Connecting 3 capacitors of .05uF, 20uF, and 40uF in series results in a total capacitance of?

Connecting 3 capacitors of .05uF, 20uF, and 40uF in series results in a total capacitance of?

• A. .0498uF
• B. 60uF
• C. .05uF
• D. 1.2uF

Answer: (A)

When capacitors are connected in series, the total capacitance can be calculated using the formula: \[ \frac{1}{C_{total}} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3} \] In this case, the three capacitors have capacitances of 0.05 µF, 20 µF, and 40 µF. Plugging in these values into the formula gives: \[ \frac{1}{C_{total}} = \frac{1}{0.05} + \frac{1}{20} + \frac{1}{40} \] Calculating each term separately: 1. For the 0.05 µF capacitor: \[ \frac{1}{0.05} = 20 \] 2. For the 20 µF capacitor: \[ \frac{1}{20} = 0.05 \] 3. For the 40 µF capacitor: \[ \frac{1}{40} = 0.025 \] Now, adding these values together: \[ \

When capacitors are connected in series, the total capacitance can be calculated using the formula:

[

\frac{1}{C_{total}} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3}

]

In this case, the three capacitors have capacitances of 0.05 µF, 20 µF, and 40 µF. Plugging in these values into the formula gives:

[

\frac{1}{C_{total}} = \frac{1}{0.05} + \frac{1}{20} + \frac{1}{40}

]

Calculating each term separately:

1. For the 0.05 µF capacitor:

[

\frac{1}{0.05} = 20

]

2. For the 20 µF capacitor:

[

\frac{1}{20} = 0.05

]

3. For the 40 µF capacitor:

[

\frac{1}{40} = 0.025

]

Now, adding these values together:

[

\