EE GATE 2012 - Online Test

Q1. In the 3-phase inverter circuit shown, the load is balanced and the gating scheme is 1800 -conduction mode. All the switching devices are ideal
If the dc bus voltage V= 300 V, the power consumed by 3-phase load is
Answer : Option D
Explaination / Solution:



Q2.
The impedance looking into nodes 1 and 2 in the given circuit is

Answer : Option A
Explaination / Solution:




Q3. The sequence components of the fault current are as follows: Ipositive = j1.5 pu, Inegative = –j0.5 pu, Izero = –j1 pu. The type of fault in the system is 
Answer : Option C
Explaination / Solution:

I1 = I2 + I0

So LLG fault

Q4.
Consider the differential equation
 The numerical value of  is
Answer : Option D
Explaination / Solution:



Q5.
The transfer function of a compensator is given as

Gc(s) is a lead compensator if
Answer : Option A
Explaination / Solution:


both option (A) and (C) satisfier but option (C) will pot polar and zero as RHS of s-plane thus not possible option (A) is right 

Q6. In the following figure, C1 and C2 are ideal capacitors. C1 has been charged to 12 V before the ideal switch S is closed at t = 0. The current i(t) for all t is 

Answer : Option D
Explaination / Solution:

Time constant = RC In the given circuit, R = 0 Rise time = 0; hence capacitor charges instantaneously and the current can be represented as impulse function

Q7. A cylindrical rotor generator delivers 0.5 pu power in the steady-state to an infinite bus through a transmission line of reactance 0.5 pu. The generator no-load voltage is 1.5 pu and the infinite bus voltage is 1 pu. The inertia constant of the generator is 5 MW-s/MVA and the generator reactance is 1 pu. The critical clearing angle, in degrees, for a three-phase dead short circuit fault at the generator terminal is
Answer : Option D
Explaination / Solution:




Q8. A fair coin is tossed till a head appears for the first time. The probability that the number of required tosses is odd, is
Answer : Option C
Explaination / Solution:



Q9. The transfer function of a compensator is given as

The phase of the above lead compensator is maximum at
Answer : Option A
Explaination / Solution:



Q10. The average power delivered to an impedance (4 - j3)Ω by a current 5cos(100πt +100) A is
Answer : Option B
Explaination / Solution: