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

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:

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