DMRC JE/AM Based quiz Electrical Engineering (miscellaneous)

Q1. Surge impedance of pure inductor is –
(a) 0
(b) 1
(c) -1
(d) ∞
Q2. Surge impedance of pure capacitor is –
(a) 0
(b) 1
(c) -1
(d) ∞
Q3. Match the following.
Wave RMS (Irma)
Sine wave (a) I peak/√3
Sawtooth wave (b) I peak/√2
Square wave (c) I peak
Triangular wave (d) Ipeak/(√2)
I II III IV
(a) b a c d
(b) b a c a
(c) a b c d
(d) d c b a
Q4. Current through the solenoid is given by –
(a) I =Bl/Mon
(b) I =B/Mon
(c) I =B/N
(d) I =Bl/N
Q5. Which of the following is the expression for quality factor of a series RLC circuit ?
(a) 1/R √(L/c)
(b) 1/R √(R/C)
(c) R√(C/L)
(d) R√(L/C)
Q6. The insulation resistance of a cable of length 10 km is 1MΩ . for a length of 100 km of the same cable, the insulation resistance will be –
(a) 1MΩ
(b) 0.1 MΩ
(c) 10 MΩ
(d) 100 MΩ
Q7. The corona loss on a particular system at 50HZ is 1kw/km per phase. The corona loss at 60 HZ will be –
(a) 1.0 kw|km|ph
(b) 2.1 kw|km|ph
(c) 2.0 kw|km|ph
(d) 1.13 kw|km|ph
Q8. A 400 kV transmission line is having per phase line inductance and capacitance 1.1 mH/km and 44 uf/km. Ignoring the length of the line, its ideal power transfer capability in MW is –
(a) 2094 MW
(b) 3094 MW
(c) 3000 MW
(d) 3200 MW
Q9. Consider two buses connected by an impedance of (0+j5) Ω. The bus – 1 voltage is 10<30° and bus-2 voltage is 100<0° V. The real power supplied by bus – 1 is –

(a) 10 w
(b) 100 w
(c) 1000 w
(d) 10000 w
Q10. In previous question, the reactive power supplied by bus 1 is –
(a) 100 VAR
(b) 200 VAR
(c) 268 VAR
(d) 368 VAR
Solutions
S1. Ans.(d)
Sol. Z_C= √(L/C)= √(L/O)=∞
S2. Ans.(a)
Sol. . Z_C= √(L/C)= √(O/L)=0
S3. Ans.(b)
S4. Ans.(a)
S5. Ans.(a)
Sol.
Q=(W_R L)/R=X_L/R=1/(W_r CR)=X/R=1/R √(L/C)

S6. Ans.(b)
Sol
Rcable α 1/(length )
⇒R_1/R_2 =l_2/l_1
⇒R_2=(R_1 l_1)/l_2
=(1×10)/100=0.1 MΩ
S7. Ans.(d)
Sol. Corona loss α (f + 25)
(Pc_2)/(Pc_1 )=((f_2+25))/((f_1+25))
⇒Pc_2=((f_2+25))/((f_1+25) )
=1×((60+25)/(50+25))
=85/75 kw|km|ph
=1.13kw|km\ph
S8. Ans.(d)
Sol. surge impedancez = √(L/C)= √((101×10^(-3))/(44×10^(-6) ))=10/2=5Ω
SIL=(V_R )^2/Z_c
=(400)^2/5
=(16×10^4)/5 MW
=3200 MW
S9. Ans.(c)
Sol.
p=|V_s ||V_R |/|×_5 | Sin S
=(100×100)/5 sin30°=(100×100×1)/(5×2)
=1000 MW