Calculate the equivalent capacitance of the three series capacitors in Figure 12-1 A) 0.060 uF B) 0.8 uF C) 0.58 uF D) 0.01 uF
1 answer:
Answer:
The correct answer is option (A) 0.060 uF
Note: Kindly find an attached image of the complete question below
Sources: The complete question was well researched from Quizlet.
Explanation:
Solution
Given that:
C₁ = 0.1 μF
C₂ =0.22 μF
C₃ = 0.47 μF
In this case, C₁, C₂ and C₃ are in series
Thus,
Their equivalent becomes:
1/Ceq = (1/C₁ + 1/C₂ +1/C₃
1/Ceq =[ (1/0.1 + 1/0.22 +1/0.47)]
1/Ceq =[(0.22 * 0.47) + (0.1 * 0.47) + (0.1 * 0.22)/(0.1 * 0.22 *0.47)]
1/Ceq =[(0.1034 + 0.047 + 0.022)/(0.01034)
1/Ceq =[(0.1724)/(0.01034)]
1/Ceq = [(16.67)]
1/Ceq =(1/16.67) = 0.059μf
Ceq = 0.059μf ≈ 0.060μf
Therefore the equivalent capacitance of the three series capacitors is 0.060μf
You might be interested in
Answer:
Explanation:
Given that:
The direction of the applied tensile stress =[001]
direction of the slip plane = [01]
normal to the slip plane = [111]
Now, the first thing to do is to calculate the angle between the tensile stress and the slip by using the formula:
where;
= directional indices for tensile stress
= slip direction
replacing their values;
i.e = 0 ,
= 0
= 1 &
= -1 ,
= 0 ,
= 1
Also, to find the angle between the stress [001] & normal slip plane [111]
Then;
replacing their values;
i.e = 0 ,
= 0
= 1 &
= 1 ,
= 1 ,
= 1
However, the critical resolved SS(shear stress) can be computed using the formula:
where;
applied tensile stress 13.9 MPa
∴
Answer:
a) 22.5number
b) 22.22 m length
Explanation:
Given data:
Bridge length = 500 m
width of bridge = 12 m
Maximum temperature = 40 degree C
minimum temperature = - 35 degree C
Maximum expansion can be determined as
where , \alpha is expansion coefficient degree C
SO,
number of minimum expansion joints is calculated as
b) length of each bridge
Answer:
This is the C++ code for the above problem:
#include<bits/stdc++.h>
using namespace std;
int computeFunLevel(int start, int duration, int prefs[], int ngames, int plan[]) {
if(start + duration > 365) { //this is to check wether duration is more than total no. of vaccation days
return -1;
}
int funLevel = 0;
for(int i=start; i<start+duration; i++) { //this loop runs from starting point till
//start + duration to sum all the funlevel in plan.
funLevel = funLevel + prefs[plan[i]];
}
return funLevel;
}
int findBestVacation(int duration, int prefs[], int ngames, int plan[]) {
int max = 0, index = 0, sum = 0 ;
for(int i=1; i<11; i++){ //this loop is to run through whole plan arry
sum = 0; //sum is initialized with zero for every call in plan ,
//in this case loop should run to 366,but for example it is 11
//as my size of plan array is 11
for(int j=0; j<duration; j++) { // this loop is for that index to index+duration to calc
//fun from that index
sum = sum + prefs[plan[i]];
}
if(sum>max) { //this is to check max funlevel and update the index from which max fun can be achieved
max = sum;
index = i;
}
}
return index;
}
int main() {
int ngames = 5;
int prefs[] = { 1,2,0,5,2 };
int plan[] = { 0,2,0,3,3,4,0,1,2,3,3 };
int start = 1;
int duration = 4;
cout << computeFunLevel(start, duration, prefs, ngames, plan) << endl;
cout << computeFunLevel(start, 555, prefs, ngames, plan) << endl;
cout << findBestVacation(4, prefs, ngames, plan) << endl;
}
The screen of the program is given below.
