What is the total capacitance when connecting three capacitors of 0.05 uF, 20 uF, and 40 uF in series?

What is the total capacitance when connecting three capacitors of 0.05 uF, 20 uF, and 40 uF in series?

• A. 10.05 uF
• B. 12 uF
• C. 0.0498 uF
• D. 0.052 uF

Answer: (C)

To determine the total capacitance when capacitors are connected in series, one must use the formula: 1/C_total = 1/C1 + 1/C2 + 1/C3, where C1, C2, and C3 are the capacitances of the individual capacitors. In this case, the capacitors have the following values: - C1 = 0.05 µF - C2 = 20 µF - C3 = 40 µF Applying the formula, we compute: 1/C_total = 1/0.05 + 1/20 + 1/40. Calculating each term separately: 1/0.05 = 20, 1/20 = 0.05, 1/40 = 0.025. Now, summing these values: 1/C_total = 20 + 0.05 + 0.025 = 20.075. Next, we take the reciprocal to find C_total: C_total = 1 / 20.075 ≈ 0.0498 µF. Thus, the total capacitance when 0.05 µF, 20 µF, and 40 µF capacitors are connected in

To determine the total capacitance when capacitors are connected in series, one must use the formula:

1/C_total = 1/C1 + 1/C2 + 1/C3,

where C1, C2, and C3 are the capacitances of the individual capacitors.

In this case, the capacitors have the following values:

• C1 = 0.05 µF

• C2 = 20 µF

• C3 = 40 µF

Applying the formula, we compute:

1/C_total = 1/0.05 + 1/20 + 1/40.

Calculating each term separately:

1/0.05 = 20,

1/20 = 0.05,

1/40 = 0.025.

Now, summing these values:

1/C_total = 20 + 0.05 + 0.025 = 20.075.

Next, we take the reciprocal to find C_total:

C_total = 1 / 20.075 ≈ 0.0498 µF.

Thus, the total capacitance when 0.05 µF, 20 µF, and 40 µF capacitors are connected in