Quiz: Electrical Engineering

Exam: GATE

Topic: Miscellaneous

Each question carries 2 mark.

Negative marking: 1/3 mark

Time: 20 Minute

Q1. A squirrel cage induction motor runs slow, probably cause could be

(a) low voltage

(b) short stator coil

(c) one phase open

(d) Any of the above

Q2. A 5 kVA, 50 V/100 V, single-phase transformer has a secondary terminal voltage of 96 V when loaded. The regulation of the transformer is

(a) 4%

(b) 8%

(c) 2%

(d) 4.5%

Q3. The Voltage wave V=V_m sin〖(ωt-15°)〗 Volt is applied across an AC circuit. If the current leads the voltage by 10° and the maximum value of current is I_m,then the equation of current is

(a) i=I_m sin〖(ωt+5°)〗

(b) i=I_m sin〖(ωt-25°)〗

(c) i=I_m sin〖(ωt+25°)〗

(d) i=I_m sin〖(ωt-5°)〗

Q4. Two identical coils each having inductance L are placed together on the same core. If an overall inductance of aL is obtained by interconnecting these two coils, the minimum value of a is

(a) 2

(b) .5

(c) 0

(d) None

Q5. The inductance of a power transmission line increases with

(a) decrease in line length

(b) increase in diameter of conductor

(c) increase in spacing between the phase conductor

(d) increase in load current carried by conductors

Q6. A current impulse of 10 δ(t), is forced through a capacitor C. The voltage Vc(t), across the capacitor is given by

(a) 10t

(b) 10 u(t) – C

(c) 10/Ct

(d) (10u(t))/C

Q7. Which of the 3-phase connection of a Transformer causes interference to nearby communication system?

(a) Star-Delta

(b) Star-star

(c) Delta-star

(d) Delta-delta

Q8. When transformer winding suffers a short circuit, the adjoining turns of the same winding experience

(a) an attractive force

(b) a repulsive force

(c) no force

(d) none of the above

Q9. A 3-phase delta star Transformer has secondary to primary turns ratio per phase of 5. For a primary line current of 10 A, the secondary line current would be

(a) 3.464 A

(b) 50 A

(c) 1.155 A

(d) 1.633 A

Q10. G (s) =((0.5s+1))/((0.05s+1)) maximum phase lead of the compensator is

(a) 52 deg at 4 rad/sec

(b) 52 deg at 10 rad/sec

(c) 55 deg at 12 rad/sec

(d) none of the above

Solution

S1. Ans. (d)

Sol. All the condition result in slow running of motor. Over load and open stator coil may be additional reason for slow running.

S2. Ans. (a)

Sol. We know that (%VR=1-V_FL/V_NL )×100

%VR=(1-96/100)×100=4%

S3. Ans. (d)

Sol. Given that, V=V_m sin〖(ωt-15°)〗, and current is leading by 10°.

So, I=I_m sin〖(ωt-15°+10°)〗= I=I_m sin〖(ωt-5°)〗

S4. Ans. (c)

Sol.

Case 1

L_eff=L_1+L_2=2L,a=2

Case 2

L_eff=(L_1 L_2)/(L_1+L_2 )=L^2/2L=0.5L=a=0.5

Case 3

If both are differentially coupled then

L_eff=0;So minimum Value=0

S5. Ans. (c)

Sol. L∝ln〖d/r’〗

S6. Ans. (d)

Sol. V_c (t) = 1/C ∫*(-∞)^t▒i(t)dt =1/C ∫*(-∞)^t▒〖10 δ(t)dt〗

= (10u(t))/C

S7. Ans. (b)

Sol. Star-star

S5. Ans. (a)

Sol. Due to same direction of current in the adjoining winding it experiences force of attraction.

S9. Ans. (c)

Sol. Given,N_Y(P) /N_∆(P) =5=I_∆(P) /I_Y(P)

I_∆(P) =5×I_Y(P)

I_∆(P) = I_(∆(L)/√3)=10/√3 A

Now, I_(Y(P) )=10/(5√3)=1.155 A

S10. Ans. (d)

Sol. G (s) =((0.5s+1))/((0.05s+1)) On comparing with,

G (s) =((1+aTs))/((1+Ts)) We get, T= 0.05 and a=10

∅_n= 〖sin^(-1) (〗〖(a-1)/(a+1))〗= 55°

And ω_m= 1/(T√a) = 6.32 rad/sec