**Quiz Electrical EngineeringExam: GATETopic: MiscellaneousDate: 02/07/2021**

Each Question carries 2 Mark

Negative Marking: 1/3

Time: 20 Minutes

Q1. in the network as shown in figure, the marked parameters are pu impedances. The bus-admittance matrix of the network is

Q2. A sample power system network is shown in the figure. Is shown in the figure. The reactance marked are in pu. What is the pu value of Y_22 of the Bus Admittance matrix (YBus )?

(a) j10.0

(b) j0.5

(c) -j0.1

(d) -j20.0

Q3. In the op-amp circuit as shown in the figure, Current I_L is

Q4. The circuit shown in the figure uses an ideal op-amp working with +5 V and -5V power supplies. The output voltage V_0 is equal to

(a) +5 V

(b) -5 V

(c) +5 V

(d) -1 V

Q6. The position and velocity error coefficients for the system of transfer function,

G(s)=50/(1+0.1s)(1+2s) are respectively

(a) zero and zero.

(b) zero and infinity.

(c) 50 and zero.

(d) 50 and infinity.

Q7. The block diagram shows a unity feedback closed – loop system. The steady-state error in the response to a unit step input is

(a) 14%

(b) 28%

(c) 42%

(d) 57%

Q8. A 10 kVA, 400 V/ 200 V, single-phase transformer with a percentage resistance of 3% and percentage reactance of 6% is supplying a current of 50 A to a resistive load. The value of the load voltage is

(a) 192 V

(b) 194 V

(c) 196 V

(d) 390 V

Q9. The synchronous speed for the 7^th space harmonic mmf wave of a 3-phase, 8 pole, 50 Hz induction machine is

(a) 5250 rpm in forward direction

(b) 5250 rpm in reverse direction

(c) 107.14 rpm in forward direction

(d) 107.14 rpm in reverse direction

Q10. In an electrodynamometer type wattmeter:

(a) the current coil is made fixed

(b) the pressure coil is fixed

(c) both the coil is fixed

(d) both the coils are movable

SOLUTIONS

S7. Ans.(d)

Sol. G(S)=15/(s+2)(s+10)

Kp=(Lt@S→0) G(s)

= (Lt@S→0) 15/(s+2)(s+10) =15/(2×10)=3/4

Steady -state error,

e_ss=A/(1+K_P )=1/(1+3\/4)=4/7=0.57 or 57%

S8. Ans.(b)

Sol. % regulation= (R cosϕ+X sinϕ)×100

For resistive load: ?=0⁰

∴% regulation=3×1+6×0=3%=200×3/100=6 V

∴Terminal voltage=200-6=194 V

S9. Ans.(c)

Sol. N_s=120f/P=(120×50)/8=750 rpm

Speed of 7^th harmonic=N_s/7=750/7=107.14 rpm

And, for7^th harmonic=7×120^0=840^0=840-360-360=+120⁰(i.e.forward direction)

S10. Ans.(a)

Sol. Electro dynamo type wattmeter: –

I. Current coil is fixed and split in to two parts

II. pressure coil is moving and inserted between fixed coils.

III. These coils are air-cored so free from eddy and hysteresis loss.