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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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If the rotational speed of a pump motor is reduced by 35%, what is the effect on the pump performance in terms of capacity, head
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Answer:

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the pump capacity is related to the speed as speed is reduces by 35%

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3 years ago
Bob and Alice are solving practice problems for CSE 2320. They look at this code: for(i = 1; i &lt;= N; i = (i*2)+17 ) for(k = i
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Answer:

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Read 2 more answers
Consider fully developed laminar flow in a circular pipe. If the viscosity of the fluid is reduced by half by heating while the
gladu [14]

Answer:

The pressure drop across the pipe also reduces by half of its initial value if the viscosity of the fluid reduces by half of its original value.

Explanation:

For a fully developed laminar flow in a circular pipe, the flowrate (volumetric) is given by the Hagen-Poiseulle's equation.

Q = π(ΔPR⁴/8μL)

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μ₁ = (μ/2)

The new pressure drop (ΔP₁) is then

ΔP₁ = Kμ₁ = K(μ/2)

Recall,

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ΔP₁ = K(μ/2) = (ΔP/μ) × (μ/2) = (ΔP/2)

Hence, the pressure drop across the pipe also reduces by half of its initial value if the viscosity of the fluid reduces by half of its value.

Hope this Helps!!!

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3 years ago
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