Compute the total capacitance in microfarads (μF) of a series circuit containing two capacitors: 5 μF and 10 μF.

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Prepare for the Biomedical Equipment Technician CDC Set A Volume 3 Exam with multiple choice questions and insightful explanations. Boost your confidence and knowledge, and get ready for your test!

In a series circuit, the total capacitance can be calculated using the formula for capacitors in series:

1/C_total = 1/C1 + 1/C2

For this specific scenario with two capacitors (5 μF and 10 μF), the calculation proceeds as follows:

  1. First, take the reciprocals of the individual capacitances:

1/C_total = 1/5 + 1/10

  1. To add these, it’s necessary to find a common denominator. The common denominator for 5 and 10 is 10:

1/5 = 2/10

1/10 = 1/10

So, adding these together:

1/C_total = 2/10 + 1/10 = 3/10

  1. Now, to find C_total, take the reciprocal of 3/10:

C_total = 10/3 μF

  1. Which can be simplified for practical purposes:

C_total ≈ 3.33 μF.

This is why the total capacitance of the capacitors connected in series is approximately 3.33 μF, making this answer the correct choice.

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