UPPCL AE Electrical Engineer Solved Paper 2019

Answer.4. 1 rad/sec

Explanation:-

Transfer function G(s) = 1/s²

In case of 0 dB crossing in Bode plot

|G(s)| = 1

|G(s)| = 1/ω² = 1

ω = 1 rad/sec

Answer.4. 2.817

Explanation:-

Given

V_BE = 0.7 V

β = 100

Also

I_E = (1 + β)I_B

Applying KVL in the input loop

V_cc − (R_B × I_B) − V_BE − [I_E × R_E] = 0

V_cc − (R_B × I_B) − V_BE − [(1 + β)I_B × R_E] = 0

12 − (300 × I_B) − 0.7 −101 × I_B = 0

11.3 − 402I_B = 0

I_B = 0.0281 mA

Collector current I_C is given by

I_C = β × I_B

I_C = 100 × 0.0281

I_C = 2.81

Answer.3. Increase in spacing between the phase conductors

Explanation:-

The inductance of transmission line is given by

[katex]L = \dfrac{{{\mu _o}}}{\pi } \times \ln \left( {\dfrac{D}{{{r^\iota }}}} \right)[/katex]

Where

D = distance between the conductor

r’ = radius of the conductor

[katex]L \propto \ln \left( {\dfrac{D}{{{r^\iota }}}} \right)[/katex]

Hence inductance is directly proportional to the distance between the conductor. If we increase the spacing between the conductors then inductance will also increase.

Answer.D. All the energy sources mentioned above

Explanation:-

Resources that can be renewed by nature again and again so that their supply is not adversely affected by the rate of their consumption are called renewable sources of energy. E.g are solar, tidal, wind, geothermal, etc.

Answer.2. 16 mH

Explanation:-

Self-inductance of the solenoid is given by

[katex]{L_1} \propto \dfrac{{{N^2}}}{l} = 2mH[/katex]………(1)

Now the number of turns in the coil is doubled and the length is halved

N = 2N²

l = l/2

[katex]{L_2} = \dfrac{{{{(2N)}^2}}}{{l/2}}[/katex]……..(2)

Dividing equation 1 and 2

[katex]\dfrac{{{L_2}}}{{{L_1}}} = \dfrac{{8{N^2}}}{l} \times \frac{l}{{{N^2}}}[/katex]

L₂ = 8 × 2 = 16 mH

Answer.1. 80 μJ

Explanation:-

The electrostatic field is given by

[katex]E = \left( {\dfrac{x}{2} + 2y} \right)\widehat i + 2x\widehat j[/katex]

The work is done in moving a point charge Q = −20 μC from the origin (0, 0, 0) to point (4, 0, 0)

W = V × Q

The  potential difference from point A to B is given by

[katex]V = – \int\limits_A^B {\overrightarrow E .\overrightarrow {dl} }[/katex]

As y ordinate doesn’t change along the path, therefore, y = 0, then dy = 0

[katex]V = – \int\limits_0^4 {\dfrac{1}{2}xdx = – \dfrac{1}{4}\left[ {{x^2}} \right]_0^4}[/katex]

V = −4 V

W = V × Q

W = −4 × −20

W = 80 μJ

Answer.1. 31.4 mA

Explanation:-

Charge flow q = 5t sin 4πt

The rate of flow of electrical charge is given by

I = dq/dt

I = d[5t sin 4πt]/dt

I = [5 sin(4πt) + 5t × 4π cos(4πt)]

t = 0.5 sec

I = [5 sin(2π) + 20π × 0.5 cos(2π)]

I = [0 +  10π × 1]

I = 10π = 10 × 3.14

I = 31.4 mA

Answer.2. f_o/2

Explanation:-

The resonant frequency for series R-L-C circuit is given by

[katex]{f_o} = \dfrac{1}{{2\pi \sqrt {LC} }}[/katex]——(1)

Now inductor and capacitor value is doubled the above equation becomes

[katex]{f_1} = \dfrac{1}{{2\pi \sqrt {2L \times 2C} }}[/katex]——(2)

Dividing equation 2 from 1 we get

[katex]\begin{array}{l}\dfrac{{{f_1}}}{{{f_o}}} = \dfrac{1}{{4\pi \sqrt {LC} }} \times 2\pi \sqrt {LC} \\\\{f_1} = \dfrac{{{f_o}}}{2}\end{array}[/katex]

Answer.3. 1.2

Explanation:-

Given inductance

[katex]L = 10 + 3\theta – \dfrac{{{\theta ^2}}}{4}\mu H[/katex]

Control spring torque (K) = 25 × 10⁶ Nm/radian

Current I = 5A

The pointer will come to rest at a position where controlling torque T_C, is equal to the deflecting torque T_d

T_C = T_d

[katex]\begin{array}{l}25{\rm{ }} \times {\rm{ }}{10^{ – 6}} \times \theta = \dfrac{1}{2}{(5)^2}\dfrac{{d\left( {10 + 3\theta – \dfrac{{{\theta ^2}}}{4}} \right)}}{{d\theta }}\\\\25{\rm{ }} \times {\rm{ }}{10^{ – 6}} \times \theta = \dfrac{1}{2}{(5)^2} \times \left( {3 – \dfrac{\theta }{2}} \right) \times {10^{ – 6}}\\\\\theta = \dfrac{1}{2}\left( {3 – \dfrac{\theta }{2}} \right)\end{array}[/katex]

5θ/2 =3

θ = 1.2 rad

Answer.1. 2000 Hz

Explanation:-

Given

Gain × bandwidth = 10,000 Hz

Applying nodal analysis at V_in and V⁻

(V_in − 0)/20 = (0 − V_o)/100

V_in /20 = − V_o/100

Gain = |V_o/V_in| = 5

∵ Gain × bandwidth = 10,000 Hz

= 5 × bandwidth = 10,000 Hz

Bandwidth = 10000/5

Bandwidth = 2000 Hz