P will be

35 Page 35 GATE EE 6 SOL 1.59 Here output of the multi vibrator is V! 1 volt Threshold voltage at positive terminal of op-amp can be obtained as following When output V + 1 V, equivalent circuit is, writing node equation at positive terminal of op-amp Vth 1 + Vth 1 1

36 Page 36 GATE EE 6 V th 6 volt (Positive threshold) So, the capacitor will charge upto 6 volt. When output V 1 V, the equivalent circuit is. node equation Vth Vth 1 5Vth + 6+ Vth V th 1 volt (negative threshold) So the capacitor will discharge upto 1 volt. At terminal P voltage waveform is. Hence (C) is correct option. MCQ 1.6 A TTL NOT gate circuit is shown in figure. Assuming VBE.7 V of both the transistors, if V i 3. V, then the states of the two transistors will be

37 Page 37 GATE EE 6 (A) Q 1 ON and Q OFF (C) Q 1 reverse ON and Q ON (B) Q 1 reverse ON and Q OFF (D) Q 1 OFF and Q reverse ON SOL 1.6 MCQ 1.61 Hence ( ) is correct Option A student has made a 3-bit binary down counter and connected to the R-R ladder type DAC, [Gain ( 1 kω/ R) ] as shown in figure to generate a staircase waveform. The output achieved is different as shown in figure. What could be the possible cause of this error? (A) The resistance values are incorrect option. (B) The counter is not working properly (C) The connection from the counter of DAC is not proper (D) The R and R resistance are interchanged SOL 1.61 Hence ( ) is correct option MCQ 1.6 A 4# 1 MUX is used to implement a 3-input Boolean function as shown in figure. The Boolean function FABC (,, ) implemented is (A) FABC (,, ) Σ( 146,,, )

38 Page 38 GATE EE 6 SOL 1.6 (B) FABC (,, ) Σ( 16,, ) (C) FABC (,, ) Σ( 456,,, ) (D) FABC (,, ) Σ( 156,, ) Function F can be obtain as, F ISS 1 + ISS ISS 1 + ISS 3 1 F AB C + A B C + 1$ BC + $ BC AB C + A BC + BC AB C + A BC + BC( A + A) AB C + A BC + ABC + A BC F Σ(, 146,, ) Hence (A) is correct option. MCQ 1.63 A software delay subroutine is written as given below : DELAY : MVI H, 55D MVI L, 55D LOOP : DCR L JNZ LOOP DCR H JNZ LOOP How many times DCR L instruction will be executed? (A) 55 (B) 51 (C) 655 (D) 6579 SOL 1.63 MCQ 1.64 MVI H and MVI L stores the value 55 in H and L registers. DCR L decrements L by 1 and JNZ checks whether the value of L is zero or not. So DCR L executed 55 times till value of L becomes. Then DCR H will be executed and it goes to Loop again, since L is of 8 bit so no more decrement possible and it terminates. Hence (A) is correct option. In an 885 A microprocessor based system, it is desired to increment the contents of memory location whose address is available in (D,E) register pair and store the result in same location. The sequence of instruction is (A) XCHG (B) XCHG INR M INX H (C) INX D (D) INR M XCHG XCHG SOL 1.64 XCHG& exchange the contain of DE register pair with HL pair So now addresses of memory locations are stored in HL pair. INR M& Increment the contents of memory whose address is stored in HL pair.

39 Page 39 GATE EE 6 Hence (A) is correct option MCQ 1.65 A single-phase inverter is operated in PWM mode generating a single-pulse of width d in the centre of each half cycle as shown in figure. It is found that the output voltage is free from 5 th harmonic for pulse width 144c. What will be percentage of 3 rd harmonic present in the output voltage ( V / ) o3 V o1 max? (A).% (B) 19.6% (C) 31.7% (D) 53.9% SOL 1.65 MCQ 1.66 SOL 1.66 In the PWM inverter output voltage of inverter V V 4 sin sin sin / n V s / nd nωt nπ 3 n 1 π So the pulse width d 144c V 4Vs 1 sin 7c sin ωt π so, V bv V 3 4V s sin 3 7 sin 3 t 3 # c ω π ^ h 3 1max 4V sin( ) 3 s # 3# 7c l π % 4Vs sin 7c π Hence (B) is correct option. A 3-phase fully controlled bridge converter with free wheeling diode is fed from 4 V, 5 Hz AC source and is operating at a firing angle of 6c. The load current is assumed constant at 1 A due to high load inductance. The input displacement factor (IDF) and the input power factor (IPF) of the converter will be (A) IDF.867; IPF.88 (B) IDF.867; IPF.55 (C) IDF.5; IPF.478 (D) IDF.5; IPF.318 Given that 4 V, 5 Hz AC source, α 6c, IL 1 A so, Input displacement factor cos α. 5 and, input power factor D.F. # cos α

