The following electrical characteristics have been determined for both intrinsic and p-type extrinsic gallium antimonide (GaSb)
at room temperature: σ (????∙m) –1 n (m–3 ) p (m–3 ) Intrinsic 8.9 × 10^4 8.7 × 10^23 8.7 × 10^23Extrinsic 2.3 × 10^5 7.6 × 10^22 1.0 × 10^25 (p-type)Calculate electron and hole mobilities.
1 answer:
Answer:
0.5m^2/Vs and 0.14m^2/Vs
Explanation:
To calculate the mobility of electron and mobility of hole for gallium antimonide we have,
(S)
Where
e= charge of electron
n= number of electrons
p= number of holes
mobility of electron
mobility of holes
electrical conductivity
Making the substitution in (S)
Mobility of electron
Mobility of hole in (S)
Then, solving the equation:
(1)
(2)
We have,
Mobility of electron
Mobility of hole is
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The question is incomplete! Complete question along with answers and explanation is provided below.
Question:
A 15 Watt desk-type fluorescent lamp has an effective resistance of 200 ohms when operating (note: the 15 Watts is only associated with the lamp). It is in series with a ballast that has a resistance of 80 ohms and an inductance of .9H. The lamp and ballast are operated at 120V, 60Hz.
a) Draw the circuit
b) Calculate the power drawn by the lamp
c) Calculate the apparent power
d) Calculate the power factor
e) Calculate the reactive power
f) Calculate the size of the capacitor necessary to provide unity power factor correction
Explanation:
a) draw the circuit
Refer to the attached image.
As you can see in the attached drawing, it is a series circuit containing two resistors and one inductor.
In a series circuit, current remains same throughout the circuit
The circuit is powered by an AC voltage source having voltage of 120 V and frequency 60 Hz.
The current flowing in the circuit can be found by ohm's law
I = V/Z
where V is the voltage and Z is the total impedance of the circuit
Z = R + XL
where XL is the inductive reactance
XL = j2 π f L
XL = j2*π*60*0.9
XL = j339.29Ω
Total resistance is
R =200 + 80 = 280 Ω
Total impedance is
Z = 280 + j339.29 Ω
b) Calculate the power drawn by the lamp
First calculate the current
I = V/Z
I = 120/(280 + j339.29)
I = 0.272<-50.46° A (complex notation)
P = I²R
P = (0.272)²200
P ≈ 15 W
Power drawn by the circuit
P=V*I*cos(50.46°)
P=20.77 W
c) Calculate the apparent power
A = VI*
A = 120*0.272<50.46°
A = 32.64<50.46° VA
d) Calculate the power factor
PF = cos(50.46)
PF = 0.63
e) Calculate the reactive power
Q = VIsin(50.46)
Q = 120*0.272<-50.46*sin(50.46)
Q = 25.13<-50.46 VAR
f) Calculate the size of the capacitor necessary to provide unity power factor correction
The required reactive compensation power is
Qc = P (tan(old) - tan(new))
Qc = 20.77 (tan(50.46) - tan(0))
Qc = 25.16 VAR
C = Qc/2πfV²
C = 25.16/2*π*60*120²
C = 4.63 uF
Hence adding a capacitor of 4.63 uF parallel to the load will improve the PF from 0.63 to 1.
Answer:
w=2.25
Explanation:
It is necessary to determine the maximum w so that the normal stress in the AB and CD rods does not exceed the permitted normal stress.
The surface of the cross-section of the stapes was determined:
A_ab= 10 mm^2
A-cd= 15 mm^2
The maximum load is determined from the condition that the normal stresses is not higher than the permitted normal stress σ_allow.
σ_ab = F_ab/A_abσ_allow
σ_cd = F_cd/A_cdσ_allow
In the next step we will determine the static size: Picture b).
We apply the conditions of equilibrium:
∑F_x=0
∑F_y=0
∑M=0
∑M_a=0 ==> -w*6*0.5*6*0.75*F_cd*6 =0
==> F_cd = 2*w*k*N
∑F_y=0 ==> F_cd+F_ab - 6*w*0.5 ==>2*w+F_ab -6*w*0.5 =0
==> F_ab = w*k*N
Now we determine the load w
<u>Sector AB: </u>
σ_ab = F_ab/A_ab σ_allow=300 KPa
= w/10*10^-6 σ_allow=300 KPa
w_ab = 3*10^-3 kN/m
<u>Sector CD: </u>
σ_cd = F_cd/A_cd σ_allow=300 KPa
= 2*w/15*10^-6 σ_allow=300 KPa
w_cd = 2.25*10^-3 kN/m
w=min{w_ab;w_cd} ==> w=min{3*10^-3;2.25*10^-3}
==> w=2.25 * 10^-3 kN/m
<u>The solution is: </u>
w=2.25 N/m
note:
find the attached graph
Answer:
Following are the program to the given question:
#include <stdio.h>//header file
double square_root(double N, double initialGuess)//defining a method square_root that takes two variable in parameters
{
double NG, LG = initialGuess,diff;//defining double variable
while(1)//use loop to calculate square root value
{
NG = 0.5 * (LG + N / LG);//using given formula
diff = NG - LG;//calculating difference
if(diff < 0)//use if to check difference is less than 0
diff = -diff;//decreaing difference
if(diff < 0.005)//use if that check difference is less than 0.005
break;//using break keyword
else//defining else block
{
LG = NG;//holding value
}
}
return NG;//return value
}
int main()//defining main method
{
double ans, n,initialguess = 1.0;//defining double variable
n = 4;//use n to hold value
ans = square_root(n, initialguess);//calculating the square root value and print its value
printf("square_root(%lf) = %lf \n", n, ans);//print calculated value with number
n = 120.5;//use n to hold value
ans = square_root(n, initialguess);//calculating the square root value and print its value
printf("square_root(%lf) = %lf \n", n, ans);//print calculated value with number
n = 36.01;//use n to hold value
ans = square_root(n, initialguess);//calculating the square root value and print its value
printf("square_root(%lf) = %lf \n", n, ans);//print calculated value with number
n = 0.25;//use n to hold value
ans = square_root(n, initialguess);//calculating the square root value and print its value
printf("square_root(%lf) = %lf \n", n, ans);//print calculated value with number
printf("\nEnter a number: ");//print message
scanf("%lf", &n);//input value
ans = square_root(n, initialguess);//calculating the square root value and print its value
printf("square_root(%lf) = %lf \n", n, ans);//print calculated value with number
}
Output:
Please find the attachment file.
Explanation:
- In this code, a method "square_root" is declared that takes two variable "N, initialGuess" in its parameters, inside the method a three double variable is declared.
- It uses the given formula and uses the diff variable to hold its value and uses two if to check its value is less than 0 and 0.005 and return its calculated value.
- In the main method, three double variables are declared that use the "n" to hold value and "ans" to call the method that holds its value and print its value.
