All categories

3 years ago

How to draw the output voltage waveform rectifier

Engineering

1 answer:

tatyana61 [14]3 years ago

7 0

Answer:

Half-wave rectifier converts an AC signal into a DC signal. It's called a half-wave because it only rectify the positive part of an AC signal.

AC Signal = An electrical signal that alternates between positive and negative voltage.

DC Signal = An electrical signal that only has positive voltage.

Rectify = A fancy word for converting something.

Adding a capacitor helps the positive part of the signal stay on longer. This work because the capacitor stores energy kinda like a battery. During the negative part of the AC signal, the energy stored in the capacitor will be drained and used, then the cycle repeats.

The load resistor is just there to prevent a short circuit from happening.

You might be interested in

Required information NOTE: This is a multi-part question. Once an answer is submitted, you will be unable to return to this part

ra1l [238]

Answer:

h1 = 290.16kj/kg

P = 1.2311

Prandil expression at 8

P=p1/p7×pr

=8(1.2311)

=9.85

Enthalpy state at 8 corresponding to 9.85

h1 = 526.13kj/kg

Now prandtl state at 9 that correspond to 1400k.

h9 = 1515.42kj/kg

Pr = 450.5

Prandtl expression at state 10

P= p10/p9×pr

=1/8(450.5)

=56.31

Enthalpy at state 10 corresponding to prandtl 56.31

h10 = 860.39kj/kg

At 520k

h11 = 523.63kj/kg

4 0

3 years ago

Steam enters an adiabatic turbine at 10MPa and 500 C and leaves at 10 kPa with a quality of 90%. Neglecting the changes in kinet

Amanda [17]

Answer:

flow ( m ) = 4.852 kg/s

Explanation:

Given:

- Inlet of Turbine

P_1 = 10 MPa

T_1 = 500 C

- Outlet of Turbine

P_2 = 10 KPa

x = 0.9

- Power output of Turbine W_out = 5 MW

Find:

Determine the mass ow rate required

Solution:

- Use steam Table A.4 to determine specific enthalpy for inlet conditions:

P_1 = 10 MPa

T_1 = 500 C ---------- > h_1 = 3375.1 KJ/kg

- Use steam Table A.6 to determine specific enthalpy for outlet conditions:

P_2 = 10 KPa -------------> h_f = 191.81 KJ/kg

x = 0.9 -------------> h_fg = 2392.1 KJ/kg

h_2 = h_f + x*h_fg

h_2 = 191.81 + 0.9*2392.1 = 2344.7 KJ/kg

- The work produced by the turbine W_out is given by first Law of thermodynamics:

W_out = flow(m) * ( h_1 - h_2 )

flow ( m ) = W_out / ( h_1 - h_2 )

- Plug in values:

flow ( m ) = 5*10^3 / ( 3375.1 - 2344.7 )

flow ( m ) = 4.852 kg/s

3 0

3 years ago

What is the purpose of the red-core yarns in the synthetic sling?

Leona [35]

An immediate strength loss. This is why Sling Webbing has red core yarns to visually reveal damage and act as a basis for sling rejection.

▞▞<☵>▆▇█▓▒░─╤╦╱彡( ⱫɆ₳Ⱡ )彡╲╦╤─░▒▓▇▆▅<☵>▞▞

5 0

3 years ago

A tank containing diesel fuel(SG = 0.8) is open to the atmosphere at the top. A U-tube manometer is connected to the bottom of t

stepan [7]

Answer:

h= 37.9 m

Explanation:

Given that

SG = 0.8 for fuel so density of fluid will be 800 kg/m³.

We know that SG = 13.6 For Hg so density will be 13600 kg/m³.

Now by balancing the pressure

$\rho_d\times g\times h +\rho_d\times g\times 1.2 =\rho_{hg}\times g\times 2.3$

$800\times 9.81\times h+ 800\times 9.81\times 1.2 =13600\times 9.81\times 2.3$

$800\times 9.81\times h =13600\times 9.81\times 2.3-800\times 9.81\times 1.2$

$h=\dfrac{297439.2}{800\times 9.81}$

h= 37.9 m

8 0

3 years ago

Read 2 more answers

Which of the following distinguishes a reservoir from an accumulator?

lozanna [386]

Answer:

Reservoirs usually have zero pressure

Explanation:

accumulators are controlled by pressure valves while reservoirs are not

accumulators "accumulate" pressure while reservoirs are used to store "overflow" fluids for later use if needed

6 0

2 years ago