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If the three capacitors of values 1.0, 1.5, and 2.0 uF each are connected in series, what is the combined capacity

1 Answer

12 votes

Answer:

Total capacitance of the three capacitors in series = 0.4615uF

Step-by-step explanation:

Take the inverse of the three capacitors (in this case you are working with µF so 10^-6) and add their inverse. The result is ~2.167*10^6. The inverse of this number is 0.4615 so the total capacitance of this circuit is 0.4615µF.

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