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Question:
Calculate the equivalent capacitance of the three series capacitors in Figure 12-1
a) 0.01 μF
b) 0.58 μF
c) 0.060 μF
d) 0.8 μF
Answer:
The equivalent capacitance of the three series capacitors in Figure 12-1 is 0.060 μF
Therefore, the correct option is (c)
Explanation:
Please refer to the attached Figure 12-1 where three capacitors are connected in series.
We are asked to find out the equivalent capacitance of this circuit.
Recall that the equivalent capacitance in series is given by
![$ \frac{1}{C_{eq}} = \frac{1}{C_{1}} + \frac{1}{C_{2}} + \frac{1}{C_{3}} $](https://tex.z-dn.net/?f=%24%20%5Cfrac%7B1%7D%7BC_%7Beq%7D%7D%20%3D%20%20%5Cfrac%7B1%7D%7BC_%7B1%7D%7D%20%2B%20%5Cfrac%7B1%7D%7BC_%7B2%7D%7D%20%2B%20%5Cfrac%7B1%7D%7BC_%7B3%7D%7D%20%24)
Where C₁, C₂, and C₃ are the individual capacitance connected in series.
C₁ = 0.1 μF
C₂ = 0.22 μF
C₃ = 0.47 μF
So the equivalent capacitance is
![$ \frac{1}{C_{eq}} = \frac{1}{0.1} + \frac{1}{0.22} + \frac{1}{0.47} $](https://tex.z-dn.net/?f=%24%20%5Cfrac%7B1%7D%7BC_%7Beq%7D%7D%20%3D%20%20%5Cfrac%7B1%7D%7B0.1%7D%20%2B%20%5Cfrac%7B1%7D%7B0.22%7D%20%2B%20%5Cfrac%7B1%7D%7B0.47%7D%20%24)
![$ \frac{1}{C_{eq}} = \frac{8620}{517} $](https://tex.z-dn.net/?f=%24%20%5Cfrac%7B1%7D%7BC_%7Beq%7D%7D%20%3D%20%20%5Cfrac%7B8620%7D%7B517%7D%20%20%24)
![$ C_{eq} = \frac{517}{8620} $](https://tex.z-dn.net/?f=%24%20C_%7Beq%7D%20%3D%20%20%5Cfrac%7B517%7D%7B8620%7D%20%20%24)
![$ C_{eq} = 0.0599 $](https://tex.z-dn.net/?f=%24%20C_%7Beq%7D%20%3D%20%200.0599%20%20%24)
Rounding off yields
![$ C_{eq} = 0.060 \: \mu F $](https://tex.z-dn.net/?f=%24%20C_%7Beq%7D%20%3D%20%200.060%20%5C%3A%20%5Cmu%20F%20%24)
The equivalent capacitance of the three series capacitors in Figure 12-1 is 0.060 μF
Therefore, the correct option is (c)
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Answer:
μ=0.329, 2.671 turns.
Explanation:
(a) ln(T2/T1)=μβ β=angle of contact in radians
take T2 as greater tension value and T1 smaller, otherwise the friction would be opposite.
T2=5000 lb and T1=80 lb
we have two full turns which makes total angle of contact=4π radians
μ=ln(T2/T1)/β=(ln(5000/80))/4π
μ=0.329
(b) using the same relation as above we will now compute the angle of contact.
take greater tension as T2 and smaller as T1.
T2=20000 lb T1=80 lb μ=0.329
β=ln(20000/80)/0.329=16.7825 radians
divide the angle of contact by 2π to obtain number of turns.
16.7825/2π =2.671 turns
Answer:
Wind energy is converted to Mechanical energy which is then converted in to electrical energy
Explanation:
In a wind mill the following energy conversions take place
a) Wind energy is converted into Mechanical energy (rotation of rotor blades)
b) Mechanical energy is converted into electrical energy (by using electric motor)
This electrical energy is then used for transmission through electric lines.
Answer:
#include <iostream>
using namespace std;
void PrintPopcornTime(int bagOunces) {
if(bagOunces < 3){
cout << "Too small";
cout << endl;
}
else if(bagOunces > 10){
cout << "Too large";
cout << endl;
}
else{
cout << (6 * bagOunces) << " seconds" << endl;
}
}
int main() {
PrintPopcornTime(7);
return 0;
}
Explanation:
Using C++ to write the program. In line 1 we define the header "#include <iostream>" that defines the standard input/output stream objects. In line 2 "using namespace std" gives me the ability to use classes or functions, From lines 5 to 17 we define the function "PrintPopcornTime(), with int parameter bagOunces" Line 19 we can then call the function using 7 as the argument "PrintPopcornTime(7);" to get the expected output.