Quiz: Electrical Engineering
Each question carries 1 mark.
Negative marking: 1/3 mark
Time: 10 Minute
Q1. The relative permeability of a medium is equal to (with M=magnetization of medium and H=magnetic field strength)
Q2. For the circuit given in figure below the power delivered by the 2-volt source is given by
(a) 4 W
(b) -4 W
(c) 2 W
(d) -2 W
Q3. Elements R, L and C are connected in parallel, the impedance of the parallel combination can be expressed as
The value of the individual elements R, L and C are
(a) 10 Ω, 40 H and 0.1 F
(b) 10 Ω, 1/40 H and 0.1 F
(c) 4 Ω, 1 H and 0.1 F
(d) 1 Ω, 40 H and 10 F
Q4. Which of the following represents the operating region slips of a 1-phase induction motor?
Q5. If the torque/weight ratio of an instrument is low, then it can be concluded that
(a) The meter will have a uniform scale
(b) The meter will have the non-uniform scale
(c) The sensitivity of the meter will be high
(d) The sensitivity of the meter will be low
Q6. The Ybus matrix of a 100-bus interconnected system is 80% sparse. Then, the number of transmission lines in the system must be
Q7. Consider the following statements in connection with frequency domain specifications of a control system:
1.Resonant peak and peak overshoot are both functions of the damping ratio ξ only.
2.The resonant frequency ω_r=ω_n for ξ > 0.707.
3.Higher the resonant peak, higher is the maximum overshoot of the step response.
Which of the statements given above are correct?
(a) 1 and 2
(b) 2 and 3
(c) 1 and 3
(d) 1, 2 and 3
Q8. An ideal transformer has an input voltage of 480V. The output current and voltage are 10A and 120V respectively. Determine the value of input current-
(a) 25 A
(b) 2.5 A
(c) 0.25 A
(d) 0.025 A
Q9. The no-load current drawn by transformer is usually what percentage of the full-load current?
(a) 0.2 to 0.5 %
(b) 12 to 15 %
(c) 20 to 30 %
(d) 2 to 5 %
Q10. The secondary winding of which of the following transformers is always kept closed?
(a) Current transformer
(b) Voltage transformer
(c) Power transformer
(d) Step-down transformer
Sol. Magnetic susceptibility(X)=M/H=µ_r-1
Where, µ_r=relative permeability
Here, i=2/1=2 A
On applying KCL at node a: I=i-1=2-1=1 A
i.e. the current delivered by the voltage source=1 A
∴power delivered by the voltage source=2×1=2 W
Sol. Given that, Z(s)=10s/(s^2+s+400)
On comparing: C=1/10 F,R=10 Ω and L=1/40 H
Sol. In a 1-phase induction motor, due to both forward and backward torque slips, the operating region lies between 0 and 2. In motoring mode slip is 0 to 1 and breaking mode 1 to 2.
Sol. The low torque/weight ratio signifies heavy weight of the moving system due to which the instrument incurs frictional losses hence the sensitivity will be low.
Sol. Number of buses, N=100
Then, the number of transmission lines is given as: N_T=(N^2 (1-s)-N)/2=(100^2 (1-0.8)-100)/2
S7. Ans. (c)
Sol. → Resonant Peak (M_r) = 1/(2ξ √(1-ξ²))
Peak overshoot = e (-ξπ)/√(1-ξ²)
Resonant frequency (wr) = Wn √(1-2ξ²)
For ξ = 0
wr = Wn.
Sol. V₁I₁ = V₂I₂
⇒ I₁ = (V₂I₂)/(V₁)
Sol. Current required to set up magnetic flux in a magnetic material is rather low. Hence no-load primary current is a very small fraction of the rated full load current
Sol. Because if CT secondary is not shorted then it can develop a very high voltage across secondary which may damage transformer insulation.