Quiz: Electrical Engineering

Exam: UPPSC-AE

Topic: Miscellaneous

Each question carries 1 mark.

Negative marking: 1/4 mark

Time: 10 Minute

Q1. A 3-phase transformer has its primary connected in delta and secondary in star.

Secondary to primary turns ratio per phase is 6. For a primary voltage of 200 V, the

secondary voltage would be

(a) 2078 V

(b) 693 V

(c) 1200 V

(d) 58 V

Q2. The iron loss in a 100 KVA transformer is 1 kW and full load copper losses are 2

kW. The maximum efficiency occurs at a load of

(a) 100 KVA

(b) 70.7 KVA

(c) 141.4 KVA

(d) 50 KVA

Q3. The iron loss per unit frequency in a ferromagnetic core, when plotted against

frequency, is a

(a) Straight line with positive slope

(b) Straight line with negative slope

(c) Parabola

(d) Constant

Q4. Following graph shows the loss characteristic of a sheet of ferromagnetic material

against varying frequency f. P_i is the iron loss at frequency f. Hysteresis and eddy current

losses of the sheet at 100 Hz are

(a) 10 W, 100 W

(b) 10 W, 50 W

(c) 1 W, 5 W

(d) 1 W, 10 W

Q5. The system with the open loop transfer function 1/s(1+s) is:

(a) Type 2 and order 1

(b) Type 1 and order 1

(c) Type 0 and order 0

(d) Type 1 and order 2

Q6. If the number of inputs to a multiplexer are 32, the number of control signals needed

are

(a) 4

(b) 5

(c) 6

(d) 8.

Q7. The figure given below represents ……… operation.

(a) AND

(b) OR

(c) EX-OR

(d) NAND.

Q8. What is the order decided by a processor or the CPU of a controller to execute an instruction?

(a) decode, fetch, execute

(b) execute, fetch, decode

(c) fetch, execute, decode

(d) fetch, decode, execute

Q9. A battery of 36V is applied across terminals A and B of the circuit shown in figure. The current in 4Ω resistor is

(a) 1.5 Amp

(b) 1.4 Amp

(c) 2 Amp

(d) 1.6 Amp

Q10. The current in 4Ω resistor is

(a) 3 Amp downward

(b) 6 Amp upward

(c) 9 Amp upward

(d) 2 Amp downward

solutions

S1. Ans.(a)

Sol. Δ /Y transformer (Given)

Phase turn ratio:

N2/N1 =6=V2/V1

V2/200=6⇒V2=1200

Secondary line voltage

=√3 V2=√3×1200

=2078 volt

S2. Ans.(b)

Sol. Load for maximum efficiency.

(S_η)max =Sfull load √(Pi/Pcufl )

=100 kVA√(1/2)=70.7 kVA

S3. Ans.(a)

Sol. Iron loss given by

Pi=af+bf²

∴Pi/f=a+bf……..STRAIGHT line with positive slope

S4. Ans.(d)

Sol. Pi/f=a+bf

From the graph, a = 0.01

And, b = 0.001

So, Hysteresis loss at f = 100 Hz

Ph=0.01×100=1 W

Pe=0.001×100×100= 10 W

S5. Ans.(d)

Sol. Type is defined as the number of poles at origin and order is defined as the total number of poles and this is calculated with the help of the transfer function from the above transfer function the type is 1 and order is 2.

S6. Ans.(b)

Sol. n = 2^m

Where n is number of inputs and m is number of control signal needed.

S7. Ans.(d)

Sol. By using binary values

If A=B=0 then Y =1

If A=0, B=1 then Y =1

If A=1, B=0 then Y =1

If A=B=1 then Y =0

S8. Ans.(d)

Sol. While any instruction is being executed, a micro-controller first fetches the instruction (captures its operand and operator). After capturing it converts these operands and operators into their corresponding hex codes. Hence after this, an instruction can be executed as now it is in the form of 0’s and 1’s (the format understood by a micro-controller).

S9. Ans.(a)

Sol. Total circuit resistance Rt=6Ω

∴I = 36/6 = 6 Amp.

Current in 12Ω resistor =36/12 = 3 Amp

∴ Voltage drop in 7Ω resistor = 7 × 3 = 21 volt.

∴ Voltage across 10Ω = 36 – 21 = 15 V

∴ Current in 10Ω = 15/10 = 1.5 Amp

Therefore, current in 4Ω is 1.5 Amp.

S10. Ans.(a)

Sol.

Applying node analysis at A,

Let voltage at node A is V

(V – 18)/6+(V-16)/2+V/4=0

2V-36+6V-96+3V=0

V=12 Volt

Therefore,

Required current I = 12/4 = 3 Amp.