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
Answer:
When the renewal period comes around (with respect to the license expiry date).
Explanation:
Answer:
K
Explanation:
For installations where the nonlinear load is huge, most consulting engineers will specify K-rated transformers.
Answer:
i believe the answer is a but i could be wrong
Explanation:
i hope it helps
Answer:
Explanation:
% Clears variables and screen
clear; clc
% Asks user for input
n = input('Total number of objects: ');
r = input('Size of subgroup: ');
% Computes and displays permutation according to basic formulas
p = 1;
for i = n - r + 1 : n
p = p*i;
end
str1 = [num2str(p) ' permutations'];
disp(str1)
% Computes and displays combinations according to basic formulas
str2 = [num2str(p/factorial(r)) ' combinations'];
disp(str2)
=================================================================================
Example: check
How many permutations and combinations can be made of the 15 alphabets, taking four at a time?
The answer is:
32760 permutations
1365 combinations
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