# UPPCL JE 2018 Electrical question paper with Explanation 27-Aug-2018

Ques.121. An alternating current is given by i = 20 sin 157t A. The frequency of the alternating current is____

1. 100 Hz
2. 75 Hz
3. 25 Hz
4. 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. 1
2. 0.98 Lagging
3. 0.88 lagging
4. 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:

1. 45,000 J
2. 450 J
3. 7500 J
4. 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____

1. Vm
2. Vm
3. Vm/√2
4. 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.

Mean value of alternating current

$\begin{array}{l}{\text{Mean of square}}\\\\ = 1/2\pi \int\limits_0^{2\pi } {{I^2}_m} {\sin ^2}\theta d\theta \\\\{\text{RMS Value}}\\\\{\rm{I = }}\sqrt {1/2\pi \int\limits_0^{2\pi } {{I^2}_m} {{\sin }^2}\theta d\theta } \\\\{\sin ^2}\theta = \dfrac{{1 – \cos 2\theta }}{2}\\\\{\rm{I = }}\sqrt {\dfrac{{{I^2}_m}}{{4\pi }}\int\limits_0^{2\pi } {(1 – \cos 2\theta } )d\theta } \\\\ = \sqrt {\frac{{{I^2}_m}}{{4\pi }}\left[ {\theta – \dfrac{{\sin 2\theta }}{2}} \right]_0^{2\pi }} \\\\ = \sqrt {\dfrac{{{I^2}_m}}{{4\pi }}} \times 2\pi \\\\ = \sqrt {\dfrac{{{I^2}_m}}{2}} \\\\I = \frac{{{I_m}}}{{\sqrt 2 }} = 0.707\end{array}$

If we calculate the r.m.s. value for half cycle, it can bc seen that we will get the same value by calculating as

$\begin{array}{l}{\text{Mean of square}}\\\\ = 1/\pi \int\limits_0^\pi {{I^2}_m} {\sin ^2}\theta d\theta \\\\{\text{RMS Value}}\\\\{\rm{I = }}\sqrt {1/\pi \int\limits_0^\pi {{I^2}_m} {{\sin }^2}\theta d\theta } \\\\{\sin ^2}\theta = \dfrac{{1 – \cos 2\theta }}{2}\\\\{\rm{I = }}\sqrt {\frac{{{I^2}_m}}{{2\pi }}\int\limits_0^\pi {(1 – \cos 2\theta } )d\theta } \\\\ = \sqrt {\dfrac{{{I^2}_m}}{{2\pi }}\left[ {\theta – \dfrac{{\sin 2\theta }}{2}} \right]_0^\pi } \\\\ = \sqrt {\dfrac{{{I^2}_m}}{{2\pi }}} \times \pi \\\\ = \sqrt {\dfrac{{{I^2}_m}}{2}} \\\\I = \dfrac{{{I_m}}}{{\sqrt 2 }} = 0.707\end{array}$

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.

1. 0.8 lagging
2. 0.3 leading
3. 0.9 lagging
4. 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

1. Star connected load
2. Delta connected load
3. Star connected without the neutral point
4. 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____

1. Ductility
2. Density
3. Volume
4. 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.

Dielectric properties:-

The following important properties of insulation are taken into consideration in electrical insulating materials.

(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_____

1. 150 V
2. 300 V
3. 100 V
4. 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) ⁄ (C+ 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

1. 28.57 A
2. 10 A
3. 11.44 A
4. 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___

1. 2.5 μF
2. 25 μF
3. 10 μF
4. 20 μF

When two capacitance are connected in parallel their equivalent capacitance

C = C1 + C2

C = 5 + 5 = 10 μF

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