Ques.121. An alternating current is given by i = 20 sin 157t A. The frequency of the alternating current is____
100 Hz
75 Hz
25 Hz✓
50 Hz
Let the expression be as follows:-
I(t) = Im sin(ωt ± φ) = Im sin (157t)
Where φ represent the concerned phase-shift
Vm = Maximum value of the voltage = 200 V
Angular frequency ω = 2πf
157 = 2πf
Hence frequency f = 157/2π = 25Hz
Ques.122. The two-wattmeter method is used to measure the power of a three-phase balanced system powered by a 415 V, three-phase, 50 Hz power supply. If the reading on both wattmeters is 8.5 kW, calculate the power factor.
1✓
0.98 Lagging
0.88 lagging
0.858 lagging
As given in the above question
W1 = 8.5 kW
W2 = 8.5 kW
VL = 415 V
The power factor of the two wattmeters is
tanØ = √3[(W1 – W2) / (W1 + W2)]
tanØ = √3[(8.5 – 8.5) ⁄ (8.5 + 8.5)]
Φ = 0°
Power factor = cosφ = cos0° = 1
Ques.123. The energy used by a 1.5 kW heater in 5 minutes is:
45,000 J✓
450 J
7500 J
None of these
Electric energy is the total amount of electrical work done in an electrical circuit. The electric energy can also be defined as the product of power and time. The S.I Unit of Electrical- Energy is joule or watt-sec.
1.5 kW = 1500 joule
time = 5 minute = 5 × 60 = 300 sec
Electric Energy = 1500 × 300 = 45,000 Joules
Ques.124. The RMS value of a sine wave having the peak value Vm over one cycle is____
Vm
Vm/π
Vm/√2✓
Zero
The RMS value of a sine wave having the peak value Vm over one cycle is Vm/√2.
The effective or r.m.s. value of alternating current is that steady current (d.c.) which when flowing through a given resistance for a given time produces the same amount of heat as produced by the alternating current when flowing through the same resistance for the same time. It is also called the virtual value of a.c. and is represented by Irms or Ieff or Iv.
For example, when we say that r.m.s. or effective value of an alternating current is 5A, it means that the alternating current will do work (or produce heat) at the same rate as 5A direct current under similar conditions.
To determine r.m.s. value of a sinusoidal alternating current, we have to first square it, then take its mean over one cycle or half-cycle and then take the square root (note that r.m.s. value is calculated by making reverse operation, that is, the first square, then take mean and then take square root).
Square of current i = Im sinθ = I2m sin2θ
Its mean over one cycle is calculated by integrating it from 0 to 2π and dividing by the time period of 2π.
Hence the r.m.s. value or effective value or virtual value of alternating current is 0.707 times the peak value of alternating current for the half-cycle as well as for the full cycle.
Ques.125. A three-phase star-connected the balanced load of (4 +j3)Ω per phase is connected across a three-phase, 50 Hz, 400 V AC supply. Determine the power factor of the load.
0.8 lagging✓
0.3 leading
0.9 lagging
0.7 leading
Given
Resistance R = 4 Ω
Line voltage VL = 400 V
Frequency f = 50 Hz
Impedance Z = (4 +j3)Ω
Impedance per phase ZP
ZP = √(R2 + X2L)
ZP = √(42 + 32)
ZP = 5Ω
Power factor cosφ = Rph/Zph = 4/5 = 0.8
Ques.126. In a three-phase system, the relation VL = Vph is applicable to a
Star connected load
Delta connected load✓
Star connected without the neutral point
Single-phase system also
In delta connected system the line voltage is equal to the phase voltage and the Line current is √3 times of Phase Current.
Ques.127. The dielectric strength of a material depends on its____
Ductility
Density
Volume
Moisture content✓
Materials that restrict or prohibit the flow of sound or heat or electricity through them are called insulators or insulating materials. They also avoid the accumulation of heat by proper distribution.
The requirement of good insulating materials can be classified as electrical, mechanical, thermal and chemical. Electrically the insulating material should have high resistivity to reduce the Leakage current and high dielectric strength to enable it to withstand higher voltage without being punctured or broken down. Further, the insulator should have a small dielectric loss.
(a) High Resistivity:- Electrical conductivity is the capacity of a material to conduct the electric current. Resistivity is the reciprocal of conductivity and is commonly measured as Insulation resistance. Thus a good insulator is that which possesses low conductivity or high resistivity.
(b) High Dielectric strength:- The Dielectric strength of an insulating material determines its ability to resist puncture or breakdown by electric potentials. Hence the limiting intensity by electric field above which the insulators get punctured or ruptured is called the dielectric strength of the dielectric and is commonly expressed in volts per mm of the thickness of the material or volts per centimeter.
The dielectric strength of a material can also be defined as the electrical field strength, that must be applied to cause a disruptive effect or discharge (arc or spark) or accumulate current through the body of the material.
The dielectric strength is affected by a number of factors such as
(a) Dielectric strength decreases as the thickness of the material increases
(b) Dielectric strength decreases as the time of the application of electric current increases
(c) Dielectric strength decreases as the frequency of the applied electric field is increased
(d) Dielectric strength decreases in presence of humidity and moisture content.
(e) Dielectric strength decreases with an increase in temperature in the case of air as an insulator.
(c) Low Dielectric loss:- These are generally caused due to absorption of electrical energy and also by the leakage of current through the insulating material. The absorption of electrical energy. under an alternating electric field gives rise to the dissipation of the electrical energy in the insulating material, while leakage takes place as a result of conduction.
The conduction is appreciable, however, only at high temperatures, otherwise, it is negligible. A good and ideal dielectric is that which does not absorb electrical energy and the charge of the capacitor is completely recovered when the electrical field is removed.
A dielectric material, however, always causes some electrical energy loss, which is measured by the phase difference between the phase angle and 90 angles of an ideal capacitor. A perfect insulator is that has a phase angle of 90° and a loss angle of zero.
Ques.128. The capacitor of capacitance C1 = 0.1μF and C2 = 0.2μF are connected in series across a 300-V source. The voltage across C1 will be_____
150 V
300 V
100 V
198 V✓
Given
Capacitance C1 = 0.1μF
Capacitance C2 = 0.2μF
Voltage = 300 V
Since the capacitance is connected in series their equivalent capacitance will be
C = (C1 × C2) ⁄ (C1 + C2)
C = (0.1 × 0.2) ⁄ (0.1 + 0.2)
C = 0.06 μF
So, the total charge flowing in the circuit is
Q = CV = 0.066 × 300 = 19.8 μC
So, the voltage drop across 0.1 μF will be
=19.8/0.1 =198 V
Ques.129. Find the phase current if a three-phase star connected system is connected to a 400 V, 50 Hz AC supply. Assume Zph = (9.8 + j10)Ω
28.57 A
10 A
11.44 A
16.5 A✓
Given
Resistance R = 9.8 Ω
Line voltage VL = 400 V
Frequency f = 50 Hz
Impedance Z = (9.8 + j10)Ω
Impedance per phase ZP
ZP = √(R2 + X2L)
ZP = √(9.82 + 102)
ZP = 14Ω
In star connected system the phase voltage is
VP = VL/√3 = 400/√3 = 230.94 V
IP = Vp/Zp
IP = 230.94/14
IP = 16.5A
Ques.130. If two 5-μF capacitor are connected in parallel, then the effective capacitance will be___
2.5 μF
25 μF
10 μF✓
20 μF
When two capacitance are connected in parallel their equivalent capacitance