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

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

• A. 0.3 μF.
• B. 3.33 μF.
• C. 15 μF.
• D. 50 μF.

Answer: (B)

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 2. 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 3. Now, to find C_total, take the reciprocal of 3/10: C_total = 10/3 μF 4. 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.

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

2. 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

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

C_total = 10/3 μF

4. 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.