UPPCL AE Electrical Engineer Solved Paper 2019

Answer.3. 100 Hz

Explanation:-

Given,

Supply frequency = 50 Hz (Indian Standard)

Instantaneous power frequency is twice the supply frequency .

F_I = 2 × 50 Hz

F_I = 100 Hz

Answer.1. 0

Explanation:-

Surge impedance = 300Ω

End line load = 300Ω

The reflection coefficient of the transmission line at the load end is given by

ρ = (Z_L − Z_o)/(Z_L + Z_o)

Where,

Z_L = Load or termination impedance at the end of the line

Z_o = Characteristics impedance or Surge impedance

ρ = (300 − 300)/(300 + 300)

ρ = 0

Answer.4. 1.77 pF

Explanation:-

Given,

Area (A) = 2 × 10⁻⁴ m

Distance (d) = 1 mm = 10⁻³ m

C = ε_oA/d

The capacitance of a parallel plate capacitor is given by

C = (8.85 × 10⁻¹² × 2 × 10⁻⁴)/10⁻³

C = 17.7 × 10⁻¹³

C = 1.77 pF

Answer.1. 20 µF

Explanation:-

As shown in the figure the capacitor 5 µF and 20 µF are connected in series therefore its equivalent capacitance

= (5 × 20)/(5 + 20) = 4µF

Now capacitance 4µF, 6µF & 20 µF are connected in parallel

= 4 + 6 + 20 = 30 µF

At last 30 µF and 60 µF capacitors is connected in series

C_eq = (30 × 60)/(30 + 60) = 20 µF

Answer.1. Low resistance

Explanation:-

The resistance with values less than or equal to 1Ω is called low resistance. Example: Armature winding of machines, cables, and ammeter shunts have a low resistance value.

Kelvin double bridge is best suited for the measurement of resistances of very low value.

Answer.3. 0.2 second

Explanation:-

Given time Constant of casual system

G(S) = 1/(s + 5)

or

G(S) = 1/5(1 + 0.2s)

Comparing the above equation with the standard time constant system i.e.

G(S) = 1/(1 + τs)

τ = 0.2 sec

Answer.2. Faraday’s and Lenz’s laws

Explanation:-

Faraday’s 1st laws of electromagnetic induction tell us about the condition under which an e.m.f. is induced in a conductor or coil when the magnetic flux linking a conductor or coil changes.

Faraday’s 2nd laws of electromagnetic induction give the magnitude of the induced e.m.f in a conductor or coil and may be stated as:

The magnitude of the e.mf induced in a conductor or coil is directly proportional to the rate of change of magnetic flux linkages.

Suppose a coil has N turns and magnetic flux linking the coil increases (i.e. changes) from φ₁ Wb to φ₂ Wb in t seconds. Now, magnetic flux linkages mean the product of magnetic flux and the number of turns of the coil.

N = e dφ/dt

Lenz Law:- Lenz’s law states: the direction of the induced e.m.f. is such as to oppose the change producing it. Therefore, the magnitude and direction of induced e.m.f. should be written as :

N = −e dφ/dt

Answer.3. S/(S + 1)²

Explanation:-

To get the transfer function from step response we need to get the impulse response of the system.

That is, the impulse response h(t) of a system is equal to the derivative of its step response s(t). Therefore, the impulse response of a system can be determined from its step response simply through differentiation.

[katex]\begin{array}{l}h(t) = \dfrac{d}{{dt}}(yt)\\\\= \dfrac{d}{{dt}}(t{e^{ – t}})\end{array}[/katex]

h(t) = e^−t − te^−t

Laplace transform of e^−t = 1/(S + 1)

Laplace transform of te^−t = 1/(S + 1)²

[katex]\begin{array}{l}h(t) = \dfrac{1}{{S + 1}} – \dfrac{1}{{{{(S + 1)}^2}}}\\\\h(t) = \dfrac{S}{{{{(S + 1)}^2}}}\end{array}[/katex]

Answer.1. 14.4Ω

Explanation:-

From the given figure resistance 6Ω & 3Ω are connected in parallel

= (6 × 3)/(6 + 3) = 2 Ω

Resistance 5Ω & 1Ω are connected in series

5Ω + 1Ω = 6Ω

Now resistance 2 Ω & 2 Ω are connected in series

= 2 + 2 =4Ω

Now resistance 4Ω and 6Ω are connected in parallel

= (4 × 6)/(4 + 6) = 2.4 Ω

Finally resistance 4Ω, 2.4Ω & 8Ω are connected in series

R_eq = 4 + 2.4 + 8

R_eq = 14.4Ω

Answer.A. 1:1

Explanation:-

Given primary-to-secondary turns ratio

[katex]\dfrac{{{N_P}}}{{{N_S}}} = \dfrac{N}{1}[/katex]

The instantaneous input power to instantaneous output power is given by

[katex]\begin{array}{l}\dfrac{{{P_i}}}{{{P_o}}} = \dfrac{{{V_i}{I_i}}}{{(N{V_i})\left( {\dfrac{1}{N}{I_i}} \right)}}\\\\\dfrac{{{P_i}}}{{{P_o}}} = 1:1\end{array}[/katex]