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Akimi4 [234]
3 years ago
10

Suppose you wanted to convert an AC voltage to DC. Your AC voltage source has Vrms= 60V. If you use a full wave rectifier direct

ly on the line voltage with Vf = 0.7V, what will be your output DC voltage in Volts (assuming a negligible ripple)?
Engineering
1 answer:
vovangra [49]3 years ago
7 0

Answer:

V_{dc}=84.15\ V

Explanation:

Given that

Vrms= 60 V

Vf= 0.7 V

We know that peak value of AC voltage given as

V_{o}=\sqrt{2}\ V_{rms}

Now by putting the values

V_{o}=60\sqrt{2}\ V

The output voltage of the DC current given as

V_{dc}=V_{o}-V_f

V_{dc}=60\sqrt{2}-0.7\ V

V_{dc}=84.15\ V

Therefore output voltage of the DC current is 84.15 V.

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A steel bolt has a modulus of 207 GPa. It holds two rigid plates together at a high temperature under conditions where the creep
VikaD [51]

Answer:

14.36((14MPa) approximately

Explanation:

In this question, we are asked to calculate the stress tightened in a bolt to a stress of 69MPa.

Please check attachment for complete solution and step by step explanation

7 0
2 years ago
Determine if the following errors are systematic or random. Justify your response. (a) Effect of temperature on the circuitry of
k0ka [10]

Answer:

a) temperature: random error

b) parallax: systematic error

c) using incorrect value: systematic error

Explanation:

Systematic errors are associated with faulty calibration or reading of the equipments used and they could be avoided refining your method.

4 0
3 years ago
A large retirement village has a total retail employment of 120. All 1600 of the households in this village consist of two nonwo
polet [3.4K]

Answer:

See explaination

Explanation:

Please kindly check attachment for the step by step solution of the given problem.

5 0
3 years ago
Five kilograms of air at 427°C and 600 kPa are contained in a piston–cylinder device. The air expands adiabatically until the pr
son4ous [18]

Answer:

The entropy change of the air is 0.240kJ/kgK

Explanation:

T_{1} =427+273K,T_{1} =700K\\P_{1} =600kPa\\P_{2} =100kPa

T_{2}  is unknown

we can apply the following expression to find T_{2}

-w_{out} =mc_{v} (T_{2} -T_{1} )

T_{2} =T_{1} -\frac{w_{out } }{mc_{v} }

now substitute

T_{2} =700K-\frac{600kJ}{5kg*0.718kJ/kgK} \\T_{2}=533K

To find entropy change of the air we can apply the ideal gas relationship

Δs_{air}=c_{p} ln\frac{T_{2} }{T_{1} } -Rln\frac{P_{2} }{P_{1} }

Δs_{air} =1.005*ln(\frac{533}{700})-0.287* in(\frac{100}{600} )

Δs_{air} =0.240kJ/kgK

4 0
3 years ago
A wood pole with a diameter of 10 in. has a moisture content of 5%. The fiber saturation point (FSP) for this wood is 30%. The w
Mekhanik [1.2K]

Answer:

a) Δd(change in wood diameter) = 5%

b) The wood would swell since the moisture content is increasing which will also led to increase in the wood's diameter

C) new diameter (D2) = 10.5 in

Explanation:

Wood pole diameter = 10 inches

moisture content = 5%

FSP = 30%

A) The percentage change in the wood's diameter

note : moisture fluctuations from 5% to 30% causes dimensional changes in the wood but above 30% up to 55% causes no change. hence this formula can be used to calculate percentage change in the wood's diameter

Δd/d = 1/5(30 - 5)

Δd/d = 5%  

Δd = 5%

B) would the wood swell or shrink

The wood would swell since the moisture content is increasing which will also led to increase in the wood's diameter

C) The new diameter of the wood

D2 = D + D( \frac{M1}{100} )

D = initial diameter= 10 in , M1 = initial moisture content = 5%

therefore D2 = 10 + 10( 5/100 )

new diameter (D2) = 10.5 in

5 0
3 years ago
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