40 Page 4 GATE EE 6 distortion factor I s(fundamental) I s 4 # 1 sin 6 # c π 1 # / so, input power factor. 955 # Hence (C) is correct option. MCQ 1.67 A voltage commutation circuit is shown in figure. If the turn-off time of the SCR is 5 μsec and a safety margin of is considered, then what will be the approximate minimum value of capacitor required for proper commutation? (A).88 μ F (C).91 μ F (B) 1.44 μf (D).7 μf SOL 1.67 MCQ 1.68 We know that So T C RC ln R T # # μf Hence (A) is correct option. A solar cell of 35 V is feeding power to an ac supply of 44 V, 5 Hz through a 3-phase fully controlled bridge converter. A large inductance is connected in the dc circuit to maintain the dc current at A. If the solar cell resistance is.5 Ω,then each thyristor will be reverse biased for a period of (A) 15c (B) 1c (C) 6c (D) 55c SOL 1.68 Let we have R solar.5 Ω, I A

41 Page 41 GATE EE 6 so V s 35 #.5 34 V ` 34 3 # 44 # cos α π cos α 55c So each thyristor will reverse biased for 18c 55c 15c. Hence (A) is correct option. MCQ 1.69 A single-phase bridge converter is used to charge a battery of V having an internal resistance of. Ω as shown in figure. The SCRs are triggered by a constant dc signal. If SCR gets open circuited, what will be the average charging current? (A) 3.8 A (C) 11.9 A (B) 15 A (D) 3.54 A SOL 1.69 In this circuitry if SCR gets open circuited, than circuit behaves like a half wave rectifier. So I avg Average value of current # π θ ( V sin t E) d 1 1 m ω θ πr θ1 a I (avg) cos ( ) 1 R Vm θ E π θ π 6

42 Page 4 GATE EE 6 θ 1 1 [ ( 3 ) cos ( )] π # # # θ π θ 1 1 sin bv E l m 1 sin c 38c.66 Rad 3 # m ` I avg) ( 1 [ 3 cos 38 (.66)] # # c π π # 11.9 A Hence (C) is correct option. MCQ 1.7 An SCR having a turn ON times of 5 μsec, latching current of 5 A and holding current of 4 ma is triggered by a short duration pulse and is used in the circuit shown in figure. The minimum pulse width required to turn the SCR ON will be (A) 51 μsec (C) 1 μsec (B) 15 μsec (D) 5 μsec SOL 1.7 Hence (B) is correct option. In this given circuit minimum gate pulse width time Time required by i a rise up to i L i 1 3 ma 5# 1 i 1 1 [ 1 e ] 4t ` anode current I I1+ I [ e 4t ] [ e 4t ] 4t 1 e 3. 5 T 15 μs

43 Page 43 GATE EE 6 Common Data for Questions 71, 7, 73: A 4-pole, 5 Hz, synchronous generator has 48 slots in which a double layer winding is housed. Each coil has 1 turns and is short pitched by an angle to 36c electrical. The fundamental flux per pole is.5 Wb MCQ 1.71 SOL 1.71 The line-to-line induced emf(in volts), for a three phase star connection is approximately (A) 88 (B) 888 (C) 14 (D) 1538 Hence (C) is correct option. Given P 4, f 5 Hz Slots 48, each coil has 1 turns Short pitched by an angle(α) to 36c electrical Flux per pole.5 Wb So, E ph 4.44 fφt K ph W Slot/Pole/ph # 3 Slot/Pole Slot angle K d K p In double layer wdg No. of coil 18 15c 1 sin( 4 15/ ) # 4sin( 15/ ). 957 cos α cos 18c. 951 No of slots MCQ 1.7 No. of turns/ph 48 # Then E ph #. 5 # 5 #. 957 #. 951 # 16 E ph 88 V E L 3 # V(approximate) E L The line-to-line induced emf(in volts), for a three phase connection is approximately (A) 1143 (B) 133

