UPPCL AE Electrical Engineer Solved Paper 2019

Answer.2. 70.7 V

Explanation:-

For Pulse width modulator (PWM) full-bridge rectifier the output voltage is given by

V_{out fundamental} = 0.707 × m × V_dc

Where

Amplitude modulation index = m = 1

Input DC voltage = V_dc = 100 V

V_{out fundamental} = 0.707 × 1 × 100

V_{out fundamental} = 70.7 V

Answer.3. R_x = 50Ω

Explanation:-

Given

R₁ = 500 Ω

R₂ = 125 Ω

R₃ = 200Ω

R_x = ?

According to Wheatstone bridge relation

R₁ R_x = R₂ R₃

500 × R_x = 125 × 200

R_x = 50 Ω

Answer.3. Critically damped

Explanation:-

[katex]G(s) = \dfrac{9}{{({s^2} + 6s + 9)}}[/katex]

Comparing the above equation with standard second-order closed-loop transfer function

[katex]G(s) = \dfrac{{\omega _n^2}}{{({s^2} + 2\xi {\omega _n}s + \omega _n^2)}}[/katex]

∴ 2ξω_n = 6

ω²_n = 9  = ω_n = 3 rad/sec

∵ 2ξω_n = 6

2ξ × 3 = 6

ξ = 1

Hence the system is critically damped.

Answer.2. 90°

Explanation:-

Given,

Average output voltage (V_o) = 103.53V

RMS voltage = 230 V

Max Voltage (V_m) = √2 × RMS voltage

V_m = 325.22 V

The average output voltage of the half-converter bridge rectifier is given by

[katex]\begin{array}{l}{V_o} = \dfrac{{{V_m}}}{\pi }(1 + \cos \alpha )\\\\103.53 = \dfrac{{325.22}}{{3.14}}(1 + \cos \alpha )\end{array}[/katex]

(1 + cosα) = 325.22/325.08

(1 + cosα) ≅ 1

cosα = 0

α = cos⁻¹ 0

α = 90°

Answer.4. When the transformation ratio for voltage is small

Explanation:-

The transformers used to step down the transmission voltage to the distribution voltage are called distribution transformers. Hence the voltage ratio of such transformers is very high. Similarly, the isolation between primary and secondary is necessary. But the autotransformers are economical only when the voltage ratio is less than 2. Similarly, the autotransformers cannot provide isolation between primary and secondary.

An autotransformer is never used as a distribution transformer because the lack of isolation can cause dangerously high-voltages in a customer’s location if the neutral opens. Autotransformers are generally used for transforming one transmission voltage to another when the ratio is 2:1 or less. They are used in a transmission substation to transform from one high-voltage to another high-voltage or from a transmission voltage to a substation-voltage. They are normally connected in wye with the neutral solidly grounded, having a delta-connected tertiary for harmonics suppression.

Answer.4. Is located on pole shoes

Explanation:-

Compensating windings are used to overcome the commutation and cross-magnetization effect of armature reaction, which causes high flux density in the trailing pole tip. Also, the coil under this tip can develop an induced voltage that is high enough to cause a flashover between nearby commutator segments.

Compensating winding is connected in series with the armature winding in such a way that the current flowing in them is directly opposite to the current flowing in the armature located just below the pole faces. Half of the conductors (say left hand side of the pole) are connected in series with the other half of the conductors (on the right side of the adjacent pole).

Physically, this coil is much closer to the commutation zone in which the air temperature can be high due to the commutation process. Therefore, the compensating windings are located on pole shoes to avoid flashover.

Answer.2. +5V and −10V

Explanation:-

Given

V_in = 10 v

V_R = 5 V

During the Positive half cycle, the diode is forward bias i.e V_in > V_r and the diode acts as a short circuit

Therefore the output voltage

V_out = V_in − V_r

V_out = 5 V (max Voltage)

During the negative half cycle, the diode is reversed biased and acts as an open circuit i.e V_in < V_r

Hence output voltage

V_out = V_in

V_out = −10 V (min voltage)

Answer.4. The system losses will be unchanged but the complex bus voltages will change.

Explanation:-

The system losses are a function of system properties that are independent of choosing a type of bus. Whereas the complex bus voltage is a function of parameters that are being calculated. When a load-flow program is run twice, in the second run, the previous reference bus gets changed to PQ bus then the losses in the system remains unchanged

Answer.1. V_ac = 100√3 ∠50V

Explanation:-

The phase sequence is called a positive sequence or abc sequence when V_an leads V_bn by 120° and V_bn leads V_cn by 120°. Line to line voltage V_an leads phase voltage V_a by 30º

The phase sequence is called a negative sequence or acb sequence when V_an leads V_cn by 120° and V_cn leads V_bn by 120°. The line to line voltage V_an lags phase voltage V_a by 30º

For Negative Phase sequence ACB, the line to line voltage V_ac is given as

V_ac = √3V_an∠−150°

V_ac = √3(100V∠20°) × (1∠−150°)

V_ac = 100√3∠−130°

or

V_ac = 100√3∠50°

Answer.3. 17.32 kVA

Explanation:-

In open delta or V-V transformer bank connection, only 86.6% or 0.866 rated capacity is available. This factor is also called as utility factor.

S_VV = 0.866 × S_DD

S_VV = 0.866 × 20

S_VV = 17.32 kVA