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DIA [1.3K]
3 years ago
9

Suppose you want to find the sum of two sinusoidal voltages, given as follows: v1(t)=V1 cos(ωt+ϕ1) and v2(t)=V2 cos(ωt+ϕ2)v1(t)=

V1 cos(ωt+ϕ1) and v2(t)=V2 cos(ωt+ϕ2).
If you stay in the time domain, you will have to use trigonometric identities to perform the addition. But if you transform to the frequency domain, you can simply add the phasors V1V1 and V2V2 as complex numbers using your calculator. Your answer will be a phasor, so you will need to inverse phasor-transform it to get the answer in the time domain. This is an example of a problem that is easier to solve in the frequency domain than in the time domain.

Use phasor techniques to find an expression for v(t) expressed as a single cosine function, where v(t)=[100cos(300t+45∘)+500cos(300t−60∘)] V. Enter your expression using the cosine function. Round real numbers using two digits after the decimal point. Any angles used should be in degrees.

Engineering
2 answers:
nydimaria [60]3 years ago
5 0

Answer:

Rectangular form V = 320.71 - j362.3

Polar form V = 483.85 < -48.48°

Phasor form V = 483.85cos(300t - 48.48°)

Explanation:

We are given a sinusoidal function

V(t) = 100cos(300t + 45°) + 500cos(300t - 60°)

We are required to find the v(t) expressed as a single cosine function using phasor technique.

In polar form,

100cos(300t + 45°) = 100 < 45°

500cos(300t - 60°)  = 500 < -60°

In rectangular form,

100 < 45° = 70.71 + j70.71

500 < -60° = 250 - j433.01

Adding the two signals

(70.71 + j70.71) +  (250 - j433.01)

In rectangular form,

V = 320.71 - j362.3

In polar form

V = 483.85 < -48.48°

Therefore, the answer is  

in rectangular form V = 320.71 - j362.3

in polar form V = 483.85 < -48.48°

in phasor form V = 483.85cos(300t - 48.48°)

Conversion from Rectangular to Polar form:

V = X + jY to Magnitude < Angle

V = 320.71 - j362.3

Magnitude = \sqrt{(320.71)^2 +(362.3)^2} = 483.85

Angle = tan⁻¹(Y/X) = tan⁻¹(-362.3/320.71) = -48.48°

V = 483.85 < -48.48°

Conversion from Polar to Rectangular form:

V = 483.85 < -48.48°

X =  Magnitude*cos(Angle) and jY = Magnitude*sin(Angle)

X = 483.85*cos(-48.48°) and jY = 483.85*sin(-48.48°)

X = 320.71 and jY = -362.3

V = 320.71 - j362.3

seropon [69]3 years ago
3 0

Answer:

Attached is the full solution.

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Find the Rectangular form of the following phasors?
almond37 [142]

Answer:

The angles are missing in the question.

The angles are :

45,     30,    60,     90,    -34,     -56,      20,     -42,  -65,    -15

P=10, P=5,  P=25, P=54, P=65, P=95, P=250, P=8, P=35, P=150

Explanation:

1. P = 10,   θ = 45°  rectangular coordinates

x = r cosθ  ,   y = r sinθ

So, rectangular form is x + iy

x = P cosθ = 10 cos 45°

  = 7.07

y =P sinθ = 10 sin 45°

  = 7.07

Therefore, rectangular form

x + iy = 7.07 + i (7.07)

2. P = 5 , θ = 30°

x = 5 cos  30° = 4.33

y = 5 sin  30° = 2.5

So, (x+iy) = 4.33 + i (2.5)

3. P = 25 , θ = 60°

x = 25 cos  60° = 12.5

y = 25 sin  60° = 21.65

So, (x+iy) = 12.5 + i (21.65)

4. P = 54 , θ = 90°

x = 54 cos  90° = 0

y = 54 sin  90° = 54

So, (x+iy) = 0+ i (54)

5. P = 65 , θ = -34°

x = 65 cos  (-34°) = 53.88

y = 65 sin  (-34°) = -36.34

So, (x+iy) = 53.88 - i (36.34)

6. P = 95 , θ = -56°

x = 95 cos  (-56)° = 53.12

y = 95 sin  (-56)° = -78.75

So, (x+iy) = 53.12 - i (78.75)

7. P = 250 , θ = 20°

x = 250 cos  20° = 234.92

y = 250 sin 20° = 85.5

So, (x+iy) = 234.92 + i (85.5)

8. P = 8 , θ = (-42)°

x = 8 cos  (-42)° = 5.94

y = 8 sin  (-42)° = -5.353

So, (x+iy) = 5.94 - i (5.353)

9. P = 35 , θ = (-65)°

x = 35 cos  (-65)° = 14.79

y = 35 sin  (-65)° = -31.72

So, (x+iy) = 14.79 - i (31.72)

10. P = 150 , θ = (-15)°

x = 150 cos  (-15)° = 144.88

y = 150 sin  (-15)° = -38.82

So, (x+iy) = 144.88 - i (38.82)

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3 years ago
A reservoir is 1 km wide and 10 km long and has an average depth of 100m. Every hour, 0.1% of the reservoir's volume drops throu
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Answer:

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Length of reservoir = 10km

Width of reservoir = 1km

Height = 100m

Volume = 10x10³x10³x100

= 10⁹m³

Next we find the volume flow rate

= 0.1/100x10⁹x1/3600

= 277.78m³/s

To get the electrical power output developed by the turbine with 92 percent efficiency

= 0.92x1000x9.81x277.78x100

= 250.7MW

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A 14 inch diameter pipe is decreased in diameter by 2 inches through a contraction. The pressure entering the contraction is 28
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Answer:

5984.67N

Explanation:

A 14 inch diameter pipe is decreased in diameter by 2 inches through a contraction. The pressure entering the contraction is 28 psi and a pressure drop of 2 psi occurs through the contraction if the upstream velocity is 4.0 ft/sec. What is the magnitude of the resultant force (lbs) needed to hold the pipe in place?

from continuity equation

v1A1=v2A2

equation of continuity

v1=4ft /s=1.21m/s

d1=14 inch=.35m

d2=14-2=0.304m

A1=pi*d^2/4

0.096m^2

a2=0.0706m^2

from continuity once again

1.21*0.096=v2(0.07)

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13.789N/m^2=rho(1.65^2-1.21^2)/2

rho=21.91kg/m^3

since the pipe is cylindrical

pressure is egh

13.789=21.91*9.81*h

length of the pipe is

0.064m

AH=volume of the pipe(area *h)

the mass =rho*A*H

0.064*0.07*21.91

m=0.098kg

(193053*0.096- 179263.6* 0.07) + 0.098(1.65 – 1.21)

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