Quiz Electrical Engineering
Each Question carries 2 Mark
Negative Marking: 1/3
Time: 20 Minutes
Q1. In the circuit below, which of the following sources is supplying no power?
(a) 4 A source
(b) 24 V source
(c) Both the sources
(d) None of the above
Q2. Consider a circuit shown in the figure below. The voltage across 5 Ω resistor is measured to be V_0. If a finite resistance is connected in series with 5 A source, then value of V_0 .
(c) May increase or decrease
(d) Remains unchanged
Q3. If the total charge stored in a capacitor is doubled, then the energy stored in the capacitor will be
(d) None of the above
Q4. In a capacitor start single-phase induction motor, the capacitor is connected
(a) In series with main winding
(b) In series with auxiliary winding
(c) In parallel with auxiliary winding
(d) In series with both the windings
Q5. A 50 KVA, 1000/200 V transformer takes 5 A current at a power factor of 0.2 lagging, when connected to 1000 V side at no load is connected at the other side. The core loss in transformer will be:
(a) 500 W
(b) 1000 W
(c) 2000 W
(d) 2200 W
Q6. For a single-phase transformer, wattmeter readings for OC test and SC test are as given below:
Wattmeter reading in OC test: 2.5 KW
Wattmeter reading in SC test: 5 KW
Find maximum efficiency of 5 KVA transformer for unity power factor.
(a) 33.34 %
(b) 40.0 %
(c) 66.7 %
(d) 70 %
Q7. A moving coil instrument having meter resistance of 5Ω is to be used as a voltmeter of range 0-100 V. If the full-scale deflection current is 10 mA, then required series resistance is………….
(d) None of these
Q8. Consider the function F(s) = ω/(s^2+ω²)
Where F(s) = Laplace transform of f(t). The final value of f(t) is equal to
(c) finite constant
(d) a value in between – 1 and + 1
Q9. The power gain in decibels can be expressed as
(a) 20 log_10(P_O/P_i )
(b) 10 log_10(V_O/V_i )
(c) 10 log_10(P_O/P_i )
(d) 20 log_10(P_O/V_i )
Q10. Negative feedback used in amplifier results in-
(a) More gain, more bandwidth
(b) More gain, less bandwidth
(c) Less gain, more bandwidth
(d) Less gain, less bandwidth
Sol. Adding a finite value of resistor in series with 5 A source does not change the node equation at top right node, so V_0 remains same.
Sol. Energy stored, E=Q^2/2C
So, if the total charge stored in a capacitor is doubled, then the energy stored in the capacitor will be four times its original value.
Sol. In a capacitor start single-phase induction motor, the capacitor is connected in series with auxiliary winding.
1-𝟇 capacitor start IM Circuit:
Sol. Core loss is constant in transformer irrespective of supply-side whether the rated supply voltage is applied on high voltage side or low voltage side.
No load core loss is given by: P_core = VI_0 cosϕ_0
Where V = rated no load supply voltage
I_0 = No load currents
Cosϕ_0 = no load power factor
∴No load copper loss = 1000 × 5 × 0.2 = 1000 watts
Sol. Wattmeter reading in OC test: 2.5 KW= core or iron loss(P_i)
Wattmeter reading in SC test: 5 KW=copper loss or variable loss(P_cu)
VA rating=5 KVA
∴η=(VI cosϕ)/(VI cosϕ+P_i+P_cu )=(5×1)/((5×1)+2.5+5)=5/12.5×100=40%
So, option (b) will be right.
Sol. Series resistance,
R_s=V/I_m -R_m=(100 V)/(10 mA)-5=10000-5=9995Ω
S8. Ans. (d)
Sol. F(s) = ω/(s^2+ω²)
Taking inverse Laplace:
f(t) = 〖sin ω〗t
Therefore, the final value for sinωt is equals to a value between -1 and +1.
Sol. Power gain =P_o/P_i
in decibel (dB):
dB = 10 log_10(P_o/P_i )
=10 log_10〖(V_o/V_i )^2 〗
=20 log_10(V_o/V_i )
Sol. Effects of Negative feedback:
In negative feedback, the feedback energy (voltage or current), is out of phase with the input signal and thus opposes it.
Negative feedback reduces gain of the amplifier. It also reduces distortion, noise and instability.
This feedback increases bandwidth and improves input and output impedances.