All categories

3 years ago

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]3 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.

You might be interested in

Question

Leto [7]

Answer:

True

Explanation:

The CNC is the primary interface between the machine operator and the machine.

4 0

3 years ago

Technician A says a solenoid is not used on modern automobiles. Technician B says fuel injectors and starter motor solenoids are

julia-pushkina [17]

Answer:

ON THE BOTTOM

Explanation:

red cord plus black cord

5 0

3 years ago

What tool should be used to loosen or tighten brake or fuel lines?

emmainna [20.7K]

The tool you would use are brake line wrenches.

4 0

4 years ago

Define the Artist class in Artist.py with a constructor to initialize an artist's information. The constructor should by default

Sunny_sXe [5.5K]

Answer:

Explanation:

As a result of the codes used to making this solved, i was unable to upload my answer, therefore i'm sending a screenshot of the codes.

thank you for your understanding,

cheers i hope this helps

5 0

3 years ago

Refrigerant-134a at 700 kPa, 70°C, and 7.2 kg/min is cooled by water in a condenser until it exists as a saturated liquid at the

alex41 [277]

Answer:

The mass flow rate of cooling water required to cool the refrigerant is $123.788\,\frac{kg}{min}$ .

Explanation:

A condenser is a heat exchanger used to cool working fluid (Refrigerant 134a) at the expense of cooling fluid (water), which works usually at steady state. Let suppose that there is no heat interactions between condenser and surroundings.The condenser is modelled after the First Law of Thermodynamics, which states:

$\dot Q_{ref} - \dot Q_{w} = 0$

$\dot Q_{ref} = \dot Q_{w}$

$\dot m_{ref}\cdot (h_{ref, in} - h_{ref,out}) = \dot m_{w}\cdot (h_{w, out} - h_{w,in})$

The mass flow rate of the cooling water is now cleared:

$\dot m_{w} = \dot m_{ref }\cdot \frac{h_{ref,in}-h_{ref,out}}{h_{w,out}-h_{w,in}}$

Given that $h_{ref,in} = 808.34\,\frac{kJ}{kg}$ , $h_{ref, out} = 88.82\,\frac{kJ}{kg}$ , $h_{w,out} = 104.83\,\frac{kJ}{kg}$ and $h_{w,in} = 62.98\,\frac{kJ}{kg}$ , the mass flow of the cooling water is:

$\dot m_{w} = \left(7.2\,\frac{kg}{min} \right)\cdot \left(\frac{808.34\,\frac{kJ}{kg}-88.82\,\frac{kJ}{kg} }{104.83\,\frac{kJ}{kg}-62.98\,\frac{kJ}{kg} } \right)$

$\dot m_{w} = 123.788\,\frac{kg}{min}$

The mass flow rate of cooling water required to cool the refrigerant is $123.788\,\frac{kg}{min}$ .

4 0

3 years ago