What is the total capacitance of three capacitors connected in series with capacitances of 0.2 microfarads, 0.05 microfarads, and 0.10 microfarads?

What is the total capacitance of three capacitors connected in series with capacitances of 0.2 microfarads, 0.05 microfarads, and 0.10 microfarads?

• A. 0.1 microfarads
• B. 0.25 microfarads
• C. 0.029 microfarads
• D. 0.05 microfarads

Answer: (C)

To find the total capacitance of capacitors connected in series, the formula used is 1/C_total = 1/C1 + 1/C2 + 1/C3, where C1, C2, and C3 are the capacitances of the individual capacitors. In this case, you have three capacitors with values of 0.2 microfarads, 0.05 microfarads, and 0.10 microfarads. Applying the formula: 1. For the first capacitor: 1/0.2 = 5 2. For the second capacitor: 1/0.05 = 20 3. For the third capacitor: 1/0.10 = 10 Now, add these values together: 1/C_total = 5 + 20 + 10 = 35 To find the total capacitance, take the reciprocal: C_total = 1/35 = 0.02857 microfarads, which can be rounded to approximately 0.029 microfarads. This calculation confirms that the total capacitance of the three capacitors in series is approximately 0.029 microfarads, making that the correct choice in this problem.

To find the total capacitance of capacitors connected in series, the formula used is 1/C_total = 1/C1 + 1/C2 + 1/C3, where C1, C2, and C3 are the capacitances of the individual capacitors.

In this case, you have three capacitors with values of 0.2 microfarads, 0.05 microfarads, and 0.10 microfarads. Applying the formula:

1. For the first capacitor: 1/0.2 = 5

2. For the second capacitor: 1/0.05 = 20

3. For the third capacitor: 1/0.10 = 10

Now, add these values together:

1/C_total = 5 + 20 + 10 = 35

To find the total capacitance, take the reciprocal:

C_total = 1/35 = 0.02857 microfarads, which can be rounded to approximately 0.029 microfarads.

This calculation confirms that the total capacitance of the three capacitors in series is approximately 0.029 microfarads, making that the correct choice in this problem.