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MrRissso [65]
2 years ago
10

why is the peak value of the rectified output less than the peak value of the ac input and by how much g

Engineering
1 answer:
Bumek [7]2 years ago
5 0

Answer:

The Peak value of the output voltage is less or lower than that of the peak value of the input voltage by 0.6V reason been that the voltage is tend to drop across the diode.

Explanation:

This is what we called HALF WAVE RECTIFIER in which the Peak value of the output voltage is less or lower than that of the peak value of the input voltage by 0.6V reason been that the voltage is tend to drop across the diode.

Therefore this is the formula for Half wave rectifier

Vrms = Vm/2 and Vdc

= Vm/π:

Where,

Vrms = rms value of input

Vdc = Average value of input

Vm = peak value of output

Hence, half wave rectifier is a rectifier which allows one half-cycle of an AC voltage waveform to pass which inturn block the other half-cycle which is why this type of rectifiers are often been used to help convert AC voltage to a DC voltage, because they only require a single diode to inorder to construct.

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Carbon dioxide steadily flows into a constant pressure heater at 300 K and 100 kPa witha mass flow rate of 9.2 kg/s. Heat transf
docker41 [41]

Answer:

Carbon dioxide temperature at exit is 317.69 K

Carbon dioxide flow rate at heater exit is 20.25 m³/s

Explanation:

Detailed steps are attached below.

8 0
2 years ago
Air at 40C flows over a 2 m long flat plate with a free stream velocity of 7 m/s. Assume the width of the plate (into the paper)
mezya [45]

Complete Question

Air at 40C flows over a 2 m long flat plate with a free stream velocity of 7m/s. Assume the width of the plate (into the paper) is 0.5 m. If the plate is at a co temperature of 100C,find:

The total heat transfer rate from the plate to the air

Answer:

q=1.7845

Explanation:

From the question we are told that:

Air Temperature T_1=40c

Length l=2m

Velocity v=7m/s

Width w=0.5

Constant temperature T_t= 100C

Generally the equation for Total heat Transfer is mathematically given by

 q=hA(T_s-T_\infty)

Where

h=Convective heat transfer coefficient

 h=29.9075w/m^2k

Therefore

 q=h(L*B)(T_s-T_\infty)

 q=29.9075*(2*0.5)(100+273-(40+273))

 q=1794.45w

 q=1.7845

5 0
3 years ago
For a bronze alloy, the stress at which plastic deformation begins is 266 MPa and the modulus of elasticity is105 GPa.
pentagon [3]

Answer:

88750 N

Explanation:

given data:

plastic deformation σy=266 MPa=266*10^6 N/m^2

cross-sectional area Ao=333 mm^2=333*10^-6 m^2

solution:

To determine the maximum load that can be applied without

plastic deformation (Fy).

Fy=σy*Ao

   =88750 N

7 0
2 years ago
How can we love our country? Not by words but by deeds. - Jose P. Laurel
Vadim26 [7]

Answer:

1. You have the courage to help without expecting a reward.

2. Because actions are more eloquent than words. Actions are far more valuable and counted than  words, and that's how she inspired me.

3. Doing simple things that can make someone grateful and happy  without knowing that someone is inspired and motivated by your good deeds, and also doing some interesting things By.

Explanation:

6 0
2 years ago
For methyl chloride at 100°C the second and third virial coefficients are: B = −242.5 cm 3 ·mol −1 C = 25,200 cm 6 ·mol −2 Calcu
bogdanovich [222]

Answer:

a)W=12.62 kJ/mol

b)W=12.59 kJ/mol

Explanation:

At T = 100 °C the second and third virial coefficients are

B = -242.5 cm^3 mol^-1

C = 25200 cm^6  mo1^-2

Now according isothermal work of one mole methyl gas is

W=-\int\limits^a_b {P} \, dV

a=v_2\\

b=v_1

from virial equation  

\frac{PV}{RT}=z=1+\frac{B}{V}+\frac{C}{V^2}\\   \\P=RT(1+\frac{B}{V} +\frac{C}{V^2})\frac{1}{V}\\

And  

W=-\int\limits^a_b {RT(1+\frac{B}{V} +\frac{C}{V^2}\frac{1}{V}  } \, dV

a=v_2\\

b=v_1

Now calculate V1 and V2 at given condition

\frac{P1V1}{RT} = 1+\frac{B}{v_1} +\frac{C}{v_1^2}

Substitute given values P_1\\ = 1 x 10^5 , T = 373.15 and given values of coefficients we get  

10^5(v_1)/8.314*373.15=1-242.5/v_1+25200/v_1^2

Solve for V1 by iterative or alternative cubic equation solver we get

v_1=30780 cm^3/mol

Similarly solve for state 2 at P2 = 50 bar we get  

v_1=241.33 cm^3/mol

Now  

W=-\int\limits^a_b {RT(1+\frac{B}{V} +\frac{C}{V^2}\frac{1}{V}  } \, dV

a=241.33

b=30780

After performing integration we get work done on the system is  

W=12.62 kJ/mol

(b) for Z = 1 + B' P +C' P^2 = PV/RT by performing differential we get  

         dV=RT(-1/p^2+0+C')dP

Hence work done on the system is  

W=-\int\limits^a_b {P(RT(-1/p^2+0+C')} \, dP

a=v_2\\

b=v_1

by substituting given limit and P = 1 bar , P2 = 50 bar and T = 373 K we get work  

W=12.59 kJ/mol

The work by differ between a and b because the conversion of constant of virial coefficients are valid only for infinite series  

8 0
2 years ago
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