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Three identical capacitors are connected in series to give an equivalent total capacitance of 30 µF, what is the capacitance of each capacitor?*1 point90 µF10 µF60 µF20 µF

Question

Three identical capacitors are connected in series to give an equivalent total capacitance of 30 µF, what is the capacitance of each capacitor?*1 point90 µF10 µF60 µF20 µF

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Solution

When capacitors are connected in series, the total or equivalent capacitance (Ct) is given by the reciprocal of the sum of the reciprocals of the individual capacitances (C1, C2, C3, etc.).

In this case, we have three identical capacitors connected in series, so we can say that each capacitor has a capacitance of C. Therefore, the total capacitance (Ct) can be calculated as follows:

1/Ct = 1/C + 1/C + 1/C

Given that Ct is 30 µF, we can substitute this into the equation:

1/30 = 3/C

To solve for C, we can rearrange the equation to find:

C = 3/1/30 = 90 µF

Therefore, each capacitor has a capacitance of 90 µF.

This problem has been solved

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