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Design a half-wave recti er which provides a peak voltage of 15 V, and anaverage voltage of 3.8 V when driven by a 120 V (rms) a

nirvana33 [79]

Answer:

You need a 120V to 24V commercial transformer (transformer 1:5), a 100 ohms resistance, a 1.5 K ohms resistance and a diode with a minimum forward current of 20 mA (could be 1N4148)

Step by step design:

Because you have a 120V AC voltage supply you need an efficient way to reduce that voltage as much as possible before passing to the rectifier, for that I recommend a standard 120V to 24V transformer. 120 Vrms = 85 V and 24 Vrms = 17V = Vin
Because 17V is not 15V you still need a voltage divider to step down that voltage, for that we use R1 = 100Ω and R2 = 1.3KΩ. You need to remember that more than 1 V is going to be in the diode, so for our calculation we need to consider it. Vf = (V*R2)/(R1+R2), V = Vin - 1 = 17-1 = 16V and Vf = 15, Choosing a fix resistance R1 = 100Ω and solving the equation we find R2 = 1.5KΩ
Finally to select the diode you need to calculate two times the maximum current and that would be the forward current (If) of your diode. Imax = Vf/R2 = 10mA and If = 2*Imax = 20mA

Our circuit meet the average voltage (Va) specification:

Va = (15)/(pi) = 4.77V considering the diode voltage or 3.77V without considering it

6 0

4 years ago

In the LC-3 data path, the output of the address adder goes to both the MARMUX and the PCMUX, potentially causing two very diffe

dangina [55]

Answer:

no need for that

Explanation:

they are not the same at all

3 0

3 years ago

Refrigerant 134a enters the evaporator of a refrigeration system operating at steady state at -16oC and a quality of 20% at a ve

Dmitry [639]

Answer:

mass flow rate = 0.0534 kg/sec

velocity at exit = 29.34 m/sec

Explanation:

From the information given:

Inlet:

Temperature $T_1 = -16^0\ C$

Quality $x_1 = 0.2$

Outlet:

Temperature $T_2 = -16^0 C$

Quality $x_2 = 1$

The following data were obtained at saturation properties of R134a at the temperature of -16° C

$v_f= 0.7428 \times 10^{-3} \ m^3/kg \\ \\ v_g = 0.1247 \ m^3 /kg$

$v_1 = v_f + x_1 ( vg - ( v_f)) \\ \\ v_1 = 0.7428 \times 10^{-3} + 0.2 (0.1247 -(0.7428 \times 10^{-3})) \\ \\ v_1 = 0.0255 \ m^3/kg \\ \\ \\ v_2 = v_g = 0.1247 \ m^3/kg$

$m = \rho_1A_1v_1 = \rho_2A_2v_2 \\ \\ m = \dfrac{1}{0.0255} \times \dfrac{\pi}{4}\times (1.7 \times 10^{-2})^2\times 6 \\ \\ \mathbf{m = 0.0534 \ kg/sec}$

$\rho_1A_1v_1 = \rho_2A_2v_2 \\ \\ A_1 =A_2 \\ \\ \rho_1v_1 = \rho_2v_2 \\ \\ \implies \dfrac{1}{0.0255} \times6 = \dfrac{1}{0.1247}\times (v_2)\\ \\ \\\mathbf{\\ v_2 = 29.34 \ m/sec}$

3 0

3 years ago

Pls answer and I will give a like!

lubasha [3.4K]

Answer:

a

Explanation:

6 0

3 years ago

A. Name the major strengthening mechanisms in metals and explain the working principle under each mechanism.Give the relevant eq

Sever21 [200]

Answer:

a) Solid solution strengthening and alloying, Precipitation hardening, work hardening

b) Absence of enough crystallographic misalignment in the grain boundary region for a small-angle

Explanation:

<u>A) strengthening mechanism</u>

i) Solid solution strengthening and alloying:

In solid solution strengthening and alloying mechanism there is an addition of one atom of solute to another during this process, there might be substitution of interstitial point defect in crystal

also the shear stress required can be represented as: Δz = Gb√Ce^3/2

where : C = solute concentration , e = strain on material

ii) Precipitation hardening:

During precipitation hardening the alloying above the concentrate will lead to the formation of a second phase also under precipitation hardening a second phase can also be created via thermal treatments

particle bowing cab be written as : Δz = Gb / L-2x

iii) work hardening :

Dislocation caused by stress fields been generated hardens metals under the work hardening mechanism

dislocation can be represented as ; Gb √ p

where : G = shear modulus , b = Burgess vector, p = dislocation density

B) The small angle grain boundaries are not effective enough because there is less crystallographic misalignment in the grain boundary region for a small-angle

3 0

3 years ago