44 Page 44 GATE EE 6 (C) 1617 (D) 1791 SOL 1.7 MCQ 1.73 SOL 1.73 line to line induced voltage, so in phase winding Slot/ pole/ph 6 T ph 48 4 Slot angle 18 # 4 15c 48 K d sin 6 ( 15/ ) # 6sin( 15/ ) K d.93 K p cos 36 b.951 l E ph #. 5 # 5 # 4 #. 951 #. 93 E ph 1143 Hence (A) is correct option. The fifth harmonic component of phase emf(in volts), for a three phase star connection is (A) (B) 69 (C) 81 (D) 88 Fifth harmonic component of phase emf So Angle 18 36c 5 the phase emf of fifth harmonic is zero. Hence (A) is correct option. Common Data for Question 74, 75: A generator feeds power to an infinite bus through a double circuit transmission line. A 3-phase fault occurs at the middle point of one of the lines. The infinite bus voltage is 1 pu, the transient internal voltage of the generator is 1.1 pu and the equivalent transfer admittance during fault is.8 pu. The 1 MVA generator has an inertia constant of 5 MJ/MVA and it was delivering 1. pu power prior of the fault with rotor power angle of 3c. The system frequency is 5 Hz. MCQ 1.74 SOL 1.74 The initial accelerating power (in pu) will be (A) 1. (B).6 (C).56 (D).4 Generator feeds power to infinite bus through double circuit line 3-φ fault at middle of line.

45 Page 45 GATE EE 6 Infinite bus voltage(v ) 1 pu Transient internal voltage of generator(e ) 11. pu Equivalent transfer admittance during fault.8 pu 1/X delivering power(p S ) 1. pu Perior to fault rotor Power angle δ 3c, f 5 Hz Initial accelerating power(p a )? P P sin P a S m δ 1 EV sin 3c X # /. # 56. pu Hence (C) is correct option. MCQ 1.75 SOL 1.75 If the initial accelerating power is X pu, the initial acceleration in elect-deg/sec, and the inertia constant in MJ-sec/elect-deg respectively will be (A) X, 18 (B) 18 X,.56 (C) X/ 18,.56 (D) X/ 31. 4, 18 Hence (B) is correct option. If initial acceleration power X pu Initial acceleration? Inertia constant? X( pu) S α P a # 18 # 5 # X# S M SH/ 18F S# S α 18 X deg / sec Inertia const Linked Answer Questions : Q.76 to Q.85 carry two marks each Statement for Linked Answer Questions 76 and 77: R VT R VT R VT S 1W S W S W P S 1 W, Q S 5W, R S 7W are three vectors. S 3 W S 9 W S 1 W T X T X T X MCQ 1.76 An orthogonal set of vectors having a span that contains P,Q,Ris R V R V R V R V R V S 6W S 4 W S 4W S 5 W S 8 W (A) S 3W S W (B) S W S 7 W S W S 6 W S 3 W S 4 W S 11 W S 3 W T X T X T X T X T X

46 Page 46 GATE EE 6 SOL 1.76 R V R V R V S 6 W S 3W S 3 W (C) S 7 W S W S 9 W S 1 W S W S 4 W T X T X T X Hence ( ) is correct Option (D) R V S 4 W S 3 W S11 W T X R V S 1 W S31W S 3 W T X R V S5W S3W S4 W T X MCQ 1.77 SOL 1.77 The following vector is linearly dependent upon the solution to the previous problem R V R V S8W S W (A) S9W (B) S 17W S3 W S 3 W RT VX TR V X S4W S 13 W (C) S4W (D) S W S5 W S 3 W T X T X Hence ( ) is correct Option Statement for Linked Answer Questions 78 and 79: It is required to design an anti-aliasing filter for an, 8 bit ADC. The filter is a first order RC filter with R 1 Ω and C 1F. The ADC is designed to span a sinusoidal signal with peak to peak amplitude equal to the full scale range of the ADC. MCQ 1.78 SOL 1.78 MCQ 1.79 SOL 1.79 The transfer function of the filter and its roll off respectively are (A) 1/(1 + RCs), db/ decade (B) (1 + RCs), 4 db/ decade (C) 1/(1 + RCs), 4 db/ decade (D) { RCs/(1 + RCs)}, db/ decade Hence ( ) is correct Option What is the SNR (in db) of the ADC? Also find the frequency (in decades) at the filter output at which the filter attenuation just exceeds the SNR of the ADC. (A) 5 db, decades (B) 5 db,.5 decades (C) 6 db, decades (D) 6 db,.5 decades Hence ( ) is correct Option

