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BabaBlast [244]
3 years ago
9

Determine the average power, complex power and power factor (including whether it is leading or lagging) for a load circuit whos

e voltage and current at its input terminals are given by: v(t) = 100cos(377t-30)v, i(t) = 2.5cos(377t-60)A

Engineering
2 answers:
Charra [1.4K]3 years ago
6 0

Answer:

Average power: 108.25 W

Complex power: 125 VA

Power factor: 0.866

Refer below for the explanation.

Explanation:

Refer to the picture for brief explanation.

Paul [167]3 years ago
4 0

Answer:

Average power = 108.25W

Complex power = 125W

PF = 0.866

Explanation:

Check attachment for step by step instructions.

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A large plate is fabricated from a steel alloy that has a plane strain fracture toughness of 55 MPa √m (50 ksi √in.). If, during
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Answer:

0.024 m = 24.07 mm

Explanation:

1) Notation

\sigma_c = tensile stress = 200 Mpa

K = plane strain fracture toughness= 55 Mpa\sqrt{m}

\lambda= length of a surface crack (Variable of interest)

2) Definition and Formulas

The Tensile strength is the ability of a material to withstand a pulling force. It is customarily measured in units (F/A), like the pressure. Is an important concept in engineering, especially in the fields of materials and structural engineering.

By definition we have the following formula for the tensile stress:

\sigma_c=\frac{K}{Y\sqrt{\pi\lambda}}   (1)

We are interested on the minimum length of a surface that will lead to a fracture, so we need to solve for \lambda

Multiplying both sides of equation (1) by Y\sqrt{\pi\lambda}

\sigma_c Y\sqrt{\pi\lambda}=K   (2)

Sequaring both sides of equation (2):

(\sigma_c Y\sqrt{\pi\lambda})^2=(K)^2  

\sigma^2_c Y^2 \pi\lambda=K^2   (3)

Dividing both sides by \sigma^2_c Y^2 \pi we got:

\lambda=\frac{1}{\pi}[\frac{K}{Y\sigma_c}]^2   (4)

Replacing the values into equation (4) we got:

\lambda=\frac{1}{\pi}[\frac{55 Mpa\sqrt{m}}{1.0(200Mpa)}]^2 =0.02407m

3) Final solution

So the minimum length of a surface crack that will lead to fracture, would be 24.07 mm or more.

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