Answer:
(a) 1/L∫Vdt; integral t [0,1]
(b) 1/L∫Vdt; integral t [ 1, infinity]
Explanation:
An Inductor current I, flowing through an inductor depends on the voltage, V, across the inductor and the inductance, L, of the inductor. The switch 1, 2 timing varies the voltage V with time t
The expression for inductor current is given as:
I= 1/L∫Vdt,
where I is equal to the current flowing through the inductor, L is equal to the inductance of the inductor, and V is equal to the voltage across the inductor.
The formula can also be written as:
I= I0 + 1/L∫Vdt, where I is inductor current at time t, and io is inductor current at t = 0. Time can be varied by controlling the switch
Answer:
//This Program is written in C++
// Comments are used for explanatory purpose
#include <iostream>
using namespace std;
enum mailbox{open, close};
int box[149];
void closeAllBoxes();
void OpenClose();
void printAll();
int main()
{
closeAllBoxes();
OpenClose();
printAll();
return 0;
}
void closeAllBoxes()
{
for (int i = 0; i < 150; i++) //Iterate through from 0 to 149 which literarily means 1 to 150
{
box[i] = close; //Close all boxes
}
}
void OpenClose()
{
for(int i = 2; i < 150; i++) {
for(int j = i; j < 150; j += i) {
if (box[j] == close) //Open box if box is closed
box[j] = open;
else
box[j] = close; // Close box if box is opened
}
}
// At the end of this test, all boxes would be closed
}
void printAll()
{
for (int x = 0; x < 150; x++) //use this to test
{
if (box[x] = 1)
{
cout << "Mailbox #" << x+1 << " is closed" << endl;
// Print all close boxes
}
}
}
Explanation:
Solution :
i. Slip plane (1 1 0)
Slip direction -- [1 1 1]
Applied stress direction = ( 1 0 0 ]
τ = 50 MPa ( Here slip direction must be perpendicular to slip plane)
τ = σ cos Φ cos λ
τ = σ cos Φ cos λ
∴
σ = 122.47 MPa
ii. Slip plane --- (1 1 0)
Slip direction -- [1 1 1]
τ = σ cos Φ cos λ
∴
σ = 122.47 MPa
iii. Slip plane --- (1 0 1)
Slip direction --- [1 1 1]
τ = σ cos Φ cos λ
∴
σ = 122.47 MPa
iv. Slip plane -- (1 0 1)
Slip direction ---- [1 1 1]
τ = σ cos Φ cos λ
∴
σ = 122.47 MPa
∴ (1, 0, -1). (1, -1, 1) = 1 + 0 - 1 = 0
Answer:
See explaination
Explanation:
Please kindly check attachment for the step by step solution of the given problem.
A. The heat transfer rate from natural gas is 2105.26 MW
B. The heat transfer rate to river is 1305.26 MW
Efficiency = (power output / power input) × 100
Power input = Power input / efficiency
Power input = 800 / 38%
Power input = 800 / 0.38
Power input = 2105.26 MW
Thus, the heat transfer from natural gas is 2105.26 MW
Heat to the river = 2105.26 – 800
Heat to the river = 1305.26 MW
Learn more about efficiency:
Answer:
heat transfer from natural gas is 2105.26 MW
heat transfer to river is 1305.26 MW
Explanation:
given data
power output Wn = 800 MW
efficiency = 38%
solution
we know that efficiency is express as
......................1
put here value we get
38% =
Qin = 2105.26 MW
so heat supply is 2105.26
so we can say
Wn = Qin - Qout
800 = 2105.26 - Qout
Qout = 2105.26 - 800
Qout = 1305.26 MW
so heat transfer from natural gas is 2105.26 MW
and heat transfer to river is 1305.26 MW
Answer:
critical stress required is 18.92 MPa
Explanation:
given data
specific surface energy = 1.0 J/m²
modulus of elasticity = 225 GPa
internal crack of length = 0.8 mm
solution
we get here one half length of internal crack that is
2a = 0.8 mm
so a = 0.4 mm = 0.4 × m
so we get here critical stress that is
...............1
put here value we get
=
= 18923493.9151 N/m²
= 18.92 MPa
Answer:
At the middle of the DC load
Explanation:
For the Q point of an amplifier to have the Largest linear output. the Q plant has to be biased at the middle of the DC load line, this is because when the input voltage is low the transistor will be in the cutoff region while when the input voltage is very high the transistor will be in the saturation, hence when the Q point is biased at the middle it is will be higher linearly in relation to the active region