Quiz: Electrical Engineering
Each question carries 1 mark.
Negative marking: 1/3 mark
Time: 10 Minute
Q1. 8 pole, DC generator has a simplex wave-wound armature containing 32 coils of 6 turns each. Its flux per pole is 0.06 Wb. The machine is running at 250 rpm. The induced armature voltage is:
Q2. How many 200 W/220V incandescent lamps connected in series would consume the same total power as a single 100 W/220V incandescent lamp?
Q3. The surge impedance of 400 km long over-head transmission line is 300 Ω. For a 200 km length of the same transmission line surge impedance will be
Q4. In a dynamometer type wattmeter, the pressure coil connected across the load terminal is
(a) Highly inductive
(b) Highly capacitive
(c) Highly resistive
(d) Non inductive
Q5. Due to the inductance in the pressure coil of dynamometer type wattmeter, the reading will be
(a) High for both leading and lagging power factors
(b) Low for both leading and lagging power factors
(c) High for lagging power factor and low for leading power factor
(d) Low for lagging power factor and high for leading power factor
Q6. One unit of electrical energy equals
Q7. If the applied voltage of a DC motor is 230 V, then back emf, for maximum power developed is
Q8. A 50-volt AC is applied across an RC (series) network. The rms voltage across the resistance is 40 V, then the potential across the capacitance will be
(a) 40 V
(b) 10 V
(c) 20 V
(d) 30 V
Q9. Which of the following option is true for resonance condition in series RLC circuits?
Q10. The efficiency of two identical transformers under loading can be determined by………test.
(d)Back to back
Where, A=number of parallel paths=2 (for wave winding)
And Z= 2×32×6=384
Sol. For 200 W/220V incandescent lamp, R=V^2/P=220^2/200=242 Ω
And for 100 W/220V incandescent lamp, R_t=V^2/P=220^2/100=484 Ω
According to question, for series connection: R_t=n×R
Sol. The surge impedance is independent of the length of the transmission line. Hence it will be same as before i.e. 300Ω.
Sol. In a dynamometer type wattmeter, the pressure coil connected across the load terminal is highly Resistive.
Sol. Because of the inductance, the current of the pressure coils lags behind the voltage. Thus, the power factor of the wattmeter becomes lagging, and the meter reads high reading and vice-versa.
Sol. One unit of electrical energy equals 1KWH.
Sol. For maximum power output, the back emf is equal to half of the applied voltage.
Hence, Eb = V/2 = 230/2 ⇒115 V. where V is the terminal voltage.
(VS )^2=(VR )^2+(VC )^2
⇒ VC = √(2500-1600)
= 30 V
Sol. In case of series Resonance, X_L = X_C and Total impedance (z) = √(R^2+(XL-XC )^2 )
= √((R)^2+(0)^2 )
⇒ ▭(Z=R )
and Power factor (cosϕ) = R/Z=R/R=1
So, in series Resonance,
Z = Z(min )=R
pf = unity,
Sol. The efficiency of two identical transformers under loading can be determined by back to back test.