Bangladesh Chemical Industries Corporation (BCIC) || Sub-Assistant Engineer (EEE) (14-02-2025) || 2025

All Written Question

1

Find Vab and Is. 

Created: 3 weeks ago | Updated: 3 weeks ago
Updated: 3 weeks ago

Va=R4R3+R4E=26+2×12=3V

Vb=R2R1+R2E=35+3×12=4.5V

Vab=Va-Vb=3-4.5=-1.5 V 

Is=I1+I3=Vb3+Va2=4.53+32=3 A

(a) Real Power, P=vrmsirmscos (θv-θi)=502×22cos{(-20°)-70°}=0W

Reactive Power, Q=vrmsirmssin(θv-θi)=502×22sin {(-20°)-70} =-50VAR

Apparent Power, S=vrmsirms=502×22=50VA

(b) Power factor, pf=cos(θv-θi)=cos{(-20°)-70°}=cos(-90°)=0

(c) Impedance, Z=VmIm(θv-θi)=502{(20°)-70}=25-90°=-j25Ω

(d) As the reactive power is negative (Q = - 50VAR) and the impedance has a negative imaginary part (- j25) the load is purely capacitive (the current leads the voltage by 90° .

The voltage source is balanced (100 V) having acb sequence. 

VAN =100°

VCN =100-120° 

VBN =100-240°  

Ia=VANZA=1000°15=6.670°

Ib=VBNZB=100120°10+j5=8.9493.44°

Ic=VCNZC=100120°6-j8=10-66.87°

Neutral Line Current, 

In=-(Ia+Ib+Ic)=(6.670°+8.9493.44°+10-66.87°)

=10.06178.4°AAns.

(i) Demand factor = Maximum Demand Connected Load = 30 50 == 0.6  (Ans)

(ii) Average Load = Units generated in one year Hours in a year = (60 x 106)(365 x 24) =6849.315 kW  6.849315 MW 

 Load factor = Average Load Maximum Load =6.849315 30 0.2283 = 22.83% Ans

Given, 

Terminal Voltage, Vt = 400V 
Speed, N = 1000 r. p. m. = 1000 60 r.p.s. 
                                             = 16.667 r. p.s. 

Angular speed, ωm = 2πN = 2π× 16.667 
                                       = 104.72 rad/sec 

Armature Resistance, RA=0.25 Ω

Armature current IA = 100A 

Back emf = E

Electrical Power Converted to mechanical power = Pm 

Torque, T = ?

 

Eb= Vt-IA RA = 400-100 × 0.25 = 375 V 

Pm=EbIA= 375 × 100 = 37500 W 

T = Pm ωm=37500 104.72 358.1 Nm Ans

Modulation index for FM, 

m=m=5010=5

The highest instantaneous frequency,  max=c+=108+0.1=108.1  MHz Ans

The lowest instantaneous frequency, min=c-=108-01=107.9 MHz Ans