47 Page 47 GATE EE 6 Statement for Linked Answer Questions 8 and 81: A 3 kva transformer has 95% efficiency at full load.8 p.f. lagging and 96% efficiency at half load, unity p.f. MCQ 1.8 The iron loss ( P i ) and copper loss ( P c ) in kw, under full load operation are (A) P 41., P 851. (B) P 659., P 91. c i (C) P 851., P 41. (D) P 1. 7, P 3. 7 c i c c i i SOL 1.8 Given that: A 3 kva transformer Efficiency at full load is 95% and.8 p.f. lagging 96% efficiency at half load and unity power factor So For I st condition for full load 95% kva #.8 kva #.8 + Wcu + Wi Second unity power factor half load 96% kva #.5 kva #.5 + Wcu + Wi So Wcu + Wi Wcu Wi 65. Then W cu 851., Wi Hence (C) is correct option...(1)...() MCQ 1.81 What is the maximum efficiency (in %) at unity p.f. load? (A) 95.1 (B) 96. (C) 96.4 (D) 98.1 SOL 1.81 Hence (B) is correct option. X p.f. kva Efficiency ( η ) # # X# kva + Wi+ Wcu# X So X η % # 1 # # #(. 6956) η 96. % Statement for Linked Answer Questions 8 and 83:

48 Page 48 GATE EE 6 For a power system the admittance and impedance matrices for the fault studies are as follows. R V S j875. j15. j5. W Y bus S j15. j65. j5. W S j5. j5. j5. W T X R V Sj16. j8. j1. W Z bus Sj8. j4. j16. W Sj1. j16. j34. W T X The pre-fault voltages are 1. pu. at all the buses. The system was unloaded prior to the fault. A solid 3-phase fault takes place at bus. MCQ 1.8 SOL 1.8 The post fault voltages at buses 1 and 3 in per unit respectively are (A).4,.63 (B).31,.76 (C).33,.67 (D).67,.33 The post fault voltage at bus 1 and 3 are. Pre fault voltage. R V V R 1 1+ c V S W S W V Bus SV W S1+ cw SV 3W S1+ cw T X T X At bus solid fault occurs Zf (), r Fault current I f Z f 1+ c 4j j4. Vr V c c Z + Z Z + Z rr f f Vi () f Vic () ZirI( f), Vic Prefault voltage V1 () f Vic Z1If 1+ c j.8( j4) 1.3 V1 () f 68. pu V3 () f V3c Z3If 1+ c j.16( j4) 1.64 V3 () f.36 pu Hence (D) is correct option MCQ 1.83 The per unit fault feeds from generators connected to buses 1 and respectively are (A) 1.,.51 (B) 1.55,.61

49 Page 49 GATE EE 6 (C) 1.66,.5 (D) 5.,.5 SOL 1.83 Hence ( ) is correct option Statement for Linked Answer Questions 84 and 85: A voltage commutated chopper operating at 1 khz is used to control the speed of dc as shown in figure. The load current is assumed to be constant at 1 A MCQ 1.84 The minimum time in μsec for which the SCR M should be ON is. (A) 8 (B) 14 (C) 7 (D) SOL 1.84 Given I L 1 A. So in the + ve half cycle, it will charge the capacitor, minimum time will be half the time for one cycle. so min time required for charging LC ω π π # # 1 # 1 Hence (B) is correct option μ sec MCQ 1.85 SOL 1.85 The average output voltage of the chopper will be (A) 7 V (B) 47.5 V (C) 35 V (D) V Hence (C) is correct option. Given T on 14 μ sec Average output T V T on # total T total 1/f msec 6 so average output 14 # 1 3 # 5 35 V 1# 1

50 Page 5 GATE EE 6 Answer Sheet 1. (B) 19. (D) 37. (B) 55. (*) 73. (A). (C). (D) 38. (B) 56. (B) 74. (C) 3. (D) 1. (A) 39. (C) 57. (A) 75. (B) 4. (*). (D) 4. (A) 58. (B) 76. (*) 5. (D) 3. (*) 41. (B) 59. (C) 77. (*) 6. (D) 4. (*) 4. (A) 6. (*) 78. (*) 7. (B) 5. (D) 43. (C) 61. (*) 79. (*) 8. (B) 6. (B) 44. (*) 6. (A) 8. (C) 9. (B) 7. (*) 45. (B) 63. (A) 81. (B) 1. (A) 8. (A) 46. (A) 64. (A) 8. (D) 11. (B) 9. (D) 47. (A) 65. (B) 83. (*) 1. (B) 3. (C) 48. (A) 66. (C) 84. (B) 13. (B) 31. (*) 49. (C) 67. (A) 85. (C) 14. (C) 3. (A) 5. (*) 68. (A) 15. (C) 33. (B) 51. (*) 69. (C) 16. (A) 34. (D) 5. (A) 7. (B) 17. (A) 35. (B) 53. (C) 71. (C) 18. (D) 36. (A) 54. (D) 7. (A) **********

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