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Colt1911 [192]
3 years ago
13

What is the total capacitance in units of mF, of the two capacitors connected in series, as shown in the diagram, when C1 = 45 m

F and C2 = 26 mF?

Physics
2 answers:
elena-14-01-66 [18.8K]3 years ago
7 0

Answer:

16.7 mF

Explanation:

The total capacitance of two capacitors connected in series is given by the formula:

\frac{1}{C}=\frac{1}{C_1}+\frac{1}{C_2}

in our problem, we have:

C1 = 45 mF is the capacitance of the first capacitor

C2 = 26 mF is the capacitance of the second capacitor

Substituting into the equation, we find:

\frac{1}{C}=\frac{1}{45 mF}+\frac{1}{26 mF}=0.06 mF^{-1}\\C=\frac{1}{0.06 mF^{-1}}=16.7 mF

goldfiish [28.3K]3 years ago
5 0
<h2>Answer:</h2>

<u>The total capacitance will be </u><u>16.12mF</u>

<h2>Explanation:</h2>

The total capacitance in series is calculated by

1/Ct = 1/C1 + 1/C2 + 1/C3 .........

1/Ct = 1/ 45 + 1/26

1/Ct = 0.022 + 0.04

1/Ct = 0.062

Ct = 16.12mF

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