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11Alexandr11 [23.1K]
3 years ago
8

The primary transformer voltage is 2.8 kV and the secondary is 230 V. The transformer is connected to a variable load (0 to 300

kW) with a lagging power factor of 0.83 and a load voltage equal to the rated transformer secondary. Determine: (a) the total input impedance of the transformer when the secondary is shorted; and (b) the input current, voltage, power and power factor at full load (150 kW). (c) Plot the voltage regulation versus load, and determine the load

Engineering
1 answer:
Black_prince [1.1K]3 years ago
6 0

Answer:

a) πa⁵ Po² / 6E

Vacuum has infinite volume that's why I didn't consider that for total energy calculations

Explanation:

Check attached images for explanation and solutions to b and c

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You are working in a lab where RC circuits are used to delay the initiation of a process. One particular experiment involves an
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Answer:

t'_{1\2} = 6.6 sec

Explanation:

the half life of the given circuit is given by

t_{1\2} =\tau ln2

where [/tex]\tau = RC[/tex]

t_{1\2} = RCln2

Given t_{1\2} = 3 sec

resistance in the circuit is 40 ohm and to extend the half cycle we added new resister of 48 ohm. the net resitance is 40+48 = 88 ohms

now the new half life is

t'_{1\2} =R'Cln2

Divide equation 2 by 1

\frac{t'_{1\2}}{t_{1\2}} = \frac{R'Cln2}{RCln2} = \frac{R'}{R}

t'_{1\2} = t'_{1\2}\frac{R'}{R}

putting all value we get new half life

t'_{1\2} = 3 * \frac{88}{40}  = 6.6 sec

t'_{1\2} = 6.6 sec

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3 years ago
Determine the output logic-levels(boolean-levels) for XNOR if the two-inputs are inverted?​
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Answer:

<em><u>1</u></em>

<em><u>1What is the output of 2 Input XNOR gate if both the inputs are same? Explanation: The output of 2 Input XNOR gate is 1 if both the inputs are same. The output of the XNOR gate is 1 if both the inputs are logic 0 or logic 1. This is why they are called as equality detector.</u></em>

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how do u do that?

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A cyclone is operated in a closed circuit with a ball mill. The cyclone is feed from a rod mill with a slurry that has a density
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Here is the flow sheet. Hope this helps have a great day!!

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A car accelerates from rest with an acceleration of 5 m/s^2. The acceleration decreases linearly with time to zero in 15 s, afte
Tpy6a [65]

Answer: At time 18.33 seconds it will have moved 500 meters.

Explanation:

Since the acceleration of the car is a linear function of time it can be written as a function of time as

a(t)=5(1-\frac{t}{15})

a=\frac{d^{2}x}{dt^{2}}\\\\\therefore \frac{d^{2}x}{dt^{2}}=5(1-\frac{t}{15})

Integrating both sides we get

\int \frac{d^{2}x}{dt^{2}}dt=\int 5(1-\frac{t}{15})dt\\\\\frac{dx}{dt}=v=5t-\frac{5t^{2}}{30}+c

Now since car starts from rest thus at time t = 0 ; v=0 thus c=0

again integrating with respect to time we get

\int \frac{dx}{dt}dt=\int (5t-\frac{5t^{2}}{30})dt\\\\x(t)=\frac{5t^{2}}{2}-\frac{5t^{3}}{90}+D

Now let us assume that car starts from origin thus D=0

thus in the first 15 seconds it covers a distance of

x(15)=2.5\times 15^{2}-\farc{15^{3}}{18}=375m

Thus the remaining 125 meters will be covered with a constant speed of

v(15)=5\times 15-\frac{15^{2}}{6}=37.5m/s

in time equalling t_{2}=\frac{125}{37.5}=3.33seconds

Thus the total time it requires equals 15+3.33 seconds

t=18.33 seconds

3 0
3 years ago
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