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What total capacitance is obtained by connecting a 5.00 µF and an 8.00 µF capacitor in series?

a. 2.86 µF
b. 13.00 µF
c. 40.00 µF
d. 1.538 µF

User Kereberos
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1 Answer

1 vote

Final answer:

When a 5.00 µF and an 8.00 µF capacitor are connected in series, the total capacitance is approximately 3.08 µF. None of the given options are exactly correct, suggesting there may be an error in the options provided to the student.

Step-by-step explanation:

The total capacitance obtained by connecting a 5.00 µF and an 8.00 µF capacitor in series is determined by the formula for capacitors in series:

1/Ctotal = 1/C1 + 1/C2

Let's plug in the values:

1/Ctotal = 1/5.00 µF + 1/8.00 µF

= 0.2 µF-1 + 0.125 µF-1

= 0.325 µF-1

Now, to find the total capacitance Ctotal, we take the reciprocal:

Ctotal = 1/0.325

Ctotal = 3.08 µF (approximately)

This means the nearest answer is (a) 2.86 µF, but none of the provided options are exact. In practice, it would be best to inform the student of the correct calculation and suggest that there might be a mistake in the options given.

User Aelor
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