Respuesta :

Answer:

Total capacitance of the three capacitors in series = 0.4615uF

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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