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Goryan [66]
3 years ago
8

Uses of P-N junction

Engineering
1 answer:
Lubov Fominskaja [6]3 years ago
4 0

Answer:

Explanation:

Two that come to mind:

  1. a semiconductor diode is essentially a PN junction
  2. a transistor is made of two pn junctions.
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Given the circuit at the right in which the following values are used: R1 = 20 kΩ, R2 = 12 kΩ, C = 10 µ F, and ε = 25 V. You clo
agasfer [191]

Answer:

a.) I = 7.8 × 10^-4 A

b.) V(20) = 9.3 × 10^-43 V

Explanation:

Given that the

R1 = 20 kΩ,

R2 = 12 kΩ,

C = 10 µ F, and

ε = 25 V.

R1 and R2 are in series with each other.

Let us first find the equivalent resistance R

R = R1 + R2

R = 20 + 12 = 32 kΩ

At t = 0, V = 25v

From ohms law, V = IR

Make current I the subject of formula

I = V/R

I = 25/32 × 10^3

I = 7.8 × 10^-4 A

b.) The voltage across R1 after a long time can be achieved by using the formula

V(t) = Voe^- (t/RC)

V(t) = 25e^- t/20000 × 10×10^-6

V(t) = 25e^- t/0.2

After a very long time. Let assume t = 20s. Then

V(20) = 25e^- 20/0.2

V(20) = 25e^-100

V(20) = 25 × 3.72 × 10^-44

V(20) = 9.3 × 10^-43 V

8 0
3 years ago
Even when you have the right-of-way, you're responsible for _____ before you drive forward.
Mazyrski [523]

Answer: look left to right twice b/c somebody may try to run the light

Explanation:

3 0
3 years ago
Read 2 more answers
Using the AASHTO 1993 Flexible Pavement Design Procedure, design a pavement cross section that will provide 10 years service. Th
natali 33 [55]

Answer:

Check the explanation

Explanation:

Single Unit Truck ESAL = 43.38 + 5.16 = 48.54

Semi Unit Truck ESAL = 43.38+ 6.00+7.4 = 56.78

So total ESAL's during design life = (400*48.54 + 350*56.78)*365*10/18000 = (19416+19873)*3650= 3939

Kindly check the attached image

Here

Reliability = 95% = 0.95, therefore ZR = -1.645, S0 = 0.4, MR = 18

Delta PSI = 4.2-2.5= 1.7

Resilient Modulus = 18000 psi, So MR = 18

Assume SN = 3.0 for flexible pavements

There W18 calculates to 0.26807

So

log10 (3939) = 9.36*log10(SN+1) -.2/(.4+1094/(SN+1)5.19)) -6.01

Structural Number SN = a1*d1 + a2*d2 *m2 +a2*d3 *m3

= a1*d1 + a2*d2 +a2*d3

5 0
3 years ago
8- Concentration polarization occurs on the surface of the.......
Rainbow [258]

Explanation:

Concentration overpotential, ηc,

I hope it helps you

5 0
3 years ago
Write a program that randomly chooses between three different colors for displaying text on the screen. Use a loop to display tw
11Alexandr11 [23.1K]

Answer:

INCLUDE Irvine32.inc

.data

msgIntro  byte "This is Your Name's fourth assembly extra credit program. Will randomly",0dh,0ah

         byte "choose between three different colors for displaying twenty lines of text,",0dh,0ah

         byte "each with a randomly chosen color. The color probabilities are as follows:",0dh,0ah

         byte "White=30%,Blue=10%,Green=60%.",0dh,0ah,0

msgOutput byte "Text printed with one of 3 randomly chosen colors",0

.code

main PROC

;

//Intro Message

       mov edx,OFFSET msgIntro  ;intro message into edx

       call WriteString         ;display msgIntro

       call Crlf                ;endl

       call WaitMsg             ;pause message

       call Clrscr              ;clear screen

       call Randomize           ;seed the random number generator

       mov edx, OFFSET msgOutput;line of text

       mov ecx, 20              ;counter (lines of text)

       L1:;//(Loop - Display Text 20 Times)

       call setRanColor         ;calls random color procedure

       call SetTextColor        ;calls the SetTextColor from library

       call WriteString         ;display line of text

       call Crlf                ;endl

       loop L1

exit

main ENDP

;--

setRanColor PROC

;

; Selects a color with the following probabilities:

; White = 30%, Blue = 10%, Green = 60%.

; Receives: nothing

; Returns: EAX = color chosen

;--

       mov eax, 10              ;range of random numbers (0-9)

       call RandomRange         ;EAX = Random Number

       .IF eax >= 4          ;if number is 4-9 (60%)

       mov eax, green           ;set text green

       .ELSEIF eax == 3         ;if number is 3 (10%)

       mov eax, blue            ;set text blue

       .ELSE                    ;number is 0-2 (30%)

       mov eax, white           ;set text white

       .ENDIF                   ;end statement

       ret

setRanColor ENDP

6 0
4 years ago
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