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Alex787 [66]
3 years ago
12

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

Engineering
1 answer:
Naddik [55]3 years ago
4 0

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

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10.0 kmol of a 40.0 mol% methanol and 60.0 mol% water mixture is processed in a normal batch distillation system with a still po
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[2]. 0.536

[c]. 4.15 kmol; 5.84 kmol.

Explanation:

Without mincing words let's dive straight into the solution to the question above.

                                                            [a